Rd sharma class 10 solutions some applications of trigonometry
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Rd sharma class 10 solutions some applications of trigonometry
Slide Content
RD Sharma Class 10 Solutions Applications of Trigonomefry „
Heionts ANARRUGESE N
4
que
Distance betünen poht of obtewalitn and deck of
tower = 20m. 280
Age del op Hep dj bu = ade
He Her Het = AB
Now From Fb. “8e
Ac Ë a mue tangle
Temp = Adjacentotde
Opel cle.
> |tane = oppesill aide &8)
Adjacert side (00
20
> AG = 2olones
> N= 208
= 208
Height of tows H = 205 on.
2
Distance befioten tor ep Ladder and wall = 9500
ge op clevatiao 0260 SES
Length of halter = 4
Ac:
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Heionts ANARRUGESE N
4
que
Distance betünen poht of obtewalitn and deck of
tower = 20m. 280
Age del op Hep dj bu = ade
He Her Het = AB
Now From Fb. “8e
Ac Ë a mue tangle
Temp = Adjacentotde
Opel cle.
> |tane = oppesill aide &8)
Adjacert side (00
20
> AG = 2olones
> N= 208
= 208
Height of tows H = 205 on.
2
Distance befioten tor ep Ladder and wall = 9500
ge op clevatiao 0260 SES
Length of halter = 4
Ac:
RD Sharma Class 10 Solutions Applications of Trigonometry
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RD Sharma Class 10 Solutions Applications of Trigonometry
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Now are forms a gt angle lngle
“ABC >
we know A
cose = Adjacent sde
tens
D Coser’= Be
Ac
FUE
ne
> Aa as lm
Jengin ch dadder & = Ian,
Distance between oct of ladder à mall = zen = Be,
Ange rade by dadder coo greed
8 = 60°
freight of walt tes = A8 A
Now Fe ABC forms a wight
angled ge
tone = paf sta
Adfacent dee
+ of watt tr
23 * Sen
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Now are forms a gt angle lngle
“ABC >
we know A
cose = Adjacent sde
tens
D Coser’= Be
Ac
FUE
ne
> Aa as lm
Jengin ch dadder & = Ian,
Distance between oct of ladder à mall = zen = Be,
Ange rade by dadder coo greed
8 = 60°
freight of walt tes = A8 A
Now Fe ABC forms a wight
angled ge
tone = paf sta
Adfacent dee
+ of watt tr
23 * Sen
RD Sharma Class 10 Solutions Applications of Trigonometry
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pi E
Height of Me elective pole = 1070 =A0
Age made by Steel coe wl Proud Chorizontat)
en 4s
dt engin ol ré sde Ae.
Ab we vepreent- akove data 8 4
de + due tm te 4 1,”
dent à ner detengo ne
Jere
5
gue
Be IE ham ground = 00.2 Ag
Bretiration of Abg tia ges
9= 60°
Length e) He d=9=4
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pi E
Height of Me elective pole = 1070 =A0
Age made by Steel coe wl Proud Chorizontat)
en 4s
dt engin ol ré sde Ae.
Ab we vepreent- akove data 8 4
de + due tm te 4 1,”
dent à ner detengo ne
Jere
5
gue
Be IE ham ground = 00.2 Ag
Bretiration of Abg tia ges
9= 60°
Length e) He d=9=4
RD Sharma Class 10 Solutions Applications of Trigonometry
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Ajout vepout The above data & $
ford due 98 da Then, he A
don a wight angled Mange, mac.
here "e
sie = oppofiln etde Ao=6 4
potro ,
shes = 4e
AC
> BS. 45
A 2
> de sx
Ya G
> b= 505m.
deg + A L = 504 en.
6
engin oh Abn betaxen poñt en grund and
Ki = aom.
Angle made >
y di ed ating coda goma & o
x is
?
ie
Oj Me Ki Le Hm
ay we vepretent de above data
et hoes They PE format
che Gor
5 figure
Qt anges Aérrgle
sh
RD Sharma Class 10 Solutions Applications of Trigonometry
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Ajout vepout The above data & $
ford due 98 da Then, he A
don a wight angled Mange, mac.
here "e
sie = oppofiln etde Ao=6 4
potro ,
shes = 4e
AC
> BS. 45
A 2
> de sx
Ya G
> b= 505m.
deg + A L = 504 en.
6
engin oh Abn betaxen poñt en grund and
Ki = aom.
Angle made >
y di ed ating coda goma & o
x is
?
ie
Oj Me Ki Le Hm
ay we vepretent de above data
et hoes They PE format
che Gor
5 figure
Qt anges Aérrgle
sh
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ue have
ane ppoctle atde
Adjacent cde
dono = AB |
Be
rc
> SCH
Es re
> e =p Ea
15
Date, by pythagerat Ionen we have
AS BCRA]
q = Ex) Be
+
9 1% = eur
Fe
Han = er
> Her 229) = Gouge
>
H = oxi)
229
(eur
ir
> pe
H= Qoxç
AUS = 39 4
o) Kile from dd eng,
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ue have
ane ppoctle atde
Adjacent cde
dono = AB |
Be
rc
> SCH
Es re
> e =p Ea
15
Date, by pythagerat Ionen we have
AS BCRA]
q = Ex) Be
+
9 1% = eur
Fe
Han = er
> Her 229) = Gouge
>
H = oxi)
229
(eur
ir
> pe
H= Qoxç
AUS = 39 4
o) Kile from dd eng,
RD Sharma Class 10 Solutions Applications of Trigonometry
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+ 1
vertical eue setrmounted by deg ff
Dietanee betwen lave and okterver = 40m. = 6c
Age of val of top eh aan a= 48°
Age of elevation of top of agit p= es
ete of digue ho
eight of tour = Ho = Ms
> iene
we vepvetent The above data \ Oy
ci A
Rom due Mn th dont
ag ges foren Area Y 4
“sete À En
à + E
se .
tap = ne
x
toed = ADtAS = hen
so to
> hero = 0(63)
> h= Holm)
= Ho(t#a34) = “Ox 0132
E Sgen
chelght of Tos tom height of fag stafp = Skagen
RD Sharma Class 10 Solutions Applications of Trigonometry
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+ 1
vertical eue setrmounted by deg ff
Dietanee betwen lave and okterver = 40m. = 6c
Age of val of top eh aan a= 48°
Age of elevation of top of agit p= es
ete of digue ho
eight of tour = Ho = Ms
> iene
we vepvetent The above data \ Oy
ci A
Rom due Mn th dont
ag ges foren Area Y 4
“sete À En
à + E
se .
tap = ne
x
toed = ADtAS = hen
so to
> hero = 0(63)
> h= Holm)
= Ho(t#a34) = “Ox 0132
E Sgen
chelght of Tos tom height of fag stafp = Skagen
RD Sharma Class 10 Solutions Applications of Trigonometry
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8
Trftlat Height of tree
het wu atume dat P
Thun | gie Tat erate made by aber Fark
wll ground oe
eight tron ground fo bole palate =.= 80
AGE AC HEL
> H= Ath S Aczli-him
DE we vaut the abe data ay due
as thos En VE ont be ad Aude
d E Mac
tom da
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8
Trftlat Height of tree
het wu atume dat P
Thun | gie Tat erate made by aber Fark
wll ground oe
eight tron ground fo bole palate =.= 80
AGE AC HEL
> H= Ath S Aczli-him
DE we vaut the abe data ay due
as thos En VE ont be ad Aude
d E Mac
tom da
RD Sharma Class 10 Solutions Applications of Trigonometry
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ns of Trigonometry
> Wlis-h) = ah,
7 1SB-hG = ah
> GnOh = 166
bE EE echan denoutter
mA a Rabinal fig efectos of
> he (en b=0) lose
==
= CG
he tas
bent broken polat tron quema =IS RD.
tefghe ep the log stapp he Ap
Angle ch clevatte, oh tr oh dire = ooh
Ange of chatte ob batten Den
tet bean + Tower be Hw. = ag.
BY un wepretent JR above data &
dem of ye Men À orm gfe
Ce whith Acc
included. ufn
£898. |
Dy waht ole tegen 4
age fe en
tuo =
RD Sharma Class 10 Solutions Applications of Trigonometry
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ns of Trigonometry
> Wlis-h) = ah,
7 1SB-hG = ah
> GnOh = 166
bE EE echan denoutter
mA a Rabinal fig efectos of
> he (en b=0) lose
==
= CG
he tas
bent broken polat tron quema =IS RD.
tefghe ep the log stapp he Ap
Angle ch clevatte, oh tr oh dire = ooh
Ange of chatte ob batten Den
tet bean + Tower be Hw. = ag.
BY un wepretent JR above data &
dem of ye Men À orm gfe
Ce whith Acc
included. ufn
£898. |
Dy waht ole tegen 4
age fe en
tuo =
RD Sharma Class 10 Solutions Applications of Trigonometry
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Tone = BD
ee
> toto ABtAn a
Be >
> (= Hts _@ ZATÍO
© > &. the
a H Pf
> 3
> mes
Hager + Howe H= arm.
lo
Y. of stevation ey tp ep ture fon
Aer point + obcervatiior (A) <=
het
13
Angle
Now
gee
ae!
she tentked Som from that pobla to
Jo AR = 50m
e) elevation rom second parte» =
det us vepretent he
data À Ir o) then
dom ge ACD wth
Kong
ker
eco AH & 2
eight 4 teuer be
Hm = co:
Be = x mı
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Tone = BD
ee
> toto ABtAn a
Be >
> (= Hts _@ ZATÍO
© > &. the
a H Pf
> 3
> mes
Hager + Howe H= arm.
lo
Y. of stevation ey tp ep ture fon
Aer point + obcervatiior (A) <=
het
13
Angle
Now
gee
ae!
she tentked Som from that pobla to
Jo AR = 50m
e) elevation rom second parte» =
det us vepretent he
data À Ir o) then
dom ge ACD wth
Kong
ker
eco AH & 2
eight 4 teuer be
Hm = co:
Be = x mı
RD Sharma Class 10 Solutions Applications of Trigonometry
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E 3% age
> HH
> nH = DR = cg
>
no) tours H = a8 m
het the degollado of tower
cohen ge ef elwahln Blase) be x wm
then accord te problere
<engte % The Ahadao wofin Angle 5) clevatin,
(p= 45) & (loto) m: = ep
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E 3% age
> HH
> nH = DR = cg
>
no) tours H = a8 m
het the degollado of tower
cohen ge ef elwahln Blase) be x wm
then accord te problere
<engte % The Ahadao wofin Angle 5) clevatin,
(p= 45) & (loto) m: = ep
RD Sharma Class 10 Solutions Applications of Trigonometry
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The, above data À
FR we pre
dee mo Geo
dtargte nec Ú Beluled
A
Le Help of Howe be Hes As
3 wt eagle Felge ow h
of Me angie 0 then CES
dene 6 Az
Pong —&
tone = 46 tapo Ae
oe eo
> toned = H > ton ger
x Ed
As > xno=u
Rey > x= ho —O
subtHtlute x= 1-108 8 ©
Ho = u
E
> EH=108=H
>. (=p. = 106
> NS og
a Ratnalde deroaBatoy
> Hz 0 Ber xalinal ita facts of
Bolero an ane
= Palen
2 edgar of tower
= San) Helge u: oe
> 260m Sen
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The, above data À
FR we pre
dee mo Geo
dtargte nec Ú Beluled
A
Le Help of Howe be Hes As
3 wt eagle Felge ow h
of Me angie 0 then CES
dene 6 Az
Pong —&
tone = 46 tapo Ae
oe eo
> toned = H > ton ger
x Ed
As > xno=u
Rey > x= ho —O
subtHtlute x= 1-108 8 ©
Ho = u
E
> EH=108=H
>. (=p. = 106
> NS og
a Ratnalde deroaBatoy
> Hz 0 Ber xalinal ita facts of
Bolero an ane
= Palen
2 edgar of tower
= San) Helge u: oe
> 260m Sen
RD Sharma Class 10 Solutions Applications of Trigonometry
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‘Trigonometry
1.
ker u oh parachute at beat port As
O ave con a por
ground there show in CD = too
Angle > clevahty tron post c= 45% Cay
Angie of elevan tro polars = 60° Cp]
ber 8 be he A dem ame
parachute
fire Accord to
a te draw above cate” ke, tee foma Tre
due ar don atten,
AOE ire and Man Beluded PB LEE Lot
“AG. teorgte Poctuded.
Maximum height ef Tarachut-
Aron ground = 482 Ho
Dittance of Pat chere
Perachut gate to fu
measeıt — obsewaltin qu = x m.
TH th mht angie Hg one of The
eluded angle do thn
Adjacent side.
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‘Trigonometry
1.
ker u oh parachute at beat port As
O ave con a por
ground there show in CD = too
Angle > clevahty tron post c= 45% Cay
Angie of elevan tro polars = 60° Cp]
ber 8 be he A dem ame
parachute
fire Accord to
a te draw above cate” ke, tee foma Tre
due ar don atten,
AOE ire and Man Beluded PB LEE Lot
“AG. teorgte Poctuded.
Maximum height ef Tarachut-
Aron ground = 482 Ho
Dittance of Pat chere
Perachut gate to fu
measeıt — obsewaltin qu = x m.
TH th mht angie Hg one of The
eluded angle do thn
Adjacent side.
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= lof mn)
=
3 X= St.
2% = 50 (aan
> X= 5004)
FRE Rémi
he sante bre Do
max ten umn heizen of qerachut from ground
HE 2366 0
Dittasce between post where parachut gal on
ground Ja oleo Br 13660
13.
Of doux, Hz AB = 156m
ER eo cect on TR ground
Na
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= lof mn)
=
3 X= St.
2% = 50 (aan
> X= 5004)
FRE Rémi
he sante bre Do
max ten umn heizen of qerachut from ground
HE 2366 0
Dittasce between post where parachut gal on
ground Ja oleo Br 13660
13.
Of doux, Hz AB = 156m
ER eo cect on TR ground
Na
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| Angie of depresión of, object ALEMAN] = p= 4e Zane,
[axila]
[Age ep depresión op kiere
x
Land =a == Zags
[ax ate]
ler Az pe
Th we vepraent the above data
fh dame To hk ao
wlth ego
A
E E Ff oe of the
freluded ongle fi © Mm
toro = ppt atte
Afneent ide
tank = 4e, tang = A8
ee! ne
> toned = 150 tongs? 160.
4 ey
Ye 5.0 q = 60 —0
E
DRE spi
Per 50x22 160
a
? X= 150-508 = 150- Sola)
50 86:6 = 63-41
Dittance beiten object ale! = 65-4 en
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an
ven LeamCBSE in
| Angie of depresión of, object ALEMAN] = p= 4e Zane,
[axila]
[Age ep depresión op kiere
x
Land =a == Zags
[ax ate]
ler Az pe
Th we vepraent the above data
fh dame To hk ao
wlth ego
A
E E Ff oe of the
freluded ongle fi © Mm
toro = ppt atte
Afneent ide
tank = 4e, tang = A8
ee! ne
> toned = 150 tongs? 160.
4 ey
Ye 5.0 q = 60 —0
E
DRE spi
Per 50x22 160
a
? X= 150-508 = 150- Sola)
50 86:6 = 63-41
Dittance beiten object ale! = 65-4 en
RD Sharma Class 10 Solutions Applications of Trigonometry
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an
RD Sharma Class 10 Solutions Applications of Trigonometry
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4e a
Angle of levain of top of tower rom
At pot À «2 30
Let we advanced through’ À toe by 150m
| Ro Ne= (50m
Ange of clevaliin of top of dower from second
Por, Bae
[3
Let Helge op “fewer Co = Hm.
Ay we vepecien! de above data
[5 qomo of ae foe 5
Age m Abm win zung. KE AE
Ah th vit angled tangle „one “of eluded ange
he / dhen
are = epale el
Adjacent de
iotx = 5-0
loro
HG + HA) 2 150
ASI
129.4
150 SHE SOKG 245K
|
|
|
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4e a
Angle of levain of top of tower rom
At pot À «2 30
Let we advanced through’ À toe by 150m
| Ro Ne= (50m
Ange of clevaliin of top of dower from second
Por, Bae
[3
Let Helge op “fewer Co = Hm.
Ay we vepecien! de above data
[5 qomo of ae foe 5
Age m Abm win zung. KE AE
Ah th vit angled tangle „one “of eluded ange
he / dhen
are = epale el
Adjacent de
iotx = 5-0
loro
HG + HA) 2 150
ASI
129.4
150 SHE SOKG 245K
|
|
|
RD Sharma Class 10 Solutions Applications of Trigonometry
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RD Sharma Class 10 Solutions Applications of Trigonometry
te
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16. A
Angle of elevattin of top of tower fon polea
Az 80°
Angle el elevation e) top of Tower ron pote
R= 60"
Distance between 428, AGE 20m
Let height of, aser co = Hm
Diitence beitucn Accord pot 8 fron doot of tewr-beiw
A à
Ep we veprerent Sa Abo data In une
dema due 0 Abm
diga En
eo
Sin wight agled Magie Ph one
of de eludes e
Je Hane
> AE) 02 Aes cox Maa
height of tour he sam
Dittanee o) toros from pot A= Bote) = 30m.
i=
RD Sharma Class 10 Solutions Applications of Trigonometry
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16. A
Angle of elevattin of top of tower fon polea
Az 80°
Angle el elevation e) top of Tower ron pote
R= 60"
Distance between 428, AGE 20m
Let height of, aser co = Hm
Diitence beitucn Accord pot 8 fron doot of tewr-beiw
A à
Ep we veprerent Sa Abo data In une
dema due 0 Abm
diga En
eo
Sin wight agled Magie Ph one
of de eludes e
Je Hane
> AE) 02 Aes cox Maa
height of tour he sam
Dittanee o) toros from pot A= Bote) = 30m.
i=
RD Sharma Class 10 Solutions Applications of Trigonometry
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+ E
i Let AG be De bufld®g and co be Be tour
hefgur of de balding fe 15m =he ae.
| Angie of a Of tp of too tem top of
balldég «= 0°
| Age of elevatiin of pe eh Oe row batten of
lufldig peter
Dittane between “toes bufldäq en ==
her high oo abore Lai Le ‘arm
| Ff we veprént the above data &
om of jue Ann Me ou
Ligure a down ue Lo
4 oleo draw Axlleo, ZAxe =.
vectargle
AG= XO =
here (Aa À à
B= bx = em 2,
a
Fight Magie
Se Noh Be eluded angte
hoe ze fie
PES
“Adjocent side:
Wade,
7 xB ans. -@
> 2tI5=a6(8)
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+ E
i Let AG be De bufld®g and co be Be tour
hefgur of de balding fe 15m =he ae.
| Angie of a Of tp of too tem top of
balldég «= 0°
| Age of elevatiin of pe eh Oe row batten of
lufldig peter
Dittane between “toes bufldäq en ==
her high oo abore Lai Le ‘arm
| Ff we veprént the above data &
om of jue Ann Me ou
Ligure a down ue Lo
4 oleo draw Axlleo, ZAxe =.
vectargle
AG= XO =
here (Aa À à
B= bx = em 2,
a
Fight Magie
Se Noh Be eluded angte
hoe ze fie
PES
“Adjocent side:
Wade,
7 xB ans. -@
> 2tI5=a6(8)
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Let AB Le Pare „ Fées
> JR tower Fe 50 le diagerahf-
Dittance
Sp fer of oltewattin trou” feck “of
tower Bp = 40
Aa (ef etuvakin of top of Magie eo
Arge oy elevaba of Lotto cf lag’ pole (a p> 307
Net height of lowes 210° = nei
«height Of pole = ESP = 80.
the obove data vepretented
0 our wlth Lama?
SpE wight ange Joe o he “
| Felude ongle 50, den
dom of dure
Hone = oppotttr aide
~Adjacent le
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Let AB Le Pare „ Fées
> JR tower Fe 50 le diagerahf-
Dittance
Sp fer of oltewattin trou” feck “of
tower Bp = 40
Aa (ef etuvakin of top of Magie eo
Arge oy elevaba of Lotto cf lag’ pole (a p> 307
Net height of lowes 210° = nei
«height Of pole = ESP = 80.
the obove data vepretented
0 our wlth Lama?
SpE wight ange Joe o he “
| Felude ongle 50, den
dom of dure
Hone = oppotttr aide
~Adjacent le
RD Sharma Class 10 Solutions Applications of Trigonometry
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RD Sharma Class 10 Solutions Applications of Trigonometry
wwnw.LeannCBSE. 20.
le + dope ae
| 76 a>
\ tonte mty zen AS
| ty = 9% LEA SB = Sem
Ye 18-86
= GIB = bx
= 10.392.
shetght =o} “tower x= 5:14 6m
height ob pole 45 Sen
19.
Let Billy tree height Le ne.
heb A, tune that the tree breken at
pop ©
Ange made by broken govt ce! wth ana
=
Pätance between tect of “ee do pot where ft
tourer ground ge E eo
her op tee sh 2 aceeel.= Actce
E
RD Sharma Class 10 Solutions Applications of Trigonometry
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wwnw.LeannCBSE. 20.
le + dope ae
| 76 a>
\ tonte mty zen AS
| ty = 9% LEA SB = Sem
Ye 18-86
= GIB = bx
= 10.392.
shetght =o} “tower x= 5:14 6m
height ob pole 45 Sen
19.
Let Billy tree height Le ne.
heb A, tune that the tree breken at
pop ©
Ange made by broken govt ce! wth ana
=
Pätance between tect of “ee do pot where ft
tourer ground ge E eo
her op tee sh 2 aceeel.= Actce
E
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|
I
RD Sharma Class 10 Solutions Applications of Trigonometry
ven LeamCBSE in
Tre above format & vepreient & Ehe form ey
déque où Ahoun.
cose = Adjacent side | dano shopper‘ side
Hupotercue Apart side,
NS Stree A ET
ket AG be the
Ange of elevation of top: of bulldite from P
220.
AB= height of tauxr = tor
Pale of elevation ch top of dagioff Aro r
B= 45%
ker heGht. of, Ylagsrajf = Bb am. à
al LA a e
Fe eve formation fi svepretented PAL
B fon of due at hewn Pr
wh ZA = ap.
Ty à he angled Arfangle Howey
Sr ong a a
ve e Peludo
4 Eee a
RD Sharma Class 10 Solutions Applications of Trigonometry
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tower and go be the lag stagy.
a
I
RD Sharma Class 10 Solutions Applications of Trigonometry
ven LeamCBSE in
Tre above format & vepreient & Ehe form ey
déque où Ahoun.
cose = Adjacent side | dano shopper‘ side
Hupotercue Apart side,
NS Stree A ET
ket AG be the
Ange of elevation of top: of bulldite from P
220.
AB= height of tauxr = tor
Pale of elevation ch top of dagioff Aro r
B= 45%
ker heGht. of, Ylagsrajf = Bb am. à
al LA a e
Fe eve formation fi svepretented PAL
B fon of due at hewn Pr
wh ZA = ap.
Ty à he angled Arfangle Howey
Sr ong a a
ve e Peludo
4 Eee a
RD Sharma Class 10 Solutions Applications of Trigonometry
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tower and go be the lag stagy.
a
RD Sharma Class 10 Solutions Applications of Trigonometry
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dan
Henze tan ste Moya!
ar
AP= 108 tota = ap
toxetsa,
a
GE 14-32-10
52m
Big ef elagitalf Wa > 152m
dütane béleem Plant foot of toner = 19:52,
2
height + dhe gu = tbm 2 CD,
bflance of ght ton damp pot = En
Legis of shodaw catted by dam pat = fem = En
Let x be te angle Aubtended” ty edge
# |
ade do top ob gl and Lamp post: ze |
Su aloe deta À prete R dom /
Ce xo
tron dy O gemelo Cats:
Inavec. Tank = co
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22
ven LearnCBSE in
dan
Henze tan ste Moya!
ar
AP= 108 tota = ap
toxetsa,
a
GE 14-32-10
52m
Big ef elagitalf Wa > 152m
dütane béleem Plant foot of toner = 19:52,
2
height + dhe gu = tbm 2 CD,
bflance of ght ton damp pot = En
Legis of shodaw catted by dam pat = fem = En
Let x be te angle Aubtended” ty edge
# |
ade do top ob gl and Lamp post: ze |
Su aloe deta À prete R dom /
Ce xo
tron dy O gemelo Cats:
Inavec. Tank = co
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22
RD Sharma Class 10 Solutions Applications of Trigonometry
ven LearnCBSE in
Hana = 48.
| tangs as
| à
| ee
|
of “amp post = AG = Am
ls Lua _prepeaty a Katar Hg
Gn Avec x ApEn
|
| Lote = LEA. Leerrenon “Angti
| Zoe = LAE 90%
| eg AA states
48m
| 2 het, damp Pott = 4 tm
22
height ep The boy PRE 150.2 Ans
| heizt op The building co 20m
Arte of euvatin From Aier poor eleyul 2 a 36°
=
|
RD Sharma Class 10 Solutions Applications of Trigonometry
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ven LearnCBSE in
Hana = 48.
| tangs as
| à
| ee
|
of “amp post = AG = Am
ls Lua _prepeaty a Katar Hg
Gn Avec x ApEn
|
| Lote = LEA. Leerrenon “Angti
| Zoe = LAE 90%
| eg AA states
48m
| 2 het, damp Pott = 4 tm
22
height ep The boy PRE 150.2 Ans
| heizt op The building co 20m
Arte of euvatin From Aier poor eleyul 2 a 36°
=
|
RD Sharma Class 10 Solutions Applications of Trigonometry
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ler he moved
de
Large dy satten trom!
do A! and eye palio changed to
be
tebance bélier chtervatfon porte
<
= ABs ate!
A eat
Meg op tl abever ce eq my
ces co-ep
> co- (ee) a le
A vf EL
arse yet
New The olive data & vepeetentea 8 Chu! took
of dau a hein
To wht angled tte
one ef The angle &
> Ye 20 ety o es
mo 2258-0958
= 148.
SAX. Sn
Dittance walked Heard bulldog = Mm.
Sa
5
a
RD Sharma Class 10 Solutions Applications of Trigonometry
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ler he moved
de
Large dy satten trom!
do A! and eye palio changed to
be
tebance bélier chtervatfon porte
<
= ABs ate!
A eat
Meg op tl abever ce eq my
ces co-ep
> co- (ee) a le
A vf EL
arse yet
New The olive data & vepeetentea 8 Chu! took
of dau a hein
To wht angled tte
one ef The angle &
> Ye 20 ety o es
mo 2258-0958
= 148.
SAX. Sn
Dittance walked Heard bulldog = Mm.
Sa
5
a
RD Sharma Class 10 Solutions Applications of Trigonometry
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RD Sharma Class 10 Solutions Applications of Trigonometry |
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23 . 7
Let height of tower -AB= "hoo
Aegle crade by edge or Hp copo Ahadew do top
of tows «= Got
Let Length of thada be 'xtm. = BD
then length of shadow E e+40)m Thee
age of elval th p= ad”
TRL aleve deta BU vepracated Bau A
br op ure at ran À
SE lo a Anlage ome Fours
at i x, q Cos
Rete kote un V7
taded agile fe E
Port, ee te = A
Adfacent ide A ne
TR à
bé
Hana = 28 Hanpe ne
eo Be
tanto = sh Hane
= he
x —O +402 HE —@
from Da ®
740 = DW)
> 202 24 Dra
2045 m.
height of taser = 2085 en
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23 . 7
Let height of tower -AB= "hoo
Aegle crade by edge or Hp copo Ahadew do top
of tows «= Got
Let Length of thada be 'xtm. = BD
then length of shadow E e+40)m Thee
age of elval th p= ad”
TRL aleve deta BU vepracated Bau A
br op ure at ran À
SE lo a Anlage ome Fours
at i x, q Cos
Rete kote un V7
taded agile fe E
Port, ee te = A
Adfacent ide A ne
TR à
bé
Hana = 28 Hanpe ne
eo Be
tanto = sh Hane
= he
x —O +402 HE —@
from Da ®
740 = DW)
> 202 24 Dra
2045 m.
height of taser = 2085 en
RD Sharma Class 10 Solutions Applications of Trigonometry
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dpt ej bullding = 20m.=46
der shelght ojo ave aleve deg = Km = ge
ergo of aos bald noo Con A
Arge dj elevation of beton tron «> ger.
Ange oh elevate Got of towers pe de
Let dittance between toute. 2 chtervation
de above data E weprumntäd fa
The dom of due at thown er
A ome oa) The eluded age >?
8 “ER ee de à e E
ff
hen ase = oppoitr
ae
RD Sharma Class 10 Solutions Applications of Trigonometry
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ae.
ven LeamCBSE in
dpt ej bullding = 20m.=46
der shelght ojo ave aleve deg = Km = ge
ergo of aos bald noo Con A
Arge dj elevation of beton tron «> ger.
Ange oh elevate Got of towers pe de
Let dittance between toute. 2 chtervation
de above data E weprumntäd fa
The dom of due at thown er
A ome oa) The eluded age >?
8 “ER ee de à e E
ff
hen ase = oppoitr
ae
RD Sharma Class 10 Solutions Applications of Trigonometry
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ae.
RD Sharma Class 10 Solutions Applications of Trigonometry
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25 2
het eight ch EA
Heights sp tale buflditg = 8m=cep
age ef depen of top of tate buat
Angie Of depresion of Labs of tat Lua po 45°
Phitance between two budget om = Ro
ABE A but garen th Mo
robin À
. € na I
Asse on + 9 1
A6 = (ar aye x
Te above honte E sepeeredey
BE dom ep que as trou
> 2° ao
MO > arerag
> ain =e Jar Bn = Alan
ay = lan)
= Han) Le arm AR Rate = Allem
height of muleta haPlding> Alardm.
Distance betucen balding = Ale nm
RD Sharma Class 10 Solutions Applications of Trigonometry
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2
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25 2
het eight ch EA
Heights sp tale buflditg = 8m=cep
age ef depen of top of tate buat
Angie Of depresion of Labs of tat Lua po 45°
Phitance between two budget om = Ro
ABE A but garen th Mo
robin À
. € na I
Asse on + 9 1
A6 = (ar aye x
Te above honte E sepeeredey
BE dom ep que as trou
> 2° ao
MO > arerag
> ain =e Jar Bn = Alan
ay = lan)
= Han) Le arm AR Rate = Allem
height of muleta haPlding> Alardm.
Distance betucen balding = Ale nm
RD Sharma Class 10 Solutions Applications of Trigonometry
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2
RD Sharma Class 10 Solutions Applications of Trigonometry
28
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26 .
der height of pedatat be ht
Male = bbe.
height el
O ej top of Mate ange
Angie of elevation of paper pe que
JR above data 5 vepueted à
dh Yorn ob are a shown.
ape abr ange Afargle oe
of te Felidea ale o
Ten at =>
FA apport stas |
HAjacent ee
ro
Hand = Be
De
paras che
be
Dee chi
be = Him.
rom 010 hera
à
PAR = bone
te
>
Ll ot)
A
Helge of qedatal = of
RD Sharma Class 10 Solutions Applications of Trigonometry
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28
www. LeamnCBSE.in
26 .
der height of pedatat be ht
Male = bbe.
height el
O ej top of Mate ange
Angie of elevation of paper pe que
JR above data 5 vepueted à
dh Yorn ob are a shown.
ape abr ange Afargle oe
of te Felidea ale o
Ten at =>
FA apport stas |
HAjacent ee
ro
Hand = Be
De
paras che
be
Dee chi
be = Him.
rom 010 hera
à
PAR = bone
te
>
Ll ot)
A
Helge of qedatal = of
RD Sharma Class 10 Solutions Applications of Trigonometry
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2. .
La height of ter = thn = 4e
Latta, + ver be Wem = Bee
angle op élévation from bank of vive 60 deme)
Angle ef Platón from 20m) away Me bank
q ve Be 86% [Asom
Thi above data 1% vepreseoted
I dons op digume oF shown
By wight angle age oo oy
Te Édudea age 0 Bien
Hove = ort atte |
= “hajacent ide)
+20 he -@
O HO + cross
7. +=
Pa 20% 2210
he M = 1028 < 108 om.
hefgne of
width of
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2. .
La height of ter = thn = 4e
Latta, + ver be Wem = Bee
angle op élévation from bank of vive 60 deme)
Angle ef Platón from 20m) away Me bank
q ve Be 86% [Asom
Thi above data 1% vepreseoted
I dons op digume oF shown
By wight angle age oo oy
Te Édudea age 0 Bien
Hove = ort atte |
= “hajacent ide)
+20 he -@
O HO + cross
7. +=
Pa 20% 2210
he M = 1028 < 108 om.
hefgne of
width of
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RD Sharma Class 10 Solutions Applications of Trigonometry
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22 given
Height of building. = two
Hee ef
“Angle
e
cable tower = ‘Wen eb
ey ckevattin of top of eme from top
of huida = co
Angle ef deprisa,
Hop of bufiding pos”
tthe above data À veprtientes
dorm d igure. as shown
à
CX = tx
bation ec) tower Arem
Let
CD= De exe = tm tat
Des Ae
opp Ade (xe) Mendo s
Eta a
EU Ge 2
> Axe Hm, |) arse
!
but Cp = +4
= AGE = FG ayn
he of cable towers = (Mr
RD Sharma Class 10 Solutions Applications of Trigonometry
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ven LeamCBSE in
22 given
Height of building. = two
Hee ef
“Angle
e
cable tower = ‘Wen eb
ey ckevattin of top of eme from top
of huida = co
Angle ef deprisa,
Hop of bufiding pos”
tthe above data À veprtientes
dorm d igure. as shown
à
CX = tx
bation ec) tower Arem
Let
CD= De exe = tm tat
Des Ae
opp Ade (xe) Mendo s
Eta a
EU Ge 2
> Axe Hm, |) arse
!
but Cp = +4
= AGE = FG ayn
he of cable towers = (Mr
RD Sharma Class 10 Solutions Applications of Trigonometry
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RD Sharma Class 10 Solutions Applications of Trigonometry
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20: given . 5
Hein cf LSM howe = 25m = am» Aw
Ange of depression of aps a = 36
Argle of deprettion ch ahip 2, pe 48"
The above data & vepruentd
fom oh E mu then.
her
Dttance
betiteen Abe be
Km = ep
In sight rSagle EY ove
e
Peluded angle &e then
hae e
Le
themes 4e
E
+86 =3456, O
020» Re
> fail
vs Distenee “betiveen hip
= ASLAN,
RD Sharma Class 10 Solutions Applications of Trigonometry
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20: given . 5
Hein cf LSM howe = 25m = am» Aw
Ange of depression of aps a = 36
Argle of deprettion ch ahip 2, pe 48"
The above data & vepruentd
fom oh E mu then.
her
Dttance
betiteen Abe be
Km = ep
In sight rSagle EY ove
e
Peluded angle &e then
hae e
Le
themes 4e
E
+86 =3456, O
020» Re
> fail
vs Distenee “betiveen hip
= ASLAN,
RD Sharma Class 10 Solutions Applications of Trigonometry
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zo. :
Angle of elevation of top of bullding dro
foot of “tower = 3072 4
“Ange of elevation of top of “lowes Aron
foot of han = ep
height of “Tower = SO mAs
Aer of bulla = ohm
es
Ti aleve A A ed à gone +
E00 do Plane = ppal aide
Adjacent arde
In aden SN re
ton p> de Hone = co
en Bo
Hlonco”= 50 | tas = h.
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Em
®
me oh Re tual dou
www LearnCBSE in
zo. :
Angle of elevation of top of bullding dro
foot of “tower = 3072 4
“Ange of elevation of top of “lowes Aron
foot of han = ep
height of “Tower = SO mAs
Aer of bulla = ohm
es
Ti aleve A A ed à gone +
E00 do Plane = ppal aide
Adjacent arde
In aden SN re
ton p> de Hone = co
en Bo
Hlonco”= 50 | tas = h.
RD Sharma Class 10 Solutions Applications of Trigonometry
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Em
®
me oh Re tual dou
RD Sharma Class 10 Solutions Applications of Trigonometry
ven LeamCBSE in os
3h
“Height of the bridge = Sem. Cael
Angle oh depreitten of banka ne «= 20° (en
Angle depression of bork a he p= 45116,
given banks ove on epale sles.
Distance beliveen banks ASE Abt 88.
JR above thfowmahin & weptieded Eu dom
ure a Ab
do ne ange ae Y on
Oj the Preludes age À e
Then
reve = were |
_ Aajaceat ae
5 (nes, a L
5 ake
tora = ae Yong Dane
ee cre
Hans = 30 30:
Be Se,
Be = 30% em 88, = 30m
= Bro
= sola)
hekween bank = 30(% +0 m,
RD Sharma Class 10 Solutions Applications of Trigonometry
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ven LeamCBSE in os
3h
“Height of the bridge = Sem. Cael
Angle oh depreitten of banka ne «= 20° (en
Angle depression of bork a he p= 45116,
given banks ove on epale sles.
Distance beliveen banks ASE Abt 88.
JR above thfowmahin & weptieded Eu dom
ure a Ab
do ne ange ae Y on
Oj the Preludes age À e
Then
reve = were |
_ Aajaceat ae
5 (nes, a L
5 ake
tora = ae Yong Dane
ee cre
Hans = 30 30:
Be Se,
Be = 30% em 88, = 30m
= Bro
= sola)
hekween bank = 30(% +0 m,
RD Sharma Class 10 Solutions Applications of Trigonometry
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RD Sharma Class 10 Solutions Applications of Trigonometry
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ES .
ker two poles As co be of equal Heine a"
ABE Cp ahi
Distance between poles BD = 80m.
Let X be pot betiueen he. From X
Ange élevatin of mae bei a = bot
Ange of elevate) be p= oe
e Piz AS
ex. Van
Distance 3
+ {rom qele co
Dictonce of (Pear wer"
dr above deta le |
dem où her
a ee À 7 À
2 o
Te cluded a X) D >
plz side ji 74 ER
bo | —6
O MO Box = xnla) > Sox =3%
> iso» = 20 he 208
het o) poles = 208 m
Diane ch peht trem polea = 20
Dirtarce of pot from pole = com.
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ES .
ker two poles As co be of equal Heine a"
ABE Cp ahi
Distance between poles BD = 80m.
Let X be pot betiueen he. From X
Ange élevatin of mae bei a = bot
Ange of elevate) be p= oe
e Piz AS
ex. Van
Distance 3
+ {rom qele co
Dictonce of (Pear wer"
dr above deta le |
dem où her
a ee À 7 À
2 o
Te cluded a X) D >
plz side ji 74 ER
bo | —6
O MO Box = xnla) > Sox =3%
> iso» = 20 he 208
het o) poles = 208 m
Diane ch peht trem polea = 20
Dirtarce of pot from pole = com.
RD Sharma Class 10 Solutions Applications of Trigonometry
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RD Sharma Class 10 Solutions Applications of Trigonometry
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33. :
Height of ee AB = 20m
“Arge
ge Of depression of poes sect es
Ange of dep of pol 2 deep = 30.
BC be ore pole 5 ES be other pole
re po axe 0 Spout ates
Lattin of Re e
8,6188,
Di above Boats di
vepresented 8 oros oh due ME
abro
To argot “régle one o) Beluded age o
oppostl MIR
Aspıeent ide
tap’: Ae
tenact= 20.
ve,
BA = 206
24 20% =
a
Lien + vives Sm.
“
RE en
3
RD Sharma Class 10 Solutions Applications of Trigonometry
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ven LeamCBSE in ss
33. :
Height of ee AB = 20m
“Arge
ge Of depression of poes sect es
Ange of dep of pol 2 deep = 30.
BC be ore pole 5 ES be other pole
re po axe 0 Spout ates
Lattin of Re e
8,6188,
Di above Boats di
vepresented 8 oros oh due ME
abro
To argot “régle one o) Beluded age o
oppostl MIR
Aspıeent ide
tap’: Ae
tenact= 20.
ve,
BA = 206
24 20% =
a
Lien + vives Sm.
“
RE en
3
RD Sharma Class 10 Solutions Applications of Trigonometry
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RD Sharma Class 10 Solutions Applications of Trigonometry
By given
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Meet ch diag Aah = Am = Be.
Met height ch tower = thm = 46
nage of clevattin of bation of doqua ue aot
Age oh levas of top oof gato pe gar
peat 0} ebtewa
Ti aba data lo vepréientes
of uve at shown
a sant ange Age
JE Retudea angle
Hon (be vp?
one
re
ans = gro!
tong ike lay aa
oe dote hey
Ar ar
Ar aD Mn
ten 0x0
he has
heey
Hav st > the x an
Bt Gn
= dl a Bslaty
het of Howe = sn,
RD Sharma Class 10 Solutions Applications of Trigonometry
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By given
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Meet ch diag Aah = Am = Be.
Met height ch tower = thm = 46
nage of clevattin of bation of doqua ue aot
Age oh levas of top oof gato pe gar
peat 0} ebtewa
Ti aba data lo vepréientes
of uve at shown
a sant ange Age
JE Retudea angle
Hon (be vp?
one
re
ans = gro!
tong ike lay aa
oe dote hey
Ar ar
Ar aD Mn
ten 0x0
he has
heey
Hav st > the x an
Bt Gn
= dl a Bslaty
het of Howe = sn,
RD Sharma Class 10 Solutions Applications of Trigonometry
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ot
35.
Let
Length of shadow be EL e A
be w= 45
engin Ad et be Ed olan
fun attitude fe PER
Let chedght oh tower be ‘hin = AB,
the above Fran À repretented 8
9 dom
o dígure a Alu er
A right ge ne ep SS)
enge fee then
fe
conti. € 24
Ty Rec An Ape.
Hana = AB Hop AE.
Be pared.
ons ch Month = sh.
Eu ara
h= à —O sata: hs. —O
D RO > ren
me a) = 22
>.
height ch Tower = alarm
RD Sharma Class 10 Solutions Applications of Trigonometry
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ot
35.
Let
Length of shadow be EL e A
be w= 45
engin Ad et be Ed olan
fun attitude fe PER
Let chedght oh tower be ‘hin = AB,
the above Fran À repretented 8
9 dom
o dígure a Alu er
A right ge ne ep SS)
enge fee then
fe
conti. € 24
Ty Rec An Ape.
Hana = AB Hop AE.
Be pared.
ons ch Month = sh.
Eu ara
h= à —O sata: hs. —O
D RO > ren
me a) = 22
>.
height ch Tower = alarm
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Se 5
het AB be hehe of free à th ik broken at
Porc. and dep touche grout até d
Angle made ty tp &= 30°
Dütanee fren Spot of tree trom point where fr
toucha ground = 80 metre
wu above frfownatfin fk reprinted om of
8
dure mu Aron
hegnt of tree = Aas Arc
= Ac + ca
Ee a
age fe am
Et oprofte_atde
| Adjacent sde
Han sc = Ac
BA
AC = 10. en:
da
Aa = CAFC = 10, 20
Ke
heft ch bee = 106
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Se 5
het AB be hehe of free à th ik broken at
Porc. and dep touche grout até d
Angle made ty tp &= 30°
Dütanee fren Spot of tree trom point where fr
toucha ground = 80 metre
wu above frfownatfin fk reprinted om of
8
dure mu Aron
hegnt of tree = Aas Arc
= Ac + ca
Ee a
age fe am
Et oprofte_atde
| Adjacent sde
Han sc = Ac
BA
AC = 10. en:
da
Aa = CAFC = 10, 20
Ke
heft ch bee = 106
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ES
hengn ch cabe comectzd to balloon = 2150 [ce]
tel tration ate wf
Age o of cate eft ground
a= ee
Height of batten fron grund = “him = AB
fa above data À weprtsented A, down
oh que © Aba
cin wight twhngle one of The
fecludea angie be Sen
[doo = oppesite sae | 2
hypatenwe
Bab > he 2
shif> Ab > ic .
de Bb BSG = 10458
Zu height of balloon shen ground = 101544 m.
38:
heéght of elf = Som = ag.
Age of elevan fren Man 2, à 30 [m]
Ange E dran mana, pe bo [mr
tance between “hie men = Man,
= amr em,
u above goma
veptiented E dom of
dee 05 hour
By sant age Monge me dj ue
focludes ange
Fes =
en
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ES
hengn ch cabe comectzd to balloon = 2150 [ce]
tel tration ate wf
Age o of cate eft ground
a= ee
Height of batten fron grund = “him = AB
fa above data À weprtsented A, down
oh que © Aba
cin wight twhngle one of The
fecludea angie be Sen
[doo = oppesite sae | 2
hypatenwe
Bab > he 2
shif> Ab > ic .
de Bb BSG = 10458
Zu height of balloon shen ground = 101544 m.
38:
heéght of elf = Som = ag.
Age of elevan fren Man 2, à 30 [m]
Ange E dran mana, pe bo [mr
tance between “hie men = Man,
= amr em,
u above goma
veptiented E dom of
dee 05 hour
By sant age Monge me dj ue
focludes ange
Fes =
en
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MM > MB BM, =
NG
Diktorce — belicccn men = OB meta,
5
3. ker
help of pele=him = kunt altttude rom Around,
déngih ef Ahada be ‘1
qe thar Lah
Ange of elevatin oh cust altflude be ©
The above data &
op dique à shown
Bn wight twang
included angle & 0 Then
ES
tone = AB
Be
vepresente E qomo
hey
7 02 tartoe4se
Angle oh tute attitude % 460
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MM > MB BM, =
NG
Diktorce — belicccn men = OB meta,
5
3. ker
help of pele=him = kunt altttude rom Around,
déngih ef Ahada be ‘1
qe thar Lah
Ange of elevatin oh cust altflude be ©
The above data &
op dique à shown
Bn wight twang
included angle & 0 Then
ES
tone = AB
Be
vepresente E qomo
hey
7 02 tartoe4se
Angle oh tute attitude % 460
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Per AB be me ‘hurtas
à
Angle of cialis an poh [Eté tatin ] à 2 be?
Aoge dj tlivalton trom pot QC Real à] B= ast
Dfitanee between Fifattahlne PQ = 20km
Be above Frfomatin tk vepractea a
Be den op iger at theres
Ta wight triangle 4 one + Th
fe Fe
ane = proie efde PEN a &
Adjacent eee. u 9 ps;
tan = ne. tanp= 46
ar aa
toast = AR ton4s?= a
AP AR
Are a AR = 4a, —
e © | 7
Ore > =
© > Arras = Bene = re)
> 205 a/a ie
(SE) > AB = 208
Natt
AB= 20%
== ILA =10 (2-6)
AR = AB: .
8 = ABE to(3- A) = 1033) = 12:64 L.,
Ars “Ae =
Be = to (Ga) = 10x 0-82 = parle
Station 4
should Aend The team and they ham
do Haut 4.22 km,
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Per AB be me ‘hurtas
à
Angle of cialis an poh [Eté tatin ] à 2 be?
Aoge dj tlivalton trom pot QC Real à] B= ast
Dfitanee between Fifattahlne PQ = 20km
Be above Frfomatin tk vepractea a
Be den op iger at theres
Ta wight triangle 4 one + Th
fe Fe
ane = proie efde PEN a &
Adjacent eee. u 9 ps;
tan = ne. tanp= 46
ar aa
toast = AR ton4s?= a
AP AR
Are a AR = 4a, —
e © | 7
Ore > =
© > Arras = Bene = re)
> 205 a/a ie
(SE) > AB = 208
Natt
AB= 20%
== ILA =10 (2-6)
AR = AB: .
8 = ABE to(3- A) = 1033) = 12:64 L.,
Ars “Ae =
Be = to (Ga) = 10x 0-82 = parle
Station 4
should Aend The team and they ham
do Haut 4.22 km,
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+
At
Hetgnt eh thip chem unter Level = tom = AB.
Angie of elewatten oh op of clit x = 45
Ange oly deprcion ef Eten o dif x
het Mp cD = thin
dittance of ship “rem (Epi aux. cé.
ES
Hee a ame am €
Ln m
Ten heuer oh elfe pater à N
Goran.
BH love dato vepeetented |
5 don Noy ue os me 5 =
La vint totengte, FE one ch The theluded ange à
Ou Mer Tre = oppartz ade
Afeenteise
das 4sh= cx darse > xp
Ax 22
do. Ds
Ax = ‘ae A = 10%
“ao
e cht) = don = inten
dûtare tetuces fp + Cth = 108.
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+
At
Hetgnt eh thip chem unter Level = tom = AB.
Angie of elewatten oh op of clit x = 45
Ange oly deprcion ef Eten o dif x
het Mp cD = thin
dittance of ship “rem (Epi aux. cé.
ES
Hee a ame am €
Ln m
Ten heuer oh elfe pater à N
Goran.
BH love dato vepeetented |
5 don Noy ue os me 5 =
La vint totengte, FE one ch The theluded ange à
Ou Mer Tre = oppartz ade
Afeenteise
das 4sh= cx darse > xp
Ax 22
do. Ds
Ax = ‘ae A = 10%
“ao
e cht) = don = inten
dûtare tetuces fp + Cth = 108.
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42 :
Hetght of Shfp above wale tel = Im = A8
Angle of elevation de Hp of chp Chey «= 60°
Angle of deprétien of lato of Hi B= zur
Hegpt ef hf = ep
Dittarce between chip & Hit = ax,
eigen of Hl above chip
xa
do A
Hebe of HW = rm à 8m
Whe above data vepacta E ®
& dom
Oh dure a hour
fin wight Mège Ff oe of Peludas ange Ro
Ten tose = oppottte «Pde
ajacent Tae
Hans ex tong = ao
AX Es
Hantot= & rt
AK. Ax
Axe à Ax= 9%
Key
2280 > A= 24m
a a
AX = 8m
ent
2 dent of (Ar) ma.
Diitonee between HUA SH = Alm
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42 :
Hetght of Shfp above wale tel = Im = A8
Angle of elevation de Hp of chp Chey «= 60°
Angle of deprétien of lato of Hi B= zur
Hegpt ef hf = ep
Dittarce between chip & Hit = ax,
eigen of Hl above chip
xa
do A
Hebe of HW = rm à 8m
Whe above data vepacta E ®
& dom
Oh dure a hour
fin wight Mège Ff oe of Peludas ange Ro
Ten tose = oppottte «Pde
ajacent Tae
Hans ex tong = ao
AX Es
Hantot= & rt
AK. Ax
Axe à Ax= 9%
Key
2280 > A= 24m
a a
AX = 8m
ent
2 dent of (Ar) ma.
Diitonee between HUA SH = Alm
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height eh temple 1 (48) = Som.
Age of dpi ep top d tere, w= 36%
Arge of deprettion of batten of tenplea, fe ect
height of temple 2 (09) = Win
Lifdin ch river = Aa,
us
fu above data ft vepretented
th dem of dique at than ©
Lo sign age A Ome du
fretuded angled ©) hen
Wut,
Het of temple 2 =
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height eh temple 1 (48) = Som.
Age of dpi ep top d tere, w= 36%
Arge of deprettion of batten of tenplea, fe ect
height of temple 2 (09) = Win
Lifdin ch river = Aa,
us
fu above data ft vepretented
th dem of dique at than ©
Lo sign age A Ome du
fretuded angled ©) hen
Wut,
Het of temple 2 =
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het Aeroplane travetted em À de B 5 16 eu
Angle of etvatin of pont Axe
Angle of elevation of port 6. B= 307
Bocomebyes.
height of arreplane diem Grand a
Détance havelled A IS tes=Ab-ra
velocity En speed. = Aüknce “havetlea fine
à
above data à vepretented & Horn
iure 027 Shown,
E Pa
Le
xQ
Hanse? = 3000
xQ
XP = 3000. m XQ = 300083
PR = XR=-KE = 3000(3-1)w
ua = Rd BeelOY - za Ty
one v —_—
200x 0.132. zn
= 1464 mlcec
ap ah o = ern tue
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het Aeroplane travetted em À de B 5 16 eu
Angle of etvatin of pont Axe
Angle of elevation of port 6. B= 307
Bocomebyes.
height of arreplane diem Grand a
Détance havelled A IS tes=Ab-ra
velocity En speed. = Aüknce “havetlea fine
à
above data à vepretented & Horn
iure 027 Shown,
E Pa
Le
xQ
Hanse? = 3000
xQ
XP = 3000. m XQ = 300083
PR = XR=-KE = 3000(3-1)w
ua = Rd BeelOY - za Ty
one v —_—
200x 0.132. zn
= 1464 mlcec
ap ah o = ern tue
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45.
Let Atwoplare traveled rom Ares fh 10 Leur
Angle of elevation of pot A= «= 68
Angle of ctevation of ¿Porres B = Se
shetght of arvoflane trom ground = thm =AP= a8
Dirane “havetted fh to lect = AB PQ.
Apeed =
Déttanée” “havetted He
JR Abe data &
ds u
Bs vn
re clics
veprüented
5 fom ch
tanso= 4 stonp= Be
Px. | xa
E IE A
Pa = fotoo | oy
XR EB ben
PR = aa = > 2
bie e
eG ae
Apted= £8 = Bote = 20 x Cong?
= ES
LE |
dores
= 24003 bal.
2 eed op Aeroplane = 20003 bag:
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45.
Let Atwoplare traveled rom Ares fh 10 Leur
Angle of elevation of pot A= «= 68
Angle of ctevation of ¿Porres B = Se
shetght of arvoflane trom ground = thm =AP= a8
Dirane “havetted fh to lect = AB PQ.
Apeed =
Déttanée” “havetted He
JR Abe data &
ds u
Bs vn
re clics
veprüented
5 fom ch
tanso= 4 stonp= Be
Px. | xa
E IE A
Pa = fotoo | oy
XR EB ben
PR = aa = > 2
bie e
eG ae
Apted= £8 = Bote = 20 x Cong?
= ES
LE |
dores
= 24003 bal.
2 eed op Aeroplane = 20003 bag:
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4
"ABs helfe of tower = Som
cre Het of UGC plo
| Angle of depa of top oh lala x= 452
| Angle Spain oh orte ofp bug pe 0
Du above data & repre ee form $
due o han.
M wt age (one Preluded
age Eo Then y
Hare = oppoft fate
apercu
Han cp Me A
tx Fe
Lane ax toners off
= ©
AK = cx. Ex = 80"
©
height oh busted (polo = BG
Diane between pole & tou = 50 m,
ime ze
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4
"ABs helfe of tower = Som
cre Het of UGC plo
| Angle of depa of top oh lala x= 452
| Angle Spain oh orte ofp bug pe 0
Du above data & repre ee form $
due o han.
M wt age (one Preluded
age Eo Then y
Hare = oppoft fate
apercu
Han cp Me A
tx Fe
Lane ax toners off
= ©
AK = cx. Ex = 80"
©
height oh busted (polo = BG
Diane between pole & tou = 50 m,
ime ze
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At 5
Dittance ketiveen treet = bom: Leo]
Height of deco thee = 20m. co)
hi Ca),
det height of Fever tree =
real of depreccton seen heand tee top An Ht bee
top 545
formato, À vepraented E dorm of fare)
|
Mi aloe A
a hon
Do Br drag Ye d te
Kane = te
Adjacent ride
draw) cxLAe, Cx = BD = Com.
Feluded ange Lo Thin
Naini
XB= cn = AB-Ax
tonus A
ex
tqs Ax à Allem
bo
KB= Co = Na-ax
= to-bo
= 20m
heat cp fecond tree = tom
bete of HEE ee = 20m
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At 5
Dittance ketiveen treet = bom: Leo]
Height of deco thee = 20m. co)
hi Ca),
det height of Fever tree =
real of depreccton seen heand tee top An Ht bee
top 545
formato, À vepraented E dorm of fare)
|
Mi aloe A
a hon
Do Br drag Ye d te
Kane = te
Adjacent ride
draw) cxLAe, Cx = BD = Com.
Feluded ange Lo Thin
Naini
XB= cn = AB-Ax
tonus A
ex
tqs Ax à Allem
bo
KB= Co = Na-ax
= to-bo
= 20m
heat cp fecond tree = tom
bete of HEE ee = 20m
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41
48
AB be the tee bean eos
stom déttance "aro trom “Wee (Angle of elevaba |
be xk ot potter
| trem dittare "eo from tree Angle off claros
| ke p atipèt a
| SR above data oR represented & me
[fom of diguie a1 thou
Bh nr hfe ff one of
le fecludea angle lio teo
— OE
Haro = oppodt aide Lt
] Adfoent far er © oi
| aw AX ee det & ='am,
Hana =
a | “top = Ax
| Pa ex
tone = An tape ax
xra ET
cobe= ura cotpe ab
| AK: TAK.
| ara 2 ax cote tbs axcotp
=9
O-O > (erb- (ata = Axcotp- ax cote
> beas aces
Hana took.
| PAK 0-0 tan tang
ara = tong |
A = reptan.te
| Heégbr of top from ground = 004 = fap =
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41
48
AB be the tee bean eos
stom déttance "aro trom “Wee (Angle of elevaba |
be xk ot potter
| trem dittare "eo from tree Angle off claros
| ke p atipèt a
| SR above data oR represented & me
[fom of diguie a1 thou
Bh nr hfe ff one of
le fecludea angle lio teo
— OE
Haro = oppodt aide Lt
] Adfoent far er © oi
| aw AX ee det & ='am,
Hana =
a | “top = Ax
| Pa ex
tone = An tape ax
xra ET
cobe= ura cotpe ab
| AK: TAK.
| ara 2 ax cote tbs axcotp
=9
O-O > (erb- (ata = Axcotp- ax cote
> beas aces
Hana took.
| PAK 0-0 tan tang
ara = tong |
A = reptan.te
| Heégbr of top from ground = 004 = fap =
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aa
Helaht oh vestical tower
FR.
Angle of elevatin of top from x, a = 60°
Arge of elevates of top! trom Yom above x),
Be 45
The above ornato epreiented & foro ofr
Pique der
Ty wget bg oe) of Te
Beluded age 40 thew
es
[Hans = oppoite ae
L___ Haren te
Draw AY PO
"RAP = 40m, AY= Px,
A = (PQ 4.0) mn.
todas? = ag tonceth re,
AY Pe
Ag = ay Pre Pg
=
> Pago ay 3
bur Ey Pa — 40» to
al
PO B~40G2 re > PRLB) = 400,
FRERE Gis
Ga SEL = 20GLAey = 20(3+ 03)
Hee of tower Sb" = 20(2+.9) m.
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aa
Helaht oh vestical tower
FR.
Angle of elevatin of top from x, a = 60°
Arge of elevates of top! trom Yom above x),
Be 45
The above ornato epreiented & foro ofr
Pique der
Ty wget bg oe) of Te
Beluded age 40 thew
es
[Hans = oppoite ae
L___ Haren te
Draw AY PO
"RAP = 40m, AY= Px,
A = (PQ 4.0) mn.
todas? = ag tonceth re,
AY Pe
Ag = ay Pre Pg
=
> Pago ay 3
bur Ey Pa — 40» to
al
PO B~40G2 re > PRLB) = 400,
FRERE Gis
Ga SEL = 20GLAey = 20(3+ 03)
Hee of tower Sb" = 20(2+.9) m.
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Let cloud be ak her Pa (Feprerented |
Avon lake déve |
From poil x, 25e0ra above fhe dake angle
of elevotitn of ep of elsa a = 158
ETES
Ange oh precis of had A e pa gee
dere Pas PQ! Gray Ay ire
Jet Ag = hm AP
Pa = bhrom
TR above data & vepracia |
& dom obs gure as chown
Do right Eve
angle Ke Jun
ay AY
> 0r66=h 2 AU = x + Chey
> ay i Den
Da 7 Ash —D
AmOsD ts hr > Shh = 2x 2600
eu > = sooo
8
SoG = 1830-8302,
shefgnt of cloud above late = bra
1820831 + 2600
= 4300-8319, m.
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Let cloud be ak her Pa (Feprerented |
Avon lake déve |
From poil x, 25e0ra above fhe dake angle
of elevotitn of ep of elsa a = 158
ETES
Ange oh precis of had A e pa gee
dere Pas PQ! Gray Ay ire
Jet Ag = hm AP
Pa = bhrom
TR above data & vepracia |
& dom obs gure as chown
Do right Eve
angle Ke Jun
ay AY
> 0r66=h 2 AU = x + Chey
> ay i Den
Da 7 Ash —D
AmOsD ts hr > Shh = 2x 2600
eu > = sooo
8
SoG = 1830-8302,
shefgnt of cloud above late = bra
1820831 + 2600
= 4300-8319, m.
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5
let x be got hr meter above tate
Ange of elevattin of cloud den x = x.
Ange of elepresitin of cloud “épléehèn Lake =p
shetght of cloud rom Lake = ra
A A els re
| chew xa, AOS. tm
Ñ APEX = Mien,
Ditarce of
2 cloud fron pote of obtervate,
xq
FR above data repented à dot Oh dl
a thaw,
Do Aaex Rava
tank = A
Era tonp le
bre lr
© No
Dos axe 20
Tr ana tonp- tana,
core = ANG gn wy
XQ beca
> xQ= 2haex
amp tora
eh Cloud from plot of obtervekin
= 2bsecaltonp-tone,
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5
let x be got hr meter above tate
Ange of elevattin of cloud den x = x.
Ange of elepresitin of cloud “épléehèn Lake =p
shetght of cloud rom Lake = ra
A A els re
| chew xa, AOS. tm
Ñ APEX = Mien,
Ditarce of
2 cloud fron pote of obtervate,
xq
FR above data repented à dot Oh dl
a thaw,
Do Aaex Rava
tank = A
Era tonp le
bre lr
© No
Dos axe 20
Tr ana tonp- tana,
core = ANG gn wy
XQ beca
> xQ= 2haex
amp tora
eh Cloud from plot of obtervekin
= 2bsecaltonp-tone,
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Let Fe be “Ambit of Araplane rom ground.
KEN to ellos où opel elder of
| te anopue. ate
Angie op deprttlon of € tron hax
Angle of depresion of Y ón =p
IRL above dato do Nepeetentet Ffm of
due a chou. ek
Lo vight Coa" one
of Reluded angle fo Mme
lane = pair &
¡EM
| Wire pe
xe
xa = fe
Fane SEP
xQret = Pa, fag ea (ord
Fong ES = al
| > in
= ip stoop”
Fase. tang
| > Pas done. top
Tana Hong
e evoplane = Tant: tong ele
“het of Sep Héron
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Let Fe be “Ambit of Araplane rom ground.
KEN to ellos où opel elder of
| te anopue. ate
Angie op deprttlon of € tron hax
Angle of depresion of Y ón =p
IRL above dato do Nepeetentet Ffm of
due a chou. ek
Lo vight Coa" one
of Reluded angle fo Mme
lane = pair &
¡EM
| Wire pe
xe
xa = fe
Fane SEP
xQret = Pa, fag ea (ord
Fong ES = al
| > in
= ip stoop”
Fase. tang
| > Pas done. top
Tana Hong
e evoplane = Tant: tong ele
“het of Sep Héron
RD Sharma Class 10 Solutions Applications of Trigonometry
www.LeamCBSE.in
ss
RD Sharma Class 10 Solutions Applications of Trigonometry
ven LeamCBSE in
Spa ë pat segu ae & tower help.
Angle of elevation À e Axor Pau
Ange of elivation of @ Home
Du above hdomatin E vepresented & dorm of
dure at thes |
By wht Andre A oe ob me F
Peludas ange fio, un
Draw GX EAB) PEE AK APs qx i
D dex By aer
Hope ex Hon « SAB
shegpt of fotos = a tans ane tags
Dittoree bf port # tuer = allims-tanp)
RD Sharma Class 10 Solutions Applications of Trigonometry
www.LearnCBSE.in
ven LeamCBSE in
Spa ë pat segu ae & tower help.
Angle of elevation À e Axor Pau
Ange of elivation of @ Home
Du above hdomatin E vepresented & dorm of
dure at thes |
By wht Andre A oe ob me F
Peludas ange fio, un
Draw GX EAB) PEE AK APs qx i
D dex By aer
Hope ex Hon « SAB
shegpt of fotos = a tans ane tags
Dittoree bf port # tuer = allims-tanp)
RD Sharma Class 10 Solutions Applications of Trigonometry
www.LearnCBSE.in
RD Sharma Class 10 Solutions Applications of Trigonometry
ven LeamCBSE in 65
Sá ler 46 be daddy aug ak an ela
a to ground
when The foot & pulled. through ditarce ‘a!
Let Bel = "Ben and AA
age of elation he = B
new
UR. above thdommatfon fh represented porn op
dure of thou
Ler ard groud Be AB = A8
Ax Bray
Ba nace:
sina = AR 2y Shas gb y
ae AB
dosa = EL = Com
ne
O-0 > exp- cosá = A —
CT
© CES b
> [Re cue cop
EAN
RD Sharma Class 10 Solutions Applications of Trigonometry
www.LeamCBSE.in
ven LeamCBSE in 65
Sá ler 46 be daddy aug ak an ela
a to ground
when The foot & pulled. through ditarce ‘a!
Let Bel = "Ben and AA
age of elation he = B
new
UR. above thdommatfon fh represented porn op
dure of thou
Ler ard groud Be AB = A8
Ax Bray
Ba nace:
sina = AR 2y Shas gb y
ae AB
dosa = EL = Com
ne
O-0 > exp- cosá = A —
CT
© CES b
> [Re cue cop
EAN
RD Sharma Class 10 Solutions Applications of Trigonometry
www.LeamCBSE.in
|
| RD Sharma Class 10 Solutions Applications of Trigonometry
www. LeamnCBSE.in
55 Leb heh ch tower be ho
| “Angle of elevation at point A or ground = x
= Pe
| het @ be fob ‘bin abe fe A
angle of depresstin of foot of “louer chow 8e p |
ES above deta & vepeeded 3 dor ob Figure
IS
craw PX LPR
Tn arex
thanx = Po
ESS
tod re
tg. ex
>. es Se bet boto cat
eight oh tour = beton cat
Se:
Heégt ch obtervey 15m
Height of “rower = Pa = 0m
E of che above The vltewer eye = 3015
Yes 285m,
Distance betüen “tous & obterver XB = 286%
9 be angle of cat op touxs tp “on eue
TR above data & vepretented
dome of fhe at than
hon di tono opc ate
Mjacent (Pe
RD Sharma Class 10 Solutions Applications of Trigonometry»
www.LeamCBSE.in
| RD Sharma Class 10 Solutions Applications of Trigonometry
www. LeamnCBSE.in
55 Leb heh ch tower be ho
| “Angle of elevation at point A or ground = x
= Pe
| het @ be fob ‘bin abe fe A
angle of depresstin of foot of “louer chow 8e p |
ES above deta & vepeeded 3 dor ob Figure
IS
craw PX LPR
Tn arex
thanx = Po
ESS
tod re
tg. ex
>. es Se bet boto cat
eight oh tour = beton cat
Se:
Heégt ch obtervey 15m
Height of “rower = Pa = 0m
E of che above The vltewer eye = 3015
Yes 285m,
Distance betüen “tous & obterver XB = 286%
9 be angle of cat op touxs tp “on eue
TR above data & vepretented
dome of fhe at than
hon di tono opc ate
Mjacent (Pe
RD Sharma Class 10 Solutions Applications of Trigonometry»
www.LeamCBSE.in
RD Sharma Class 10 Solutions Applications of Trigonometry
vena LeamCBSE in
tono= OK > 265
ES
Age elevation = qe
ae
> 0e tala = 400
SF het de be heu Of éd = 15
deg à Pete 6 ti gio ate
ler Pe be
The above Hgormattin Wk vepretented E form of
| dique at chown,
equal Akttance Ken
BC = Leg of dee
Sic Me y Be Le
a 2 Ba
> Rc
LSO - dam
a
cha Fe AE , @zuaa., x4 LCA, Waite
a
cz = fi
&
e
Y
CE= y
D XC= 11011. HS
ss
For bey a
deg dj iq e+ won
Angle made by Ag ut ground =
For boy 2
height oh buttding CD = 10m.
RD Sharma Class 10 Solutions Applications of Trigonometry
www.LeamCBSE.in
AC = 060, Ag
51
vena LeamCBSE in
tono= OK > 265
ES
Age elevation = qe
ae
> 0e tala = 400
SF het de be heu Of éd = 15
deg à Pete 6 ti gio ate
ler Pe be
The above Hgormattin Wk vepretented E form of
| dique at chown,
equal Akttance Ken
BC = Leg of dee
Sic Me y Be Le
a 2 Ba
> Rc
LSO - dam
a
cha Fe AE , @zuaa., x4 LCA, Waite
a
cz = fi
&
e
Y
CE= y
D XC= 11011. HS
ss
For bey a
deg dj iq e+ won
Angle made by Ag ut ground =
For boy 2
height oh buttding CD = 10m.
RD Sharma Class 10 Solutions Applications of Trigonometry
www.LeamCBSE.in
AC = 060, Ag
51
RD Sharma Class 10 Solutions Applications of Trigonometry
ven LeamCBSE in 58
Ange made by ththg win taa] hp Bo 45°
engin ch KR Stead of bey >
a! hi meet much be ‘De
E wepracated & dor ob
Pp both Te
ae above Bematter
qe o chown:
draw &L Ac, YP4 ee
La sex
An ae = ex
Ax à
Shae
Be, et Bx en
CNT eae
BY = BR-xy = So-10m
40m
Hem 4% 4e eu
es
ze > Bb=40Gm
+=
ee
length 4 read or Shing + boys = ola
sa
“Heght oh Horses AB Som
heigpt op hi ep = "bien
Fogle of agotó of top ch Mil Arm foot of
Howe «= 60!
te ch elvan) dep oh trie fon toot of
Wi pe aoe
de abe formalen wepretented dom of
igure u than
RD Sharma Class 10 Solutions Applications of Trigonometry
| wmw.LeamCBSE.in
ven LeamCBSE in 58
Ange made by ththg win taa] hp Bo 45°
engin ch KR Stead of bey >
a! hi meet much be ‘De
E wepracated & dor ob
Pp both Te
ae above Bematter
qe o chown:
draw &L Ac, YP4 ee
La sex
An ae = ex
Ax à
Shae
Be, et Bx en
CNT eae
BY = BR-xy = So-10m
40m
Hem 4% 4e eu
es
ze > Bb=40Gm
+=
ee
length 4 read or Shing + boys = ola
sa
“Heght oh Horses AB Som
heigpt op hi ep = "bien
Fogle of agotó of top ch Mil Arm foot of
Howe «= 60!
te ch elvan) dep oh trie fon toot of
Wi pe aoe
de abe formalen wepretented dom of
igure u than
RD Sharma Class 10 Solutions Applications of Trigonometry
| wmw.LeamCBSE.in
RD Sharma Class 10 Solutions Applications of Trigonometry
vena LearnCBSE in
fon 49. ;
hades
done opp = na
od La
LL: 50 =
Ee 82 ec son
peep
| ve: op date
adj ce) 6
B= of to = Soxs= 1
chs Be SOK: ion
sof o
Rest eb Het = Icom.
bo:
Me be boat 4 and & be : Boat so
bergen op AGH oe be can
Distance beliveen 6,4 = 1000
Ange of elevation oh Ados, « 2 26
Age d elevatiin of A from &. pee
Bix above Adornatn Pe repre ted
Be dome of due at chown
here I
à acer
donzo'= opp -
oy
BB Mic
Monge AG > Ans eg,
CA à 0
O+® > Berea
RD Sharma Class 10 Solutions Applications of Trigonometry
www.LeamCBSE.in
5
vena LearnCBSE in
fon 49. ;
hades
done opp = na
od La
LL: 50 =
Ee 82 ec son
peep
| ve: op date
adj ce) 6
B= of to = Soxs= 1
chs Be SOK: ion
sof o
Rest eb Het = Icom.
bo:
Me be boat 4 and & be : Boat so
bergen op AGH oe be can
Distance beliveen 6,4 = 1000
Ange of elevation oh Ados, « 2 26
Age d elevatiin of A from &. pee
Bix above Adornatn Pe repre ted
Be dome of due at chown
here I
à acer
donzo'= opp -
oy
BB Mic
Monge AG > Ans eg,
CA à 0
O+® > Berea
RD Sharma Class 10 Solutions Applications of Trigonometry
www.LeamCBSE.in
5
|
| D Agox
| :
RD Sharma Class 10 Solutions Applications of Trigonometry
ven LeamCBSE in
hr)
Sé 100 4a
Gr ar Gy
= Jools = soso
hefgue of Ah house = 5068-1) mts
eig op bullae ae Gorn.
chetgnt oh damp pat ep = hm,
Angle of depretefon of top of lamp pat from
top of bula fog «= 0.
Angle eh cepas Of halo of tony pet drove
Ao ef bullditg p= of
The, aboe fagormatíen À vepreceated th tke fom
o) figure a thowa
draw DX-LAG , PX ZAC, CD =AK
AC = Go-h) & m —O
D aeca Loup = 48 => ong,
‘ac
= Lo
ae
? ACE Lo = 2080 —@
RD Sharma Class 10 Solutions Applications of Trigonometry
www.LeamCBSE.in
co
| D Agox
| :
RD Sharma Class 10 Solutions Applications of Trigonometry
ven LeamCBSE in
hr)
Sé 100 4a
Gr ar Gy
= Jools = soso
hefgue of Ah house = 5068-1) mts
eig op bullae ae Gorn.
chetgnt oh damp pat ep = hm,
Angle of depretefon of top of lamp pat from
top of bula fog «= 0.
Angle eh cepas Of halo of tony pet drove
Ao ef bullditg p= of
The, aboe fagormatíen À vepreceated th tke fom
o) figure a thowa
draw DX-LAG , PX ZAC, CD =AK
AC = Go-h) & m —O
D aeca Loup = 48 => ong,
‘ac
= Lo
ae
? ACE Lo = 2080 —@
RD Sharma Class 10 Solutions Applications of Trigonometry
www.LeamCBSE.in
co
RD Sharma Class 10 Solutions Applications of Trigonometry
ven LeamCBSE in u
trom 0:68
Goh 20%
bo-h = 20
be 40m
shetght of lamp port =4om
Diitance between damp FAA bufldrg Ac = 20m.
Difference betivern hedghtt of bulldihg & Vomp pour
= Bk = 60-h= 60-40 = 200
62:
Hedge Pop Aa howe = 'h metres, = AB
| sas be tuo thie en oppaite Rdn of ht boue
Arle of depneitfn ob £, from top oh LB tomes
“=
Ange of depresión of & tom top of Light heute=p
wequived to prove that
E AN
ane ang 7
the above Format Ri represented fh The dom
of Figure a thous |
A
[Ln nass, F
fone = opp. Ag h |
| ay Se
se= b o à
Hana 5 | |
RD Sharma Class 10 Solutions Applications of Trigonometry
www.LeamCBSE.in
ven LeamCBSE in u
trom 0:68
Goh 20%
bo-h = 20
be 40m
shetght of lamp port =4om
Diitance between damp FAA bufldrg Ac = 20m.
Difference betivern hedghtt of bulldihg & Vomp pour
= Bk = 60-h= 60-40 = 200
62:
Hedge Pop Aa howe = 'h metres, = AB
| sas be tuo thie en oppaite Rdn of ht boue
Arle of depneitfn ob £, from top oh LB tomes
“=
Ange of depresión of & tom top of Light heute=p
wequived to prove that
E AN
ane ang 7
the above Format Ri represented fh The dom
of Figure a thous |
A
[Ln nass, F
fone = opp. Ag h |
| ay Se
se= b o à
Hana 5 | |
RD Sharma Class 10 Solutions Applications of Trigonometry
www.LeamCBSE.in
RD Sharma Class 10 Solutions Applications of Trigonometry
www LearnCBSE in ba
|, ames,
taps AB + 65,=h.-0
Es, top. ©
O+® > Bree bh 4b
tom Gon
> Sy = aj
Pilkonce between cana =
Cana tong)
Tone. tn
hltan rt
Rhone T tan), metres. |
Yana. tong, |
68 Heigpt of tous AB= Some
GAG be ten car
Angle oh drpracfin dj €, From top of “lorry
Angie op \deprrition of & rom top ef lowes p
30
= bee
Ditante “between cove je.
Whe above information fe veprciented Po form of
jure a thou,
Zo AARC
boop = qe
E
tante
ES
2
RD Sharma Class 10 Solutions Applications of Trigonometry
www.LeamCBSE.in
www LearnCBSE in ba
|, ames,
taps AB + 65,=h.-0
Es, top. ©
O+® > Bree bh 4b
tom Gon
> Sy = aj
Pilkonce between cana =
Cana tong)
Tone. tn
hltan rt
Rhone T tan), metres. |
Yana. tong, |
68 Heigpt of tous AB= Some
GAG be ten car
Angle oh drpracfin dj €, From top of “lorry
Angie op \deprrition of & rom top ef lowes p
30
= bee
Ditante “between cove je.
Whe above information fe veprciented Po form of
jure a thou,
Zo AARC
boop = qe
E
tante
ES
2
RD Sharma Class 10 Solutions Applications of Trigonometry
www.LeamCBSE.in
Pätance ef tara
RD Sharma Class 10 Solutions Applications of Trigonometry
www LearnCBSE in &
Atane between cart GG = 100.5 mir
vom tawer
Sosa mir
Pltarce of Cars trom done = Sif my
4
Height ch “lowes Ae = 1000
ho:
He of voce ep
Age oh clivatlon „op top of meet Arm top of tous
A Ke ae
“Arge of clivaltiny dp, Bop oh roch tom bation of tous
Beast
the above data & veprcentes Jorn Ob dure
0 hour €
Drow AXLCD
402 AB = 100m
XA = pe.
Mn Acxa, tone = ox
Ax
> tue Ce
as
> be: AD
In Ace, tan
Ge, tonp= dass ootex |,
ve
Taro tor ex
ES D bes wor ©
Hem D1® torero cea à ext. neo
> ex = 100
ER x Gary
sale, Nea
height of Mil = 107 olmo = 150 (ar) mil
RD Sharma Class 10 Solutions Applications of Trigonometry
www.LearnCBSE.in 11
RD Sharma Class 10 Solutions Applications of Trigonometry
www LearnCBSE in &
Atane between cart GG = 100.5 mir
vom tawer
Sosa mir
Pltarce of Cars trom done = Sif my
4
Height ch “lowes Ae = 1000
ho:
He of voce ep
Age oh clivatlon „op top of meet Arm top of tous
A Ke ae
“Arge of clivaltiny dp, Bop oh roch tom bation of tous
Beast
the above data & veprcentes Jorn Ob dure
0 hour €
Drow AXLCD
402 AB = 100m
XA = pe.
Mn Acxa, tone = ox
Ax
> tue Ce
as
> be: AD
In Ace, tan
Ge, tonp= dass ootex |,
ve
Taro tor ex
ES D bes wor ©
Hem D1® torero cea à ext. neo
> ex = 100
ER x Gary
sale, Nea
height of Mil = 107 olmo = 150 (ar) mil
RD Sharma Class 10 Solutions Applications of Trigonometry
www.LearnCBSE.in 11
RD Sharma Class 10 Solutions Applications of Trigonometry
ven LeamCBSE in “4
65. Hear cp E ‘boue -AB= 150 me
Let G4 be wo abia appnonchifg fach other
Angle of depreicion ch &, à = 20°
Angle oh depeettion oh 4 p= 4°
Díitaree — belioces cba = 9%
Wir above data À wepretentet €
Be ho oh Fe a theron
In 4%,
e
tanp = na.
es,
tas 45° 150
Es,
gs, =150m | BS = o
Si = 696 = ISOC mate
Dittone — betiveen aha = hf) mir
Hecht of Tower = AB = Sm
Hedpt oh Flogitafj ec = na
rele of eval of top op qiero pp = sr
Angle of clivallin Labs oh flagitoty BE 45°
Th above data
% vepretented & foun of
at homo
Kars Ape Baw Woe
rt
> bas Sm
RD Sharma Class 10 Solutions Applications of Trigonometry
www.LearnCBSE.in
ven LeamCBSE in “4
65. Hear cp E ‘boue -AB= 150 me
Let G4 be wo abia appnonchifg fach other
Angle of depreicion ch &, à = 20°
Angle oh depeettion oh 4 p= 4°
Díitaree — belioces cba = 9%
Wir above data À wepretentet €
Be ho oh Fe a theron
In 4%,
e
tanp = na.
es,
tas 45° 150
Es,
gs, =150m | BS = o
Si = 696 = ISOC mate
Dittone — betiveen aha = hf) mir
Hecht of Tower = AB = Sm
Hedpt oh Flogitafj ec = na
rele of eval of top op qiero pp = sr
Angle of clivallin Labs oh flagitoty BE 45°
Th above data
% vepretented & foun of
at homo
Kars Ape Baw Woe
rt
> bas Sm
RD Sharma Class 10 Solutions Applications of Trigonometry
www.LearnCBSE.in
RD Sharma Class 10 Solutions Applications of Trigonometry
www LearnCBSE in
LA 1
Lo año, tore = Ac |
AS
ane = AS tee - hee
À 5
Gs ber
y
hres G8 heslo = Soria = 3ormh
eut oy plagtop cut
64: Heeght of bower age mt
ber pit © ke dott tom 8, -Angle oh elevation Leu
Pofet D be 9 mis) som 8, “Angle of elevatin be p.
de sp ya menthe | peau op =a4
véquéea fo prove that he Gate
Te (Above, date Yepracated 2 EE tom! ofp dee
treten.
Bo 4nec, tanx =
tana
4lane x 9 cota
36 “Cone. cota)
We 3
be WB = Gente
: hero q tower = bate
RD Sharma Class 10 Solutions Applications of Trigonometry
www.LeamCBSE.in
www LearnCBSE in
LA 1
Lo año, tore = Ac |
AS
ane = AS tee - hee
À 5
Gs ber
y
hres G8 heslo = Soria = 3ormh
eut oy plagtop cut
64: Heeght of bower age mt
ber pit © ke dott tom 8, -Angle oh elevation Leu
Pofet D be 9 mis) som 8, “Angle of elevatin be p.
de sp ya menthe | peau op =a4
véquéea fo prove that he Gate
Te (Above, date Yepracated 2 EE tom! ofp dee
treten.
Bo 4nec, tanx =
tana
4lane x 9 cota
36 “Cone. cota)
We 3
be WB = Gente
: hero q tower = bate
RD Sharma Class 10 Solutions Applications of Trigonometry
www.LeamCBSE.in
RD Sharma Class 10 Solutions Applications of Trigonometry
wwnw.LeannCBSE.in
48: Hehe ch UGK boue Ag = Dm
het GA S be ah détarce hekueen Che Ge
Ange of dépit of & Gee se] sac
| ge paie à 5 Cpe pr]
‘She above data
os than
Dam =
oops Ae .
tS, Wat
tangos ho
ES
Bho E A i
em
Bo EN de» de
Be,
tms
CA
= ha ©
O-0 => Rs,-85 > hey
Res = hia)
> be 200 , wal
Ter Sl root
WO (Waa gy = 2133 ate
ede E boue = ates m
RD Sharma Class 10 Solutions Applications of Trigonometry
www.LeamCBSE.in
wwnw.LeannCBSE.in
48: Hehe ch UGK boue Ag = Dm
het GA S be ah détarce hekueen Che Ge
Ange of dépit of & Gee se] sac
| ge paie à 5 Cpe pr]
‘She above data
os than
Dam =
oops Ae .
tS, Wat
tangos ho
ES
Bho E A i
em
Bo EN de» de
Be,
tms
CA
= ha ©
O-0 => Rs,-85 > hey
Res = hia)
> be 200 , wal
Ter Sl root
WO (Waa gy = 2133 ate
ede E boue = ates m
RD Sharma Class 10 Solutions Applications of Trigonometry
www.LeamCBSE.in
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