Calc Techniques for Boards Exam Purposes

Board exam calculator techniques

1. CALCULATOR
TECHNIQUES

2. CALCULATOR
101

3. Mode 1:
COMP
(Computation)
This is the IDLE mode of your
calculator once
you turn it ON.
This MODE is commonly used for
GENERAL CALCULATIONS.

4. Mode 2 :
CMPLX
(Complex)
This is the MODE used in computing
problems involving COMPLEX
NUMBERS.
To CONFIGURE the SETTINGS on MODE 2:
COMPLEX
PRESS:
[ SHIFT ] –[ 2]
OPTIONS (after pressing):
[ 1 ] –ARG (ARGUMENT)
[ 2] –CONJG (CONJUGATE)
[ 3 ] –r ‹ θ
[ 4 ] –a + bi

5. Mode 3 : STAT
(Statistics)
This is the MODE used in computing problems
involving STATISTICS and REGRESSION.
VARIATION [ 1 ] , LINEAR [ 2 ], QUADRATIC [ 3 ],
LOGARITHMIC [ 4 ], EXPONENTIAL [ 5 ],
To CONFIGURE the SETTINGS on MODE 3: STAT
PRESS:
[ SHIFT ] –[ 1]
OPTIONS (after pressing):
[ 1 ] –TYPE
[ 2] –DATA
[ 3 ] –SUM
[ 4 ] –VAR (Variation)
[ 5 ] –DISTR (Distribution)
[ 6 ] -MinMax

6. Mode 5: EQN
(Equations)
This is the MODE used in computing
problems involving EQUATIONS and
POLYNOMIALS.
OPTIONS (after pressing [ MODE ] –[ 5 ] ) :
[ 1 ] –TWO –UNKNOWNS
[ 2] –THREE UNKNOWNS
[ 3 ] –QUADRATIC
[ 4 ] –CUBIC

7. Mode 6:
MATRIX
This is the MODE used in computing problems
involving MATRIX ANALYSIS.
Mat A [ 1 ], Mat B [ 2 ] & Mat C [ 3 ]
To CONFIGURE the SETTINGS on MODE 6:
MATRIX
PRESS:
[ SHIFT ] –[ 4]
OPTIONS (after pressing):
[ 1 ] –DIM (Dimension)
[ 2] –DATA
[ 3 ] –MATRIX A
[ 4 ] -MATRIX B
[ 5 ] –MATRIX C
[ 6 ] –MatAns(Matrix Answer)
[ 7 ] –Det(Determinant)
[ 8 ] –Trn(Transpose)

8. Mode 7:
TABLE
This is the MODE used in GENERATING A
NUMBER TABLE based on ONE OR TWO
FUNCTIONS.

9. Mode 8:
VECTOR
This is the MODE used in computing problems
involving VECTOR ANALYSIS.
VctA [ 1 ], VctB [ 2 ] & VctC [ 3 ]
To CONFIGURE the SETTINGS on MODE 8:
VECTOR
PRESS:
[ SHIFT ] –[ 5]
OPTIONS (after pressing):
[ 1 ] –DIM (Dimension)
[ 2] –DATA
[ 3 ] –VECTOR A
[ 4 ] –VECTOR B
[ 5 ] –VECTOR C
[ 6 ] –VctAns(Vector Answer)
[ 7 ] –DOT

10. Configuring
Your
Calculator’s
SETUP
To CONFIGURE the SETUP of your
CALCULATOR
ENTER:
[SHIFT ] –[ MODE ]
1
st
OPTIONS ( after pressing)
[ 1 ] –MthIO [ 2] -LineIO
[ 3 ] –Deg [ 4 ] –Rad
[ 5 ] –Gra [ 6 ] –Fix
[ 7 ] –Sci [ 8 ] –Norm
2
nd
OPTIONS (scroll down)
[ 1 ] -ab/c [ 2 ] –d/c
[ 3 ] –CMPLX [ 4 ] -STAT
[ 5 ] –Disp [ 6 ] -Contrast

11. SEMICOLON
? : ?

12. U“ING THE “EMICOLON ^:_ FUNCTION OF YOU‘ CALCULATO‘
Past CE Board Exam Problem
A block of copper having a mass of 50kg is drawn out to make 500m of wire of
uniform cross-section. Given that the density of copper is 819 g/cm
2
.
1.What is thevolume of steel in cubic centimeters?
2.What is the cross-sectional area of the wire in square millimeters?
3.What is the of the wire in millimeters?
Solution:
ρ= M/V , V = AL A =
Π
???
(D
2
)
Input: (50000/8.91) : Ans/(500 x 100) :
?????????????
Π
Press: [ = ] Display: 5611.67 cm
3
Press: [ = ] Display: 0.1122 cm
2
or 11.22 mm
2
Press: [ = ] Display: 0.378 cm or 3.78 mm

13. ENGINEERING
MECHANICS

14. RIGHT ANGLE TECHNIQUE
The ^‘e_ avd ^Pol_ Fuvtiovs
PRESS:
[ SHIFT ] –[ + ] Fo? ^POL_ Fuvtiov
PRESS
[ SHIFT ] –[ -] Fo? ^‘EC_ Fuvtiov
The ^‘e_ fuvtiov ov?e?ts POLA‘ COO‘DINATE“
(r<θ)into RECTANGULAR COORDINATES(x,y).
The ^Pol_ function converts RECTANGULAR
COORDINATES(x,y).into POLAR COORDINATES(r<θ).

15. RIGHT TRIANGLE TECHNIQUE
Example.
In the figure shown, Find the remaining side and
angles if A = 4 and B = 7.
Solution:
Let: x = 4 and y = 7
Input:
[ SHIFT ] –[ + ] –[SHIFT] –[ 4 ] –[SHIFT] –[ ( ] –[ 7 ] –
[ ) ]
Display:
Pol(4,7)
Press [ = ] Display: r = 8.062257748 C
θ= 60.2551187 β
PRESS:
[ SHIFT ] –[ + ] For ^POL_ Fuvtiov
[ SHIFT ] –[ -] For ^‘EC_ Fuvtiov

16. COMPONENT OF A 2D FORCE
Example.
In the figure shown, Find the x and y component of
the force F if F = 4000N and θ= 32.3
o
Solution:
Applying the RIGHT ANGLE TECHNIQUE
Let: r = 4000N and θ= 32.3
o
Input:
[ SHIFT ] –[ -] –[SHIFT] –[ 4000 ] –[SHIFT] –[ ( ] –
[ 32.3 ] –[ ) ]
Display: Rec(4000,32.3)
Press [ = ] Display: x = 3381.047333 Fx
y= 2137.409398 Fy
PRESS:
[ SHIFT ] –[ + ] For ^POL_ Fuvtiov
[ SHIFT ] –[ -] For ^‘EC_ Fuvtiov

17. MODE 2: COMPLEX
Argand Diagram (Complex Plane)
Where:
r = absolute value or modulus
Θ= argument or amplitude
a is the real axis and b is the imaginary axis
Similarly with Cartesian Coordinate Plane the
Complex Plane has the following forms too;
r < θis the POLAR FORM
a + bi is the RECTANGULAR FORM
NOTE:
ΘIS ALWAYS MEASURED FROM THE ORIGIN
(COUNTER CLOCKWISE)

18. MODE 2: COMPLEX
NOTE:
All angles (θ) must be referred from the x-axis.
1
st
Quadrant: As is
2
nd
Quadrant: 180 –Angle
3
rd
Quadrant: 180 + Angle
4
th
Quadrant: -Angle or 360 –angle
Where:
Angle is solved by either theRIGHT ANGLE
TECHNIQUE or tan
-1
(slope)

19. Example.
In the figure shown, Find the x and y component of
the force F if F = 4000N and θ= 32.3
o
Solution: Applying Complex Mode
Let: r = 4000N and θ= 32.3
o
Press: [Mode] –[ ? ] ZThe Calulato? vo? is iv
CMPLX MODE[
Input: [ 4000 ] –[ SHIFT ] –[ ( -) ] –[ 32.3 ]
Display: 4000 < 32.3
Press [ = ] Display: 3381.047333 + 2137.409398i
a = 3381.047333 Fx
b = 2137.409398Fy
COMPONENTS OF A 2D FORCE

20. PAST CE BOARD:
For the forces in the figure shown.
a.Find the magnitude of the horizontal
component of the resultant.
b. Find the magnitude of the resultant.
c. Find the angle that the resultant
makes with the x-axis.
CONCURRENT –COPLANAR FORCE SYSTEM

21. Resultant<θ = Summation of r<θ[?
CONCURRENT –COPLANAR FORCE SYSTEM
Solution:
Enter: [ MODE ] –[ ? ] ZCMPLX MODE[
To get the resultant:
Input: 86.6<(30) + 70.7<(180-45) + 68.6<(270)
Press: [ = ]
Output: 25.00535054 + 24.69244943i
Press: [ SHIFT ] –[ 2 ] –[ 3 ] –[ = ]
Display: Ansr<θ
Output: 35.14234788<44.6392662
a.Rx = 25.00535054 & Ry = 24.69244943 in Newtons
b.R = 35.14234788 in Newtons
c.θx = 44.6392662
o

22. NONCONCURRENT –COPLANAR FORCE SYSTEM
PAST CE BOARD:
Consider the coplanar force system
shown.
a.Find the magnitude of the resultant.
b.Compute the location of the
resultant from the y –axis.
c.Compute the location of the
resultant from the x –axis.

23. NONCONCURRENT –COPLANAR FORCE SYSTEM
Solution:
Enter: [ MODE ] –[ ? ] ZCMPLX MODE[
To get the resultant:
Input: 10<(90) + 6.7< (tan
-1
1/2)
Press: [ = ]
Display: 5.9927 + 12.9963i
Press: [ SHIFT ] –[ 2 ] –[ 3 ] –[ = ]
Display: Ansr<θ
Output: 14.31142<65.24538
Rx = 5.9927 & Ry = 12.9963 in Newtons
R = 14.31142in Newtons
θx =65.24538
o

24. NONCONCURRENT –COPLANAR FORCE SYSTEM
Solution: CALCULATING THE MOMENT USING STATISTICS
Enter: [ MODE ] –[ 3 ] – [ ? ] Z“TAT MODE – LINEA‘[
Enter: [ SHIFT ] –[ MODE ] –[ ] –[ 4 ] – [ ? ] ZF‘EQUENCY ON[
NOTE: X is the FORCE, Y is the MOMENT ARM and Freqis 1
for CLOCKWISE and -1 if COUNTERCLOCKWISE ROTATION.
Moment @ 0
Input:
X Y Freq
10 5 -1
2.9968 -1
5.9932 1
30 1 1
Enter: [AC] - [ SHIFT ] –[ 1 ] –[ 3 ] –[ 5 ] –[ = ]
Output: ??= moment at 0 = -31.982N-m
(Ry = 12.9963 )(Ix) = 31.982 Ix = 2.461m x-intercept
(Rx = 5.9927 ) (Iy) = 31.982 Iy= 5.34m y-intercept

25. NONCONCURRENT –COPLANAR FORCE SYSTEM
PAST CE BOARD:
The force system shown consist of a
Couple C and the 4 Forces.
The resultant of this system is 500kN-
m counter-clockwise couple.
a.What is the value of P?
b.What is the value of Q?
c.What is the value of C?

26. NONCONCURRENT –COPLANAR FORCE SYSTEM
Solution:
Enter: [ MODE ] – [ ? ] ZCMPLX MODE[
To get the resultant:
Resultant Force = 0, since Resultant is just a Couple
P<(tan
-1
3/4) + 20<(270) + Q(180 + tan
-1
(5/12)) + 80<0 = 0
P<(tan
-1
3/4) + Q(180 + tan
-1
(5/12)) = - ( 20<270 + 80<0)
Simplifying:
P(4/5 + 3/5i) + Q( -12/13 –5/13i) = -80 + 20i
Use [ MODE ] –[ 5 ] – [ ? ] ZEQN – TWO UNKNOWN“[
a b c
1 4/5 -12/13 - 80
2 3/5 -5/13 20
X = P = 200kN and Y = Q = 260kN

27. NONCONCURRENT –COPLANAR FORCE SYSTEM
Solution:
Enter: [ MODE ] –[ 3 ] – [ ? ] Z“TAT MODE – LINEA‘[
Enter: [ SHIFT ] –[ MODE ] –[ ] –[ 4 ] – [ ? ] ZF‘EQUENCY ON[
Mo = 500 kN-m counterclockwise
Input:
Enter: [ AC ] –[ SHIFT ] –[ 1 ] –[ 3 ] –[ 5 ]
Output: ??= -1940 kN-m
Mo = C –1940 = -500
C = 1440 kN-m
X Y Freq
80 2 1
120 6 -1
20 3 1
240 6 -1

28. Example.
In the figure shown, Find the x, y and z component of the force F if
F = 100N and passes from point A(0,0,0) to the point B(3,4,5). Its
direction cosines and the angle that it makes with the coordinate axes.
Solution:
Enter: [ MODE ] –[ 8 ] –[ 1 ] – [ ? ] ZVECTO‘ MODE – ?D[
Input: [ 3 4 5] Press: [ AC ]ZVECTO‘ DIFFE‘ENCE BET. B & A[
To get direction cosines or unit vector:
Enter: [ SHIFT ] –[ 5 ] –[ 3 ] –[ ÷] –[ SHIFT ] –[ HYP ] –
[ SHIFT ] –[ 5 ] –[ 3 ] –[ = ]
Display: VctA÷Abs(VctA)
Output: [ 0.424260.56568 0.70711 ]
To get the components:
Enter: [ x ] –[ 100 ] –[ = ] Display: [ 42.42655.56870.711 ]
Fx= 42.426 NFy= 55.568 NFz= 70.711 N
To get the angles it makes with the coordinate axes just take the inverse
cosine of the direction cosines above.
COMPONENTS OF A 3D FORCE

29. CONCURRENT –NONCOPLANAR FORCE SYSTEM
Past CE Board Exam
A concurrent force system in space is composed of three forces
described as follows.
P1 has a magnitude of 100kN and acts through the origin and the
points x = 3, y = 4 and z = 2.
P2 has the magnitude of 60kN and acts through he origin and the
points x = 4, y = 1 and z = -2
P3 has the magnitude of 80kN and acts through the origin and the
points x = 2, y = -3 and z = 3
Find the components of the resultant, its direction cosines and
magnitude of the resultant.

30. CONCURRENT –NONCOPLANAR FORCE SYSTEM
Solution:
P1 = 100kN (3, 4, 2)
P2 = 60kN (4, 1, -2)
P3 = 80kN (2, -3, 3)
Enter: [ MODE ] –[ 8 ] –[ 1 ]
Input: [ 3 4 2 ] for VctA
Enter: [ AC ]
Enter: [ SHIFT ] –[ 5 ] –[ 1 ] –[ 2 ] –[1 ]
Input: [ 4 1 -2 ]for VctB
Enter: [ AC ]
Enter: [ SHIFT ] –[ 5 ] –[ 1 ] –[ 3 ] –[ 1 ]
Input: [ 2 -3 3 ]for VctC
Enter [ AC ]
To get the COMPONENTS:
Input: 100VctA ÷Abs(VctA) + 60VctB ÷
Abs(VctB) + 80VctA ÷Abs(VctC)
Enter: [ = ]
Output: [ 142.913 36.203 62.121 ]
Rx = 142. 913 kN
Ry = 36.203 kN
Rz= 62.121 kN
This Vector is automatically stored to VctAns
Input: Abs(VctAns)
Enter: [ = ]
Output: R = 159.3378 kN
Input: VctAns÷Abs(VctAns)
Enter: [ = ]
Output: [ 0.8924 0.2272 0.3898 ]

31. PARALLEL –NONCOPLANAR FORCE SYSTEM
Find the resultant of the four
forces acting on the plane
shown. Also find its position.

32. PARALLEL –NONCOPLANAR FORCE SYSTEM
Solution:
Enter: [ MODE ] –[ 3 ] –[ 2 ] Z “TAT MODE[
Enter: [ SHIFT ] –[ MODE ] –[ ] –[ 4 ] – [ ? ] ZF?e?uev? ON[
Using Right Hand Rule the above data is obtained
X is the MOMENT ARM for the Moment about the x-axis
Y is the MOMENT ARM for the Moment about the y-axis
Freqis the FORCE
To get Mx, Enter: [SHIFT] –[ 1 ] –[ 3 ] –[ = ]
Output: ∑X = 9?? lb-ft= Mx
To Get My, Enter: [ SHIFT ] –[ 3 ] –[ 4 ] –[ = ]
Output: ∑Y = -804 lb-ft= My
Then Mo = 286yi –286xj = 956i –804j y = 3.343, x = 2.881
Resultant, R = 286 lb
X Y Freq
2 -6 70
6 -6 64
6 0 72
0 0 80

33. APPLICATIONS

34. EQUILIBRIUM OF COPLANAR FORCES
CALCULATOR TECHNIQUE USING MODE 5: EQN (TWO –UNKNOWNS)
NOTE:
For REACTIVE FORCES all components to the RIGHT and UPWARD
are POSITIVE. All components to the LEFT and DOWNWARD are
NEGATIVE
For APPLIED FORCES the sign convention of the REACTIVE FORCES
are REVERSED.
Example:
Determine the Tensile Forces in each of the chords that support
the 200 lbweight as shown.

35. EQUILIBRIUM OF COPLANAR FORCES
CALCULATOR TECHNIQUE USING MODE 5: EQN (TWO –UNKNOWNS)
Solution:
Enter: [ MODE ] –[ 5 ] –[ ? ] Z EQUATION MODE –? UNKNOWN“[
Display and Input:
Enter: [ = ]
Output: X = BC = 100lb AND Y = BA = 173.205lb
BC BA AppliedLoads
a b c
Cos30 -cos60 0 X-Component
Sin30 Sin60 200 Y-Component

36. TRUSS ANALYSIS
PROBLEM:
Determine the forces in all
members of the Crane Truss
shown. Using Method of Joints.

37. TRUSS ANALYSIS
Solution:
For Triangle ABC:
AC = 13.077 ft. [ SOLVED BY COSINE LAW]
By Sine Law :
α= 23.412
0
β= 36.586
0
For Triangle CBD
Θ= tan
-1
12/6 = 63.435
0
While, γ= 180 –( α+β+θ) = 56.567
0

38. TRUSS ANALYSIS
Solution:
Enter: [ MODE ] –[ 5 ] –[ 1 ]
Let AC = X and AB = Y , where α= 23.412
0
Input:
Enter: [ = ]
Output:
AC = 13,087.55 lb(Tension)
AB = -12,010.16 lb(Compression)
@ Joint A
AppliedLoads
a b c
-cos23.412 -1 O
X-
Component
Sin23.412 0 5200 Y-Component

39. TRUSS ANALYSIS
Solution:
Enter: [ MODE ] –[ 5 ] –[ 1 ]
Let BC = X and BD = Y
Input:
Enter: [ = ]
Output:
BC = -6005.08 lb(Compression)
BD = -10,401.104lb (Compression)
@ Joint B
AppliedLoads
a b c
-cos60 -cos30 12,010.16 X-Component
Sin60 -sin30 0 Y-Component

40. TRUSS ANALYSIS
Solution:
Enter: [ MODE ] –[ 5 ] –[ 1 ]
Let CE = X and CD = Y , where α= 23.412
0
,
β= 36.586
0
, Θ= 63.435
0
& γ= 56.567
0
Input:
a b c
-1 -cos56.567 [-13,087.55cos23.412 + 6,005.08cos59.998]
0 -sin56.567 [13,087.55sin23.412 - 6,005.08sin59.998]
Enter: [ = ]
Output:
CE = 9007.19 lb(Tension)
CD = 0.285 lb(Tension)
@ Joint C

41. BEAMS
Solution:
Enter: [ MODE ] –[ 5 ] –[ 1 ]ZEQN-? UNKNOWN“[
Let Ra = X and Rb= Y
Input:
Enter: [ = ]
Output: X = Ra = 49.6875kN
Y = Rb= 82.8125kN
CHECK: UPWARD FORCES = DOWNWARD FORCES
49.6875 + 82.8125 = 10 + .5(15)(3) + 25(4)
132.5 = 132.5 okay!!!
Find the magnitude of the reaction in the
simply supported beam as shown in the
figure.
a b c
8 0 397.5Ma or Mb = 0
1 1 132.5 Fy= 0

42. CABLES
Past CE Board
For the cable shown, determine
the angle β1 & β2 and the
forces in each segment and the
length of the cable

43. CABLES
Solution:
Enter: [ MODE ] –[ 2 ] ZCouple? Mode[
BC<35 + CD<(180 –β2) + 2000< 270 = 0
Input(Right Side of the Equation to your Calculator)
CD < (180 –β2) = -( 2333.95<35 + 2000<270 )
CD < (180 –β2) = 2023 < 160.92
Thus,
CD = 2023 lb
180 -β2 = 160.92β2 = 19.08
0
@ Joint C

44. CABLES
Solution:
Enter: [ MODE ] –[ 2 ] ZCouple? Mode[
CD<(-β2 or 360 –β2) + AD<(180 –β1) +
1600< 270 = 0
Input(Right Side of the Equation to your Calculator)
AD < (180 –β1) = -( 2023<(-19.08) + 1600<270 )
AD < (180 –β1) = 2961.19 < 130.214
0
Thus,
AD = 2961.19 lb
180 –β1= 130.214β1= 49.8
0
@ Joint D

45. CABLES
Solution
To get the length of the cable:
Levgth of the Cale = ∑L/osβ
Where β= 35
0
, β1 = 49.8
0
and β2 = 19.08
Length of the Cable = 7/cos35 + 11/cos 19.08
+ 6/cos49.8
Length of the Cable = 29.48 feet

46. CENTROID AND MOMENT OF INERTIA
Solution:
Enter: [ MODE ] –[ 3 ] –[ ? ] Z“TAT MODE –LINEA‘[
Enter: [ SHIFT ] –[ MODE ] –[ ] –[ 4 ] –[ ? ] ZF‘EQUENCY ON[
To get the CENTROID Ῡfrom the X-AXIS
Enter: [AC]
To get Ῡ: Input: [ SHIFT] –[ 1 ] –[ 3 ] –[ 4 ] –[ ÷] –[ SHIFT ] –
[ 1 ] –[ 4 ] –[ 1 ] –[ = ]
Displa?: ∑? ÷n = Ῡ= 100.4545 mm
STANDBY
C.G. OF EACH AREA
FROM X-AXIS
AREA
X Y Freq
30 60 X 200 AREA 1
60 + 250/2 250 X 40 AREA 2

47. CENTROID AND MOMENT OF INERTIA
Y MINUS A
C.G. OF EACH AREA
FROM X-AXIS
AREA
X Y Freq
30 –A 30 60 X 200 AREA 1
60+ 250/2 - A 60 + 250/2 250 X 40 AREA 2
Solution:
Enter: [ MODE ] –[ 3 ] – [ ? ] Z“TAT MODE – LINEA‘[
Enter: [ SHIFT ] –[ MODE ] –[ ] –[ 4 ] – [ ? ] ZF‘EQUENCY ON[
Store Ῡ to ZVA‘IABLE – A[
Enter: [ SHIFT ] –[ RCL ] Display: AnsA
Edit Data
Enter: [ SHIFT ] –[ 1 ] –[ 2 ]
To get the CentroidalMoment of Inertia ( Igg)
Ivput: ∑h
3
/?? + ∑?
2
Z∑?
2
is [ SHIFT ] –[ 1 ] –[ 3 ] – [ ? ][
Output: 40(250)
3
/ 12 + 200(60)
2
/ ?? + ∑?
2
Enter: [ = ]
Display: Igg = 186.7287879 x 10
6
mm
4

48. FRICTION
A 200lb block is in contact with a
plane inclined at 30
0
with the
horizontal. A force P parallel to and
acting up the plane is applied to the
body. If the coefficient of static
friction is 0.2.
a.Find the value of P to just cause
the motion to impend up the
plane
b.Find the value of P to just
prevent the motion down the
plane.

49. FRICTION
A.Solution:
Enter: [ MODE ] –[ 5 ] –[ ? ] ZEQN MODE –? UNKNOWN“[
Input:
Enter: [ = ]
Output:
P = X = 134.64 Newtons
N = Y = 173.2 Newtons
P N
Applied
Load
a b c
1 - 0.2 200sin30 X-component
0 1 200cos30 Y-component

50. FRICTION
A.Solution:
Enter: [ MODE ] –[ 5 ] –[ ? ] ZEQN MODE –? UNKNOWN“[
Input:
Enter: [ = ]
Output:
P = X = 65.35 Newtons
N = Y = 173.2 Newtons
P N
Applied
Load
a b c
1 0.2 200sin30 X-component
0 1 200cos30 Y-component

51. END OF SLIDES
Prepared By:
Engr. Niño Gem M. Ngo Lee

52. WEIGHTED OBSERVATION
The difference of elevation between B and C was taken by a survey party using different trials and paths.
Compute the probable elevation of point C if it is above point B. Elevation of point B is 825m.
Trial Distance Diff. in Elevation
1 2 0.89m
2 6 0.67m
3 4 0.78m
4 12 1.02m
Enter: [ MODE ] –[ 3 ] –[ 2 ] –[ SHIFT ] –[ MODE ] –[ 1 ] –[ ] –[ 4 ] –[ 1 ] Z“TAT MODE –F‘EQUENCY ON[
X Y
½ 0.89
1/6 0.67
¼ 0.78
1/12 1.02

53. WEIGHTED OBSERVATION
The difference of elevation between B and C was taken by a survey party using different trials and paths.
Compute the probable elevation of point C if it is above point B. Elevation of point B is 825m.
Enter: [ AC ]
Input: [ SHIFT ] –[ 1 ] –[ 3 ] –[ 5 ] –[÷] –[ SHIFT ] –[ 1 ] –[ 3 ] –[ 2 ]
Displa?: ∑xy÷∑?
Enter: [ = ]
X average = 0.83667 Thus, Elev. At C = Elev. At B + 0.83667 = 825 + 0.83667 = 825.83667m
X Y
½ 0.89
1/6 0.67
¼ 0.78
1/12 1.02

54. CLOSED COMPASS TRAVERSE
Given the following traverse notes taken by a survey party, find:
a.Linear Error of Closure
b.Relative Error
c.Correct using Compass Rule
d.Correct using Transit Rule
LINE DISTANCE BEARING
AB 410m N 45
0
??[ W
BC 605m S65
0
??[ E
CD 600m N 80
0
??[E
DA 280m S 55
0
??[ W

55. SURVEYING

56. MISSING SIDES
CASE 1: Length and Bearing of One Side Completely Missing
LINE DISTANCE BEARING
AB 300m S 37
0
??[ E
BC 400m S 43
0
??[ W
CD 250m N 73
0
??[ W
DE 350m N 12
0
??[E
EA ? ?

57. MISSING SIDES
CASE 2: Length of Two Adjacent Sides Missing
CASE 5: Length of Two Non-Adjacent Sides Missing
LINE DISTANCE BEARING
AB 300m S 37
0
??[ E
BC 400m S 43
0
??[ W
CD 250m N 73
0
??[ W
DE X N 12
0
??[E
EA Y N 65
0
??[E
LINE DISTANCE BEARING
AB 200m N 73
0
??[ W
BC X N 12
0
??[E
CD 300m S 37
0
??[ E
DE Y N 65
0
??[E
EA 400m S 43
0
??[ W

58. MISSING SIDES
CASE 3: Length of One Side and Bearing of its Adjacent Side Missing
LINE DISTANCE BEARING
AB 300m S 37
0
??[ E
BC 400m S 43
0
??[ W
CD 250m N 73
0
??[ W
DE X N 12
0
??[E
EA 277.96m βea

59. MISSING SIDES
CASE 4: Bearing of Two Adjacent Sides Missing
LINE DISTANCE BEARING
AB 300m S 37
0
??[ E
BC 400m S 43
0
??[ W
CD 250m N 73
0
??[ W
DE 350m ?
EA 277.96m ?