bc754b20-fda5-4aa5-84bc-7a7122e84bb0.pptx

MATH

1. Law of Cosines

2. Objective

* Use the Law of Cosines to solve triangles
and problems

3. Law of Cosines
Previously, we learned the Law of Sines, which as
some theorems can, it does have its limitations. To
use the Law of Sines we had to know the measures
of two angles and any side (AAS or ASA) OR the
measures of two sides and an opposite angle
(SSA).

If we don't have either of these scenarios, but
instead we have the measures of two sides and
the included angle (SAS) or the measures of all the
sides and no angles (SSS), we must use the Law of

Cosines.
a
=
6

2 b

4. Law of Cosines

© Law of Cosines:

In AABC,

5. Exomole |:
Find x if y=11,z=25, and m/X =45.

6. Example:

X2=y° +Z° =2yz cos X Law of Cosines

x? =11 4+25* -2(11)(25) cos 45° y =11 z=25, m2X =45

x?=746-550 cos 45° Simplify.

7. Your Turn:
Find rif s=15, t=32, and mZR=40.

8. Example 2:

Find mZL.

9. Example 2:

-270

-178=-270 cos L

ee,

Subtract 754 from each side.

Divide each side by -270.

- Solve for L.

a

10. Your Turn:

Find mZF.

11. CHOOSE THE METHOD To solve a triangle that is oblique, or having no right
angle, you need to know the measure of at least one side and any two other parts. If
the triangle has a solution, then you must decide whether to begin solving by using the
Law of Sines or by using the Law of Cosines. Use the chart below to help you choose,

;' - en
Concept Summary Solving an Oblique Triangle |

Given Begin by Using

two angles and any side Law of Sines

two sides and an angle opposite one of them Law of Sines

two sides and their included angle Law of Cosines

three sides Law of Cosines

12. Example 3:

Determine whether the Law of Sines or the Law of
Cosines should be used first to solve ADEF. Then
solve ADEF. Round angle measures to the nearest
degree and side measures to the nearest tenth.

D
17

32 = E

Since we know the measures of two sides and the
included angle, use the Law of Cosines.

13. Example 3:

f? =d* +e* -2de cos F Law of Cosines

f? = 32? +17 -2(32)(17) cos 145° d = 32,e=17,mZF = 145

f = 1313-1080 cos 145° Take the square root
of each side.
f = 46.9 Use a calculator.

Next, we can find mZD or mZE. If we decide to find mZD,
we can use either the Law of Sines or the Law of Cosines
to find this value. In this case, we will use the Law of
Sines.

14. EIN

sinF sinD

—= Law of Sines
fa d
sin 145° sin D
Fo ET f = 46.9,d = 32,m/F =145

32 sin 145° = 46.9 sin D Cross products

15. Example 3:
Use the Angle Sum Theorem to find mZE.
mZD+mZE +mZF =180 Angle Sum Theorem
23+mZE +145 =180 mZD = 23, mZF =145

mZE ~12 tract 168 from each

à

16. MOMIE:

Determine whether the Law of Sines or the Law of
Cosines should be used first to solve AABC. Then
solve AABC. Round angle measures to the nearest
degree and side measures to the nearest tenth.

B

Answer: b=17.5, m/A=35, mZC = 53

17. Example 4:

AIRCRAFT From the diagram of the plane shown,
determine the approximate exterior perimeter of each
wing. Round to the nearest tenth meter.

9m

Since AKIJ is an isosceles triangle, K/ = 20 and
MZIKJ = ee =65.

18. Fxample 4:

Use the Law of Sines to find KJ.

sin ZKIJ _ sin ZIKJ
KJ Id

sin 50° sin 65°

Law of Sines

mZKIJ = 50, mZIKJ = 65, IJ = 20

KJ 20

19. Fxample 4:

Use the Law of Sines to find m£H.
sin ZHKI _ sin H

Law of Sines
HI Kl
ine El mZHKI = 25, HI=9, Ki = 20
9 20

5 products

20. Example 4:

Use the Angle Sum Theorem to find mZHIK.
mZHKI+mZHIK+mZH=180 Angle Sum Theorem

25+mZHIK+70=180 mZHKI= 25, mZH ~70

21. Example 4:

Use the Law of Sines to find HK.
sin ZHKI sin ZHIK
H HK
sin 25° sin 85°

9 HK

Law of Sines

mZHKI = 25, mZHIK = 85, HI=9

22. Example 4:

The perimeter of the wing is equal to HK + H/ +/J+ KJ.

Answer: The perimeter is about 21.2+9+20+16.9 or
about 67.1 meters.

23. MOI

The rear side window of a station wagon has the
shape shown in the figure. Find the perimeter of the
window if the length of DB is 31 inches.

Answer: about 93.5 in.

24. Assignment