1. Law of Cosines 10.5

2. The Law of Sines cannot be used to solve every triangle
The Law of Sines cannot be used to solve every triangle. Instead, you can apply the Law of Cosines.
You can use the Law of Cosines to solve a triangle if you are given
• two side lengths and the included angle measure (SAS) or
• three side lengths (SSS).

3. Law of Cosines
Standard Form
Alternative Form

4. The angle referenced in the Law of Cosines is across the equal sign from its corresponding side.
Helpful Hint
Do not round your answer until the final step of the computation. If a problem has multiple steps, store the calculated answers to each part in your calculator.
Helpful Hint

5. Law of Cosines – SAS
Find the measure of XZ.
XZ2 = XY2 + YZ2 – 2(XY)(YZ)cos Y
Law of Cosines
XZ2 = – 2(35)(30)cos 110°
Substitution.
Simplify.
XZ2 
Take the square
root of both sides.
XZ  53.3

6. Law of Cosines – SSS
Find mT.
RS2 = RT2 + ST2 – 2(RT)(ST)cos T
Law of Cosines
72 = – 2(13)(11)cos T
Substitution.
49 = 290 – 286 cosT
Simplify.
Subtract 290 from both sides.
–241 = –286 cosT
Division.
Use inverse cosine
to find mT.

7. Law of Cosines – SAS
Find the measure of DE.
DE2 = EF2 + DF2 – 2(EF)(DF)cos F
Law of Cosines
DE2 = – 2(18)(16)cos 21°
Substitution.
DE2 
Simplify.
Take the square root of both sides.
DE  6.5

8. Law of Cosines – SSS
Find mK.
JL2 = LK2 + KJ2 – 2(LK)(KJ)cos K
Law of Cosines
82 = – 2(15)(10)cos K
Substitution.
64 = 325 – 300 cosK
Simplify.
Subtract 325 both sides.
–261 = –300 cosK
Division.
Use inverse cosine
to find mK.

9. Law of Cosines – SAS
Find the measure of YZ.
YZ2 = XY2 + XZ2 – 2(XY)(XZ)cos X
Law of Cosines
YZ2 = – 2(10)(4)cos 34°
Substitution.
Simplify.
YZ2 
Take the square root of both sides.
YZ  7.0

10. Law of Cosines – SSS
Find mR.
PQ2 = PR2 + RQ2 – 2(PR)(RQ)cos R
Law of Cosines
9.62 = – 2(5.9)(10.5)cos R
Substitution.
92.16 = – 123.9cosR
Simplify.
Subtract both sides.
–52.9 = –123.9 cosR
Solve for cosR.
Use inverse cosine
to find mR.

11. Law of Cosines – SAS
Solve the following triangle.
BC2 = AB2 + AC2 – 2(AB)(AC)cos A
Law of Cosines
BC2 = – 2(3.9)(3.1)cos 45°
Substitution.
BC2 
Simplify.
Take the square
root of both sides.
BC  2.8 mi
Law of Sines
Substitution.
Multiply both
sides by 3.9.
Use inverse sine to find mC.

12. AC2 = AB2 + BC2 – 2(AB)(BC)cos B Substitution.
Law of Cosines – SAS
An engineer suggested using a cable attached from the top of the tower to a point 31 m from the base. How long would this cable be, and what angle would it make with the ground? Round the length to the nearest tenth and the angle measure to the nearest degree.
Law of Cosines
31 m
AC2 = AB2 + BC2 – 2(AB)(BC)cos B
Substitution.
AC2 = – 2(31)(56)cos 100°
Simplify.
AC2 
Take the square root of both sides.
AC  68.6 m

13. Find the measure of the angle the cable would make with the ground.
Law of Sines
Substitution.
Multiply both
sides by 56.
Use inverse sine to
find mA.

14. Use ΔABC for a – c. Round lengths to the nearest tenth and angle measures to the nearest degree.
a. mB = 20°, mC = 31° and b = 210. Find a.
b. a = 16, b = 10, and mC = 110°. Find c.
c. a = 20, b = 15, and c = 8.3. Find mA.
477.2
21.6
115°

15. An observer in tower A sees a fire 1554 ft away at an angle of depression of 28°. To the nearest foot, how far is the fire from an observer in tower B? To the nearest degree, what is the angle of depression to the fire from tower B?
1212 ft; 37°

16. Assignment
Geometry:
10.1
Law of Cosines

17. Definition: Heron’s Area Formula
Given any triangle with sides of lengths a, b, and c, the area of the triangle is given by 5 10 8
Find the area of the triangle.
Definition: Heron’s Area Formula

18. EXAMPLE 4
The intersection of three streets forms a piece of land called a traffic triangle. Find the area of the traffic triangle shown.
Find the semiperimeter s.
s = (a + b + c ) 1 2 1 2
= ( )
= 380
Area =
s (s – a) (s – b) (s – c)
380 (380 – 170) (380 – 240) (380 – 350) =
Area
18,300
The area of the traffic triangle is about 18,300 square yds.

19. GUIDED PRACTICE
Find the area of ABC.
Find the semiperimeter s.
s = (a + b + c ) 1 2 1 2
= ( )
= 12
Use Heron’s formula to find the area of ABC.
Area =
s (s – a) (s – b) (s – c)
12 (12 – 8) (12 – 11) (12 – 5) =
Area
18.3
The area is about 18.3 square units.

20. GUIDED PRACTICE
Find the area of ABC.
Find the semiperimeter s.
s = (a + b + c ) 1 2 1 2
= ( )
= 10
Use Heron’s formula to find the area of ABC.
Area =
s (s – a) (s – b) (s – c)
10 (10– 4) (10 – 9) (10 – 7) =
Area
13.4
The area is about 13.4 square units.

21. Find the area of ABC.
Find the semiperimeter s.
s = (a + b + c ) 1 2 1 2
= ( )
= 25
Use Heron’s formula to find the area of ABC.
Area =
s (s – a) (s – b) (s – c)
25 (25– 15) (25 – 23) (25 – 12) =
Area
80.6
The area is about 80.6 square units.

22. Assignment
Geometry:
10.5
Heron’s Law