The document discusses the Law of Cosines and Heron's Formula. It begins by explaining that the Law of Cosines can be used to find side lengths and angle measures of a triangle when given side-angle-side or side-side-side information. It then provides examples of using the Law of Cosines to solve for missing values of triangles. It also introduces Heron's Formula, which can be used to find the area of a triangle based on its side lengths. Examples are given of using both the Law of Cosines and Heron's Formula to solve multi-step word problems involving triangles.

1. UNIT 10.5 LAW OFUNIT 10.5 LAW OF
COSINESCOSINES

2. Use the Law of Cosines to find the side
lengths and angle measures of a
triangle.
Use Heron’s Formula to find the area
of a triangle.
Objectives

3. In the previous lesson, you learned to solve
triangles by using the Law of Sines. However, the
Law of Sines cannot be used to solve triangles for
which side-angle-side (SAS) or side-side-side (SSS)
information is given. Instead, you must use the
Law of Cosines.

4. To derive the Law of Cosines, draw ∆ABC with
altitude BD. If x represents the length of AD, the
length of DC is b – x.

5. Write an equation that relates the side lengths
of ∆DBC.
a2
= (b – x)2
+ h2
a2
= b2
– 2bx + x2
+ h2
a2
= b2
– 2bx + c2
a2
= b2
– 2b(c cos A) + c2
a2
= b2
+ c2
– 2bccos A
Pythagorean Theorem
Expand (b – x)2.
In ∆ABD, c2
= x2
+ h2
.
Substitute c2
for x2
+ h2
.
The previous equation is one of the formulas for the
Law of Cosines.
In ∆ABD, cos A = or x =
cos A. Substitute
c cos A for x.

7. Example 1A: Using the Law of Cosines
Use the given measurements to solve ∆ABC.
Round to the nearest tenth.
a = 8, b = 5, m∠C = 32.2°
Step 1 Find the length of the third side.
c2
= a2
+ b2
– 2ab cos C
c2
= 82
+ 52
– 2(8)(5) cos 32.2°
c2
≈ 21.3
c ≈ 4.6
Law of Cosines
Substitute.
Use a calculator to
simplify.
Solve for the positive
value of c.

8. Step 2 Find the measure of the smaller angle, ∠B.
Law of Sines
Substitute.
Solve for sin B.
Solve for m B.
Example 1A Continued

9. Example 1A Continued
Step 3 Find the third angle measure.
m∠A + 35.4° + 32.2° ≈ 180°
m∠A ≈ 112.4°
Triangle Sum Theorem
Solve for m A.

10. Example 1B: Using the Law of Cosines
Use the given measurements to solve ∆ABC.
Round to the nearest tenth.
a = 8, b = 9, c = 7
Step 1 Find the measure of the largest angle, B.
b2
= a2
+ c2
– 2ac cos B
92
= 82
+ 72
– 2(8)(7) cos B
cos B = 0.2857
m B = Cos-1
(0.2857) ≈ 73.4°
Law of cosines
Substitute.
Solve for cos B.
Solve for m B.

11. Example 1B Continued
Use the given measurements to solve ∆ABC.
Round to the nearest tenth.
Step 2 Find another angle measure
c2
= a2
+ b2
– 2ab cos C Law of cosines
72
= 82
+ 92
– 2(8)(9) cos C Substitute.
cos C = 0.6667 Solve for cos C.
m C = Cos-1
(0.6667) ≈ 48.2° Solve for m C.

12. Example 1B Continued
Use the given measurements to solve ∆ABC.
Round to the nearest tenth.
m A + 73.4° + 48.2° ≈ 180°
m A ≈ 58.4°
Triangle Sum Theorem
Solve for m A.
Step 3 Find the third angle measure.

13. Check It Out! Example 1a
Use the given measurements to solve ∆ABC.
Round to the nearest tenth.
b = 23, c = 18, m A = 173°

14. Check It Out! Example 1b
Use the given measurements to solve ∆ABC.
Round to the nearest tenth.
a = 35, b = 42, c = 50.3

15. The largest angle of a triangle is the angle
opposite the longest side.
Remember!

16. Example 2: Problem-Solving Application
If a hiker travels at an
average speed of 2.5 mi/h,
how long will it take him to
travel from the cave to the
waterfall? Round to the
nearest tenth of an hour.
11 Understand the Problem
The answer will be the number of hours that
the hiker takes to travel to the waterfall.

17. List the important information:
• The cave is 3 mi from the cabin.
• The waterfall is 4 mi from the cabin. The path
from the cabin to the waterfall makes a 71.7°
angle with the path from the cabin to the cave.
• The hiker travels at an average speed of 2.5 mi/h.
11 Understand the Problem
The answer will be the number of hours that
the hiker takes to travel to the waterfall.

18. 22 Make a Plan
Use the Law of Cosines to find the distance d
between the water-fall and the cave. Then
determine how long it will take the hiker to
travel this distance.

19. d2
= c2
+ w2
– 2cw cos D
d2
= 42
+ 32
– 2(4)(3)cos 71.7°
d2
≈ 17.5
d ≈ 4.2
Law of Cosines
Substitute 4 for c,
3 for w, and
71.7 for D.
Use a calculator to simplify.
Solve for the positive
value of d.
Solve33
Step 1 Find the distance d between the waterfall
and the cave.

20. Step 2 Determine the number of hours.
The hiker must travel about 4.2 mi to
reach the waterfall. At a speed of 2.5
mi/h, it will take the hiker ≈ 1.7 h
to reach the waterfall.
Look Back44
In 1.7 h, the hiker would travel 1.7  2.5 =
4.25 mi. Using the Law of Cosines to solve
for the angle gives 73.1°. Since this is
close to the actual value, an answer of 1.7
hours seems reasonable.

21. Check It Out! Example 2
A pilot is flying from
Houston to Oklahoma City.
To avoid a thunderstorm,
the pilot flies 28° off the
direct route for a distance
of 175 miles. He then
makes a turn and flies
straight on to Oklahoma
City. To the nearest mile,
how much farther than the
direct route was the route
taken by the pilot?

22. The Law of Cosines can be used to derive a formula
for the area of a triangle based on its side lengths.
This formula is called Heron’s Formula.

23. Example 3: Landscaping Application
A garden has a triangular flower bed with
sides measuring 2 yd, 6 yd, and 7 yd. What is
the area of the flower bed to the nearest
tenth of a square yard?
Step 1 Find the value of s.
Use the formula for half of
the perimeter.
Substitute 2 for a, 6 for b,
and 7 for c.

24. Example 3 Continued
Step 2 Find the area of the triangle.
A =
A =
A = 5.6
Heron’s formula
Substitute 7.5
for s.
Use a calculator
to simplify.
The area of the flower bed is 5.6 yd2
.

25. Example 3 Continued
Check Find the measure of the largest angle, ∠C.
c2
= a2
+ b2
– 2ab cos C
72
= 22
+ 62
– 2(2)(6) cos C
cos C ≈ –0.375
m C ≈ 112.0°
Find the area of the triangle by using
the formula area = ab sin c.
area
Law of Cosines
Substitute.
Solve for cos C.
Solve for m c.


26. Check It Out! Example 3
The surface of a hotel swimming pool is shaped
like a triangle with sides measuring 50 m, 28 m,
and 30 m. What is the area of the pool’s surface to
the nearest square meter?

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