Warm Up

Find each measure to the nearest tenth.

1. my 2. x

3. y

4. What is the area of ∆XYZ ? Round to the

nearest square unit. 60 square units

104° ≈ 8.8

≈ 18.3

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

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.

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.

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.

Example 1A: Using the Law of Cosines

Use the given measurements to solve ∆ABC. Round to the nearest tenth.

a = 8, b = 5, mC = 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.

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

Example 1A Continued

Step 3 Find the third angle measure.

mA + 35.4° + 32.2° 180°

mA 112.4°

Triangle Sum Theorem

Solve for m A.

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.

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.

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.

Check It Out! Example 1a

Use the given measurements to solve ∆ABC. Round to the nearest tenth.

Step 1 Find the length of the third side.

a2 = b2 + c2 – 2bc cos A

a2 = 232 + 182 – 2(23)(18) cos 173°

a2 ≈ 1672.8

a ≈ 40.9

Law of Cosines

Substitute.

Use a calculator to simplify.

Solve for the positive value of c.

b = 23, c = 18, m A = 173°

Law of Sines

Substitute.

Solve for sin C.

Solve for m C.

Step 2 Find the measure of the smaller angle, C.

m C = Sin-1

Check It Out! Example 1a Continued

Step 3 Find the third angle measure.

m B + 3.1° + 173° 180°

m B 3.9°

Triangle Sum Theorem

Solve for m B.

Check It Out! Example 1a Continued

Check It Out! Example 1b

Use the given measurements to solve ∆ABC. Round to the nearest tenth.

a = 35, b = 42, c = 50.3

Step 1 Find the measure of the largest angle, C.

c2 = a2 + b2 – 2ab cos C

50.32 = 352 + 422 – 2(35)(50.3) cos C

cos C = 0.1560

m C = Cos-1 (0.1560) ≈ 81.0°

Law of cosines

Substitute.

Solve for cos C.

Solve for m C.

Check It Out! Example 1b Continued

Step 2 Find another angle measure

a2 = c2 + b2 – 2cb cos A Law of cosines

352 = 50.32 + 422 – 2(50.3)(42) cos A Substitute.

cos A = 0.7264 Solve for cos A.

m A = Cos-1 (0.7264) ≈ 43.4° Solve for m A.

Use the given measurements to solve ∆ABC. Round to the nearest tenth.

a = 35, b = 42, c = 50.3

Step 3 Find the third angle measure.

m B + 81° + 43.4° 180°

m B 55.6° Solve for m B.

Check It Out! Example 1b Continued

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

Remember!

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.

1 Understand the Problem

The answer will be the number of hours that the hiker takes to travel to the waterfall.

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.

1 Understand the Problem

The answer will be the number of hours that the hiker takes to travel to the waterfall.

2 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.

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.

Solve3

Step 1 Find the distance d between the waterfall and the cave.

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 Back4

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.

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?

1 Understand the Problem

The answer will be the additional distance the pilot had to fly to reach Oklahoma City.

List the important information:

• The direct route is 396 miles.

• The pilot flew for 175 miles off the course at an angle of 28° before turning towards Oklahoma City.

2 Make a Plan

Use the Law of Cosines to find the distance from the turning point on to Oklahoma City. Then determine the difference additional distance and the direct route.

b2 = c2 + a2 – 2ca cos B

b2 = 3962 + 1752 – 2(396)(175)cos 28°

b2 ≈ 65072

b ≈ 255

Law of Cosines

Substitute 396 for c, 175 for a, and 28° for B.

Use a calculator to simplify.

Solve for the positive value of b.

Solve3

Step 1 Find the distance between the turning point and Oklahoma City. Use side-angle-side.

255 + 175 = 430

430 – 396 = 34

Total miles traveled.

Additional miles.

Step 2 Determine the number of additional miles the plane will fly.

Add the actual miles flown and subtract from that normal distance to find the extra miles flown

Look Back4

By using the Law of Cosines the length of the extra leg of the trip could be determined.