Solving Right Triangles

M2 Unit 2: Day 6

How do you solve right triangles?

To solve a right triangle you need…..

1 side length and 1 acute angle measure-or-

2 side lengths

Every right triangle has one right angle, two acute angles, one hypotenuse, and two legs.

To SOLVE A RIGHT TRIANGLE means to find all 6 parts.

Given one acute angle and one side:

•To find the missing acute angle, use the Triangle Sum Theorem.

•To find one missing side length, write an equation using a trig function.

•To find the other side, use another trig function or the Pythagorean Theorem

GUIDED PRACTICE

Example 1

Find m∠ B by using the Triangle Sum Theorem.

180o = 90o + 42o + m∠ B

48o = m∠ B

A

C B48o

7042o

Solve the right triangle. Round decimal answers to the nearest tenth.

Approximate BC by using a tangent ratio.

tan 42o =BC 70

70 tan 42o = BC

70 0.9004 BC

63.0 ≈ BC

Approximate AB by using a cosine ratio.

cos 42o =70 AB

AB cos 42o = 70

AB 70 cos 42o=

AB 70 0.7431

AB 94.2

ANSWER

The angle measures are 42o, 48o, and 90o. The side lengths are 70 feet, about 63.0 feet, and about 94.2 feet.

GUIDED PRACTICESolve a right triangle that has a 40o angle and a 20

inch hypotenuse.

Example 2

Find m∠ X by using the

Triangle Sum Theorem.

180o = 90o + 40o + m∠ X50o = m∠ X

X

YZ

Approximate YZ by using a sine ratio.

sin 40o =XY

2020 ● sin 40o = XY

20 ● 0.6428 ≈

XY

12.9 ≈ BC

Approximate AB by using a cosine ratio.

cos 40o =YZ 20

20 ● cos 40o = YZ20 ● 0.7660 ≈ YZ

15.3 ≈ YZ

The angle measures are 40o, 50o, and 90o. The side lengths are 12.9 in., about 15.3 in., and 20 in.

ANSWER

40o

20 in50o

Solve the right triangle. Round to the nearest tenth.

53

90

30

m Q

m R

m P

PQ

PR

QR

cos53

30

p sin5330

q

37°

18.1

24.0

Example 3

18.1p 24.0q

If you know the sine, cosine, or tangent of an acute angle measure, you can use the inverse

trigonometric functions to find the measure of the angle.

Calculating Angle Measures from

Trigonometric Ratios

Use your calculator to find each angle measure to the nearest tenth of a degree.

A. cos-1(0.87) B. sin-1(0.85) C. tan-1(0.71)

cos-1(0.87) 29.5° sin-1(0.85) 58.2° tan-1(0.71) 35.4°

Example 4

Inverse trig functions:

Ex: Use a calculator to approximate the measure of the acute angle. Round to the

nearest tenth.

26.6° 20.5° 50.2°

1tan (0.5)m A 1sin (0.35)m A 1cos (0.64)m A

1. tan A = 0.5 2. sin A = 0.35 3. cos A = 0.64

EXAMPLE 2 Use an inverse sine and an inverse cosine

Let ∠ A and ∠ B be acute angles in a right triangle. Use a calculator to approximate the measures of ∠ A and ∠ B to the

nearest tenth of a degree.

a. sin A = 0.87 b. cos B = 0.15

SOLUTION

a. m ∠ A = sin –1 0.87 ≈ 60.5o

b. m ∠ B = cos –1 0.15≈ 81.4o

Example 5

Solving Right Triangles

Find the unknown measures. Round lengths to the nearest hundredth and angle measures to the nearest degree.

Method 1: By the Pythagorean Theorem,

Since the acute angles of a right triangle are complementary, mT 90° – 29° 61°.

RT2 = RS2 + ST2

(5.7)2 = 52 + ST2

Method 2:

Since the acute angles of a right triangle are complementary, mT 90° – 29° 61°.

, so ST = 5.7 sinR.

Example 6

Solve the right triangle. Round decimals the nearest tenth.

3

2

90

AB

BC

AC

m A

m B

m C

Use Pythagorean Theorem to find c…

2 2 32 3

3.6

c

c

3.6Use an inverse trig function to find a missing acute angle…

1 3tan ( ) 56.3

2m A

Use Triangle Sum Theorem to find the other acute angle…

90 56.3 33.7m B

56.3°

33.7°

Example 7

2 2 211 18

21.9

PN

PN

1 11tan ( ) 31.4

18m N

90 31.4 58.6m P

Example 8

2 2 223 7

21.9

TU

TU

1 7cos ( ) 72.3

23m S

90 72.3 17.7m U

Example 9

Solve the right triangle. Round decimals to the nearest tenth.

90 37 53m P

sin3722

13.2

PQ

PQ

cos3722

17.6

QR

QR

90 24 66m T

tan243314.7

TR

TR

33cos24

36.1AT

AT

Homework:Pg 174 (#4-22 even)