Right Triangle Trigonometry Section 1.4 Right Triangle Trigonometry

Objectives: 1. To solve right triangles. 2. To solve word problems involving right triangles.

To solve a right triangle involves finding the measures of all the sides and angles.

EXAMPLE 1 Solve ABC A c b 42º C 3 B

Solving right triangles requires the Pythagorean theorem and the trigonometric ratios. To solve a right triangle you must find the measures of all the sides and angles.

EXAMPLE 1 Solve ABC A C B c b 3 42º mA = 90° - mB = 90° - 42° = 48°

EXAMPLE 1 Solve ABC A C B c b 3 42º b 3 2.7 = b tan 42º = 3 tan 42º = b 2.7 = b

EXAMPLE 1 Solve ABC A C B c b 3 42º 3 c c = 4.0 cos 42º = c = 3

mA = 48 c ≈ 4.0 b ≈ 2.7 a = 3 mB = 42 mC = 90 EXAMPLE 1 Solve ABC A C B c b 3 42º mA = 48 c ≈ 4.0 b ≈ 2.7 a = 3 mB = 42 mC = 90

Practice: Solve ABC A 33º mB = 57° c b C 2.8 B mB = 90° - mA = 90° - 33° = 57° b C 2.8 B

Practice: Solve ABC A 33º b = 4.3 c b 2.8 b b = 4.3 C 2.8 B tan 57º =

Practice: Solve ABC A 33º c = 5.1 c 2.8 c b c = 5.1 C 2.8 B cos 57º =

Practice: Solve ABC a = 2.8 b ≈ 4.3 c ≈ 5.1 mA = 33º mB = 57º mC = 90º c b C 2.8 B

Definition The angle of elevation is the angle that is formed by a horizontal line and the line of sight as the observer looks at an object that is above the horizontal. angle of elevation horizontal line of sight

Definition The angle of depression is the angle that is formed by a horizontal line and the line of sight as the observer looks at an object that is below the horizontal. line of sight horizontal angle of depression

EXAMPLE 2 There are two lifeguards stands located 100 meters apart on a north-south coast. One guard spots an oil derrick at a bearing of S 42º40 E. If the oil derrick is due east of the other guard, how far is it from the second stand?

cos 42º40 = x 100 O1 O2 100 m S x 42º40 100 cos 42º40 = x x ≈ 92.2 m

To solve a right triangle given two sides. 1. Use the Pythagorean theorem to find the 3rd side. 2. Use the two given sides and the appropriate trig ratio to find one of the acute angles. 3. Subtract the acute angle found in step 2 from 90 to find the other acute angle.

To solve a right triangle given an acute angle and a side. 1. Subtract the given acute angle from 90 to find the other acute angle. 2. Use an acute angle, the given side, and the appropriate trig ratio to find another side. 3. Repeat step 2 to find the 3rd side.

Homework: pp. 20-22

►A. Exercises A C B 5 9 c b a = + 9 b 5 = + 81 b 25 = + 56 b = 56 b = 1. Solve ∆ABC, given a = 5, c = 9. A C B 5 9 2 c b a = + 2 9 b 5 = + 81 b 25 2 = + 56 b 2 = 56 b = 2 14 b =

o h sin A = 5 9 sin A = mA = sin-1 5 9 mA = 33.7 ►A. Exercises 1. Solve ∆ABC, given a = 5, c = 9. A C B 5 9 o h sin A = 5 9 sin A = 2 14 mA = sin-1 5 9 mA = 33.7

a h cos B = 5 9 cos B = mB = cos-1 5 9 mB = 56.3 ►A. Exercises 1. Solve ∆ABC, given a = 5, c = 9. A C B 5 9 a h cos B = 33.7° 5 9 cos B = 2 14 mB = cos-1 5 9 mB = 56.3

mA ≈ 33.7 c = 9 14 b = 2 a = 5 mB ≈ 56.3 mC = 90 ►A. Exercises 1. Solve ∆ABC, given a = 5, c = 9. A C B 5 9 mA ≈ 33.7 c = 9 14 b = 2 a = 5 mB ≈ 56.3 mC = 90 33.7° 2 14 56.3°

►B. Exercises 13. 3 ft 12 feet 8°

►B. Exercises 13. 3 ft 12 feet 9 feet 8° x feet x 9 tan 8° =

►B. Exercises 15. 1 mile = 5280 feet x 5° ; 5280 x sin 5°= ≈ 460.18 ft.

■ Cumulative Review 22. Give the midpoint between (a, b) and (-a, -b).

■ Cumulative Review 23. Give sin .  3x2 6x

■ Cumulative Review 24. Find sin 77.

■ Cumulative Review 25. Find cos 500°.

■ Cumulative Review 26. Find cot 219°.

Review: Solve ∆ABC, given b = 12, B = 37°. h o sin B = c 12 sin 37° = c = 12 sin 37° c ≈ 19.9

Review: Solve ∆ABC, given b = 12, B = 37°. tan B = a 12 tan 37° = 19.9 a = 12 tan 37° a ≈ 15.9

Review: Solve ∆ABC, given b = 12, B = 37°. 19.9 15.9

Review: Solve ∆ABC, given b = 12, B = 37°. 53° 19.9 15.9