﻿ Trigonometry Worksheets | Problems & Solutions

# Trigonometry Worksheets

Trigonometry Worksheets
• Page 1
1.
Find the measure of angle θ in the figure.

 a. 48°42′ b. 41°48′ c. 33°41′ d. 56°19′

#### Solution:

The measure of the angle of elevation from point A is θ.

In right triangle APQ,

tan θ = 20 / 30

θ = 33°41′.

2.
The angle of elevation of the top of a hill from the foot of a tower is 62° and the angle of elevation of the top of the tower from the foot of the hill is 28°. If the tower is 40 ft high, then find the height of the hill.
 a. 141ft b. 11.3 ft c. 101 ft. d. 39.5 ft

#### Solution:

Draw the diagram.

Let x represent the height of the hill and let y represent the distance between the hill and the tower.

In right triangle QPA,

tan 28° = 40y

y = 40tan 28° = 75

In right triangle PAB, tan 62° = xy =x75

x = 75 tan 62° = 141

The height of the hill is 141 ft.

3.
Tom and Sam are on either side of a tower of height $h$ meters.They measure the angle of elevation of the top of the tower as $\theta$ and $\phi$ respectively. Find the distance through which Tom and Sam are seperated. [Given $h$ = 160, $\theta$ = 50° and $\phi$ = 35°.]
 a. 302.71 meters b. 160.61 meters c. 84.57 meters d. 363 meters

#### Solution:

Height of the pole is AB = h = 160 meters, C and D are the positions of Tom and Sam as shown.
[Draw the diagram for the given data.]

ACB = θ = 50° and ADB = φ = 35°
[Write the angles of elevation of A from Tom, Sam.]

In the right triangle ADB, tan φ = tan 35° = 160BD
[tan φ = AB / BD.]

BD = 160tan 35° » 228.571 meters
[Substitute the value of tan 35° and find BD.]

In the right triangle ABC, tan θ = tan 50° = 160BC
[tan θ = AB / BC.]

BC = 160tan 50° » 134.340 meters
[Substitute the value of tan 50° and find BC.]

CD = CB + BD = 228.571 +134.340 » 363 m
[Use CD = CB + BD to find CD.]

So, the distance between Tom and Sam is 363 m.

4.
The angle of depression $\phi$ of the top of a tower of height $h$ meters from the top of another tower of height H meters is 25°. Find the horizontal distance between the two towers when $h$ = 93 and H = 125.
 a. 15 ft b. 70 ft c. 58 ft d. 43 ft

#### Solution:

Height of the first tower is CD = h = 93 meters, height of the second tower is AB = H = 125 meters as shown
[Draw the diagram for the given data.]

Let the distance between the two towers, BC = ED = x meters and the difference between the heights of the two towers, AE = y meters

The angle of depression of D from A is φ = 25°
[Write the angle of depression of D from A.]

In the right triangle ADE, tan ADE = tan 25° = AE / ED = yx and hence y = 0.46x
[Substitute the value of tan 25° and find h in terms of x.]

y = AE = AB - BE = 125 - 93 = 32 meters
[From the figure AE = AB - BE = h.]

32 = 0.46x
[From step 6 and step 7.]

x » 70
[Solve for x.]

So, the distance between two towers is 70 meters.