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

In right triangle APQ,

tan θ = 20 / 30

θ = 33°41′.

Correct answer : (3)

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 = 40tan28° = 75

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

x = 75 tan 62° = 141

The height of the hill is 141 ft.

Correct answer : (1)

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 = 160tan35° » 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 = 160tan50° » 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.

Correct answer : (4)

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.]

∠ ADE = φ = 25° [∠ ADE = φ as ∠ ADE, φ are alternate angles.]

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.]