Test: Applications of Heights & Distances - Class 10 MCQ

A tower of height 40 m is given, and the angles of depression from the top of the tower to:

• The top of the tree = 45°
• The bottom of the tree = 60°

We need to find the height of the tree (h).

Step 1: Use trigonometry for the angles of depression
Let:

• The distance between the base of the tower and the base of the tree = d
• The height of the tree = h

From the 45° angle (to the top of the tree):

The formula is:

tan(45°) = (Height of the tower - Height of the tree) / Distance (d)

Since tan(45°) = 1:

1 = (40 - h) / d

d = 40 - h (1)

From the 60° angle (to the bottom of the tree):

The formula is:

tan(60°) = Height of the tower / Distance (d)

Since tan(60°) = √3:

√3 = 40 / d

d = 40 / √3 (2)

Step 2: Solve the equations

Equate d from equations (1) and (2):

40 - h = 40 / √3

Rearrange to solve for h:

h = 40 - (40 / √3)

Rationalize the denominator:

h = 40 - (40√3 / 3)

h = (120 / 3) - (40√3 / 3)

h = (40 (3 - √3)) / 3

The height of the tree is:

b) (40 / 3) (3 - √3)

The second tower is smaller than the first tower so let the height of the first tower is h So a part of it is equal to 60m so the remaining height is h-60 m
Tan 30=perpendicular/base

To solve for the height of the tower, we use the tangent function. The angle of elevation is 45°, and the horizontal distance from the observer to the tower is 28.5 m.

Let the height of the tower be h. The observer's eye level is 1.5 m, so the difference in height between the top of the tower and the observer's eyes is h−1.5h.