# The shadow of a tower standing on a level plane is found to be 50 m longer when Sun’s elevation is 30° than when it is 60°. Find the height of the tower

**Solution:**

Given, the shadow of a tower standing on a level plane is found to be 50 m longer when the sun's elevation is 30° than when it is 60°

We have to find the height of the tower.

Let SQ be the height of the tower.

SQ = h m

Angle of elevation at first, ∠SRQ = 60°

At first, the length of the shadow = x m

The shadow of the tower is increased by 50 m when the angle of elevation ∠SPQ = 30°

In triangle SRQ,

tan 60° = SQ/RQ

√3 = h/x

x = h/√3 m --------------- (1)

In triangle SPQ,

tan 30° = SQ/PQ

We know that, PQ = PR + RQ

PQ = 50 + x

1/√3 = h/(50+x)

50 + x = √3h

x = √3h - 50 ------------ (2)

From (1) and (2),

h/√3 = √3h - 50

By grouping,

h/√3 - √3h = -50

√3h - h/√3 = 50

On simplification,

(3h-h)/√3 = 50

2h/√3 = 50

2h = 50√3

h = 50√3/2

h = 25√3 m

Therefore, the height of the tower is 25√3 m.

**✦ Try This: **The shadow of a tower standing on a level plane is found to be 40 m longer when Sun’s elevation is 30° than when it is 45°. Find the height of the tower.

**☛ Also Check:** NCERT Solutions for Class 10 Maths Chapter 8

**NCERT Exemplar Class 10 Maths Exercise 8.4 Problem 7**

## The shadow of a tower standing on a level plane is found to be 50 m longer when Sun’s elevation is 30° than when it is 60°. Find the height of the tower

**Summary:**

The shadow of a tower standing on a level plane is found to be 50 m longer when Sun’s elevation is 30° than when it is 60°. The height of the tower is 25√3 m

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