# A pole 6 m high casts a shadow 2√3 m long on the ground, then the Sun’s elevation is

a. 90°

b. 30°

c. 60°

d. 45°

**Solution:**

Given, length of pole = 6 m

Pole casts a shadow of 2√3 m long on the ground.

We have to find the sun’s elevation.

Let θ be the angle of elevation.

From the figure,

Opposite = 6 m

Adjacent = 2√3 m

tan θ = opposite/hypotenuse

tan θ = 6/2√3

tan θ = 3/√3

tan θ = √3

θ = tan⁻¹(√3)

Using the trigonometric ratios of angles,

tan 60° = √3

So, θ = 60°

Therefore, the angle of elevation is 60°

**✦ Try This: **A pole 12 m high casts a shadow 4√3 m long on the ground, then the Sun’s elevation is

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

**NCERT Exemplar Class 10 Maths Exercise 8.1 Problem 15**

## A pole 6 m high casts a shadow 2√3 m long on the ground, then the Sun’s elevation is a. 90°, b. 30°, c. 60°, d. 45°

**Summary:**

A pole 6 m high casts a shadow 2√3 m long on the ground, then the Sun’s elevation is 60°

**☛ Related Questions:**

- The value of sinθ + cosθ is always greater than 1. Write ‘True’ or ‘False’ and justify your answer
- The value of tanθ (θ < 90°) increases as θ increases. Write ‘True’ or ‘False’ and justify your answe . . . .
- tanθ increases faster than sinθ as θ increases. Write ‘True’ or ‘False’ and justify your answer

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