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