Length of the shadow of a 15 meter high pole is '15√3' meters at 8 O'clock in the morning. Then, what is the angle of elevation of the Sun rays with the ground at the time?
Answers
Step-by-step explanation:
Given:-
Length of the shadow of a 15 meter high pole is '15√3' meters at 8 O'clock in the morning.
To find:-
What is the angle of elevation of the Sun rays with the ground at the time?
Solution:-
Height of the pole = 15 m
Length of the shadow of the pole at 8 O' clock
= 15√3 m
Consider the above information as a right angled triangle ABC.
Right angle at B = 90°
Height of the pole (AB)=15m
Length of the shadow of the pole at 8 O' clock
BC = 15√3 m
Let the angle of elevation of the Sun rays with the ground at the time be θ
angle ACB = θ
Tan θ = Opposite side to θ / Adjacent side to θ
=> Tan θ = AB / BC
=> Tan θ = 15 / 15√3
=> Tan θ = 1/√3
=> Tan θ = Tan 30°
=> θ = 30°
The required angle = 30°
Answer:-
The angle of elevation of the Sun rays with the ground at the time = 30°
Used formulae:-
- Tan θ = Opposite side to θ / Adjacent side to θ
- Tan 30° = 1/√3
