Question

Question 15
A pole 6m high casts a shadow $2\sqrt{3}m$ long on the ground, then the Sun’s elevation is

(A) ${60}^{\circ }$
(B) ${45}^{\circ }$
(C) ${30}^{\circ }$
(D) ${90}^{\circ }$

Answer 1

The answer is (A).
Let BC = 6m be the height of the pole and $AB=2\sqrt{3}m$ be the length of the shadow on the ground.
Let the Sun makes an angle $\theta$ on the ground.

Now, in $\Delta BAC$,

$\tan \theta =\frac{BC}{AB}$

$\Rightarrow \tan \theta =\frac{6}{2\sqrt{3}}=\frac{3}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}$

$\Rightarrow \tan \theta =\frac{3\sqrt{3}}{3}=\sqrt{3}=\tan {60}^{\circ }$

$\therefore \theta ={60}^{\circ }$

Hence, the Sun’s elevation is ${60}^{\circ }$