A pole \( 6 \mathrm{~m} \) high casts a shadow \( 2 \sqrt{3} \mathrm{~m} \) long on the ground, then the Sun's elevation is
(A) \( 60^{\circ} \)
(B) \( 45^{\circ} \)
(C) \( 30^{\circ} \)
(D) \( 90^{\circ} \)

Given: 

A pole $6\ m$ high casts a shadow $2\sqrt{3}\ m$ long on the ground.

To do:

We have to find the sun’s elevation.

Solution:

Let height $=6\ m$

length of shadow $=2\sqrt{3}\ m$

$\theta$ is angle of elevation

$tan\theta=\frac{height}{shadow-length}$

$=\frac{6}{2\sqrt{3}}$

$=\sqrt{3}$

$=tan60^o$

$\therefore \theta=60^o$

Thus, angle of elevation is $60^o$

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Updated on: 10-Oct-2022

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