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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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