A pole 6, m high casts a shadow 2 sqrt{3}, m | vedclass

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(C) Let $BC = 6\, m$ be the height of the pole and $AB = 2\sqrt{3}\, m$ be the length of the shadow on the ground. Let the Sun's elevation angle be $\

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(C) Let $BC = 6\, m$ be the height of the pole and $AB = 2\sqrt{3}\, m$ be the length of the shadow on the ground. Let the Sun's elevation angle be $	heta$.In the right-angled $	riangle ABC$,we have:$	an 	heta = \frac{	ext{Perpendicular}}{	ext{Base}} = \frac{BC}{AB}$Substituting the given values:$	an 	heta = \frac{6}{2\sqrt{3}}$$	an 	heta = \frac{3}{\sqrt{3}}$Rationalizing the denominator:$	an 	heta = \frac{3 	imes \sqrt{3}}{\sqrt{3} 	imes \sqrt{3}} = \frac{3\sqrt{3}}{3} = \sqrt{3}$Since $	an 60^{\circ} = \sqrt{3}$,we get:$	heta = 60^{\circ}$Thus,the Sun's elevation is $60^{\circ}$.

$A$ pole $6\, m$ high casts a shadow $2 \sqrt{3}\, m$ long on the ground. Then,the Sun's elevation is: (in $^{\circ}$)

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