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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 $^{\circ}$)

Similar Questions

In $\Delta ABC$,$\angle B$ is a right angle. If $AC = 20$ and $BC = 10$,then $m\angle ACB = \ldots$ (in $^{\circ}$)

An erect tree is $300\, m$ away from a hill. If the height of the hill is $300\, m$,find the angle of depression of the base of the tree from the top of the hill. (in $^{\circ}$)

An observer $1.5 \ m$ tall is $20.5 \ m$ away from a tower $22 \ m$ high. Determine the angle of elevation of the top of the tower from the eye of the observer. (in $^{\circ}$)

In $\Delta ABC$,$\angle B$ is a right angle. If $AC = 16$ and $BC = 8$,then $m \angle C = \dots$ (in $^{\circ}$)

Watching from the top of an $80 \, m$ high hill,the angle of elevation of the top of a tower is $30^\circ$ and the angle of depression of the base of the tower is $45^\circ$. Find the height of the tower in $m$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module