From a point 20 ; m away from the foot of a t | vedclass

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(B) Let $AB$ be the height of the tower and $C$ be the point on the ground.Given that the distance from the foot of the tower $B$ to the point $C$ is

Vedclass · Accepted Answer

$\frac{20}{\sqrt{3}} \; m$

Vedclass · Answer

(B) Let $AB$ be the height of the tower and $C$ be the point on the ground.Given that the distance from the foot of the tower $B$ to the point $C$ is $BC = 20 \; m$.The angle of elevation $\angle ACB = 30^{\circ}$.In the right-angled triangle $	riangle ABC$,we have:$	an(30^{\circ}) = \frac{	ext{Perpendicular}}{	ext{Base}} = \frac{AB}{BC}$Since $	an(30^{\circ}) = \frac{1}{\sqrt{3}}$,we get:$\frac{1}{\sqrt{3}} = \frac{AB}{20}$$AB = \frac{20}{\sqrt{3}} \; m$.Therefore,the height of the tower is $\frac{20}{\sqrt{3}} \; m$.

From a point $20 \; m$ away from the foot of a tower,the angle of elevation of the top of the tower is $30^{\circ}$. The height of the tower is:

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