The angle of elevation of a tower from a dist | vedclass

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(B) Let $AB$ be the height of the tower,$h \; m$.Let $C$ be the point at a distance of $50 \; m$ from the foot of the tower $B$.Thus,$BC = 50 \; m$.Th

Vedclass · Accepted Answer

$\frac{50}{\sqrt{3}} \; m$

Vedclass · Answer

(B) Let $AB$ be the height of the tower,$h \; m$.Let $C$ be the point at a distance of $50 \; m$ from the foot of the tower $B$.Thus,$BC = 50 \; m$.The angle of elevation is given as $\angle ACB = 30^{\circ}$.In the right-angled triangle $	riangle ABC$,we have:$	an 30^{\circ} = \frac{	ext{Perpendicular}}{	ext{Base}} = \frac{AB}{BC}$Substituting the values,we get:$\frac{1}{\sqrt{3}} = \frac{h}{50}$$h = \frac{50}{\sqrt{3}} \; m$.Therefore,the height of the tower is $\frac{50}{\sqrt{3}} \; m$.

The angle of elevation of a tower from a distance $50 \; m$ from its foot is $30^{\circ}$. The height of the tower is

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