Question

$\huge\color{pink}\boxed{\colorbox{Black}{★Question★}}$
The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower is 30°. Find the height of the tower.

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Answer 1

Answer:

$10\sqrt{3} m$

Step-by-step explanation:

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❒ Let the tower be AB

❒ Let point be C

❒ Distance of point C from foot of tower = $30m$

Hence,BC = $30m$

❒ Angle of elevation = $30^o$

$So\: \angle ACB =30^o$

$\bold{Since\:tower\:is\:vertical,}$

$\angle ABC= 90^o$

❒ We need to find height of tower AB

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$\hookrightarrow\boxed{tanC = \frac{side\:opposite\:to\:angle\:C}{side\:adjacent\:to\:angle\:C} }$

$\hookrightarrow tan\:30^o = \frac{AB}{AC}$

$\hookrightarrow \frac{1}{\sqrt{3}} = \frac{AB}{30}$

$\hookrightarrow \frac{30}{\sqrt{3}} = AB$

$\hookrightarrow AB = \frac{30}{\sqrt{3} }$

$\bold{Multiplying\:\sqrt{3}\:in\:numerator\:and\:denominator}$

$\hookrightarrow AB = \frac{30}{\sqrt{3} } \times \frac{\sqrt{3} }{\sqrt{3} }$

$\hookrightarrow AB = \frac{30\sqrt{3} }{3}$

$\hookrightarrow AB = 10\sqrt{3}$

$\orange{\hookrightarrow}\red{\boxed{AB = 10\sqrt{3}}}$

$\mathbb{HEIGHT\:OF\:TOWER=}10\sqrt{3}m$