The angle of depression of a ship from the to | vedclass

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(C) Let $h = 30 \, m$ be the height of the tower and $x$ be the distance of the ship from the base of the tower.Since the angle of depression is $60^\

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$10 \sqrt{3} \, m$

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(C) Let $h = 30 \, m$ be the height of the tower and $x$ be the distance of the ship from the base of the tower.Since the angle of depression is $60^\circ$,the angle of elevation from the ship to the top of the tower is also $60^\circ$.In the right-angled triangle formed by the tower and the distance on the ground,we have:$	an 60^\circ = \frac{	ext{Height}}{	ext{Distance}} = \frac{30}{x}$Since $	an 60^\circ = \sqrt{3}$,we get:$\sqrt{3} = \frac{30}{x}$$x = \frac{30}{\sqrt{3}} = \frac{30 	imes \sqrt{3}}{\sqrt{3} 	imes \sqrt{3}} = \frac{30 \sqrt{3}}{3} = 10 \sqrt{3} \, m$.Thus,the distance of the ship from the base of the tower is $10 \sqrt{3} \, m$.

The angle of depression of a ship from the top of a tower $30 \, m$ high is $60^\circ$. Find the distance of the ship from the base of the tower.

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