The horizontal distance between two towers is | vedclass

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(C) Let the height of the first tower be $h \ m$ and the height of the second tower be $H = 150 \ m$.The horizontal distance between them is $d = 60 \

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$150 - 20\sqrt{3} \ m$

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(C) Let the height of the first tower be $h \ m$ and the height of the second tower be $H = 150 \ m$.The horizontal distance between them is $d = 60 \ m$.From the top of the second tower,the angle of depression to the top of the first tower is $30^\circ$. This implies the angle of elevation from the top of the first tower to the top of the second tower is also $30^\circ$.Considering the right-angled triangle formed by the difference in heights and the horizontal distance:$	an(30^\circ) = \frac{H - h}{d}$$\frac{1}{\sqrt{3}} = \frac{150 - h}{60}$$150 - h = \frac{60}{\sqrt{3}}$$150 - h = 20\sqrt{3}$$h = 150 - 20\sqrt{3} \ m$.Thus,the height of the first tower is $(150 - 20\sqrt{3}) \ m$.

The horizontal distance between two towers is $60 \ m$. The angle of depression of the top of the first tower as seen from the top of the second tower is $30^\circ$. If the height of the second tower is $150 \ m$,then the height of the first tower is:

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