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The angle of elevation of the top of a vertical tower from a point on the ground is 60°. From another point, 10 m vertically above the first, its angle of elevation is 45°. Find the height of the tower.

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Solution.         According to question

Let h is the height of the tower
In \bigtriangleupABE
\tan \theta = \frac{P}{B}
\tan 60^{\circ} = \frac{h}{AB}             \left ( \because \theta = 60^{\circ} \right )
\sqrt{3}= \frac{h}{AB}       \left ( \because \tan 60^{\circ}= \sqrt{3} \right )
AB= \frac{h}{\sqrt{3}}\: \cdots \left ( 1 \right )
In \bigtriangleupEDC
\tan \theta = \frac{P}{B}
\tan 45^{\circ} = \frac{h-10}{DC}
1= \frac{h-10}{AB}    \left ( \because AB-DC,\tan 45^{\circ}= 1 \right )
AB= h-10\: \cdots \left ( 2 \right )
from equation (1) and (2)
h-10= \frac{h}{\sqrt{3}}
\sqrt{3}h-\sqrt{3}10= h
\sqrt{3}h-h= \sqrt{3}\times 10
h\left ( \sqrt{3}-1 \right )= 10\sqrt{3}
h= \frac{10\sqrt{3}}{\sqrt{3}-1}
Hence the height of the tower
h= \frac{10\sqrt{3}}{\sqrt{3}-1}m

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