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7. From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.

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Suppose BC = h is the height of transmission tower and the AB be the heoght of the building and  AD is the distance between bulding and the observer point (D).
We have,
AB = 20 m, BC = h m and AD = x m
\angle CDA = 60^o and \angle BDA = 45^o

According to question,
In triangle \Delta BDA,
\tan 45^o = \frac{AB}{AD}=\frac{20}{x}
So, x = 20 m

In triangle \DeltaCAD,

\\\Rightarrow \tan 60^o = \frac{AB+BC}{AD}=\frac{20+h}{20}\\\\\Rightarrow \sqrt{3}= 1+\frac{h}{20}\\\\\Rightarrow h=20(\sqrt{3}-1)\\\\\Rightarrow 20(0.732) = 14.64 m

Answer- the height of the tower is 14.64 m

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