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Question

From a point on the ground the angles of elevation of the bottom and top of a transmission tower fixed at the top of $$20$$ m high building are $$45^\circ$$ and $$60^\circ$$ respectively. Find the height of the transmission tower.


Solution

Given that the Height of the High Building $$BD=q=20\ m$$

Now, Let the Height of the Transmission Tower be $$p$$
and Let the Distance of the point be $$x$$

From $$\triangle DBC$$

$$\Rightarrow$$$$\tan 45^o=\dfrac{q}{x}\Rightarrow q=x=20$$

From $$\triangle ABC$$

$$\Rightarrow$$$$\tan 60^o=\dfrac{p+q}{x}=\dfrac{20+p}{20}$$

$$\Rightarrow 20+p=20\sqrt3$$

$$\Rightarrow p=20(\sqrt3-1)=20\times0.73=14.6$$
Therefore, Height of Transmission Tower $$=14.6\ m$$


943826_971474_ans_68d42ea1ca95437c9978cfa23da473a1.png

Mathematics

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