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.

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Solution

Given that the Height of the High Building $B D = q = 20 m$

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

From $△ D B C$

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

From $△ A B C$

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

$\Rightarrow 20 + p = 20 \sqrt{3}$

$\Rightarrow p = 20 (\sqrt{3} - 1) = 20 \times 0.73 = 14.6$
Therefore, Height of Transmission Tower $= 14.6 m$

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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 $45^{\circ} a n d 60^{\circ}$ respectively. Find the height of the transmission tower.

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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∘ and 60∘ respectively. Find the height of the transmission tower.

Given that the Height of the High Building BD=q=20 mNow, Let the Height of the Transmission Tower be pand Let the Distance of the point be xFrom △DBC⇒tan45o=qx⇒q=x=20From △ABC⇒tan60o=p+qx=20+p20⇒20+p=20√3⇒p=20(√3−1)=20×0.73=14.6Therefore, Height of Transmission Tower =14.6 m

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.