9. The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

Answers (1)
M manish


It is given that, the height of the tower (AB) is 50 m. \angle AQB = 30^o and \angle PBQ = 60^o
Let the height of the building be h m

According to question,
In triangle PBQ,
\tan 60^o = \frac{PQ}{BQ} = \frac{50}{BQ}
\\\sqrt{3} = \frac{50}{BQ}\\ BQ = \frac{50}{\sqrt{3}}.......................(i)

In triangle ABQ,

\tan 30^o = \frac{h}{BQ}
{BQ}=h\sqrt{3}.........................(ii)
On equating the eq(i) and (ii) we get,

\frac{50}{\sqrt{3}}=h\sqrt{3}
therefore, h = 50/3 = 16.66 m = height of the building.

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