Question

A tower in a city is 150 m high and a multi storeyed hotel at the city centre is 20 m high. The angle of elevation of the top of the tower at the top of hotel is 30 degree

Answer 1

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Answer 2

The distance between the tower and the building is 225.17 meters.

Step-by-step explanation:

See the diagram attached.

DE is the building and AB is the tower and ∠ CEB = 30°, the angle of elevation of the top of tower B from the top of the building E.

Now, from Δ BCE, BC = AB - AC = 150 - 20 = 130 m.

CE = AD = Distance between the tower and the building = x ( Say ).

Now, $\tan 30^{\circ} = \frac{BC}{CE} = \frac{130}{x}$

⇒ $x = \frac{130}{\tan 30^{\circ}} = 225.17$ meters.

Therefore, the distance between the tower and the building is 225.17 meters. (Answer)