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the angle of elevation of the top of top of a tower at a distance 150m from its foot on a horizontal plane is found to be 30 find the high

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the angle of elevation of the top of top of a tower at a distance 150m from its foot on a horizontal plane is found to be 30 find the high

Question

the angle of elevation of the top of top of a tower at a distance 150m from its foot on a horizontal plane is found to be 30 find the hight of the tower

## Answers ( )

Answer:Height of the tower is

51.96 m.Step-by-step explanation:Let AB be the tower and O be the point of observation.

Then OA = 90 cm, ∠OAB = 90°.

Let AB = h meters. Then from right Δ AOB, ∠AOB = 30°.

We have,

∴ The height of the tower is

51.96 m.Answer:50

Step-by-step explanation:The distance from the foot of the tower is 150m and the angle of elevation is 30 degrees.

= tan30 degree=Perpendicular/Base

= 1/root3=P/150

= 150/root3=P

Then, by rationalising the denominator the answer will be

= P=50root3

= So the height of the tower is 50root3

Hope it helps.