Question

From a point on the ground, the angles of elevation of bottom and the top of a transmission tower fixed

at the top of a 20m high building are 45°

and 60°

respectively. Find the height of the transmission tower. Use

√3 = 1.732​

Answer 1

Let DC be the tower and BC be the building. Then,

∠ CAB=45° ,∠DAB=60° , BC=20 m,

______________________________

Let height of the tower, DC=h m.

In right △ABC,

$tan45 = \frac{bc}{ab}$

$1 = \frac{20}{ab}$

Therefore,

$ab = 20m$

In right △ABD,

$tan60 = \frac{bd}{ab}$

$\sqrt{3} = \frac{h + 20}{20}$

h =20( √3−1) m.

Answer 2

Answer:

14.64 is answer to your question

Thanks

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