Question

from a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. find the height of the tower. ​

Answer 1

SOLUTION:-

════════════

Given:

From a point on the ground, the angle of elevation of the bottom & the top of a transmission tower fixed at the top of a 20m high building are 45° & 60° respectively.

To find:

══════

The height of the tower.

Explanation:

═════════

Firstly, draw a figure attachment,

Assume AB & BC be the transmission tower & building & D be the point of observation.

We have,

BC= 20m & AB= h m

angle BDC= 45°

angle BDC= 45° angle ADC= 60°

In ∆BDC,

$\tan( \theta) = \frac{Perpendicular}{Base}$

$tan45 \degree = \frac{BC}{DC} \\ \\ 1 = \frac{20}{DC} \\ \\ DC = 20m$

&

In ∆ADC,

$tan60 \degree = \frac{AC}{DC} \\ \\ \sqrt{3} = \frac{h + 20}{20} \\ [Cross \:multiplication]\\ \\ 20 \times \sqrt{3} = h + 20 \\ \\ 20 \sqrt{3} = h + 20 \\ \\ h = 20 \sqrt{3} - 20 \\ [Taking \: common] \\ \\ h = 20( \sqrt{3} - 1) \\ \\ h = 20(1.732 - 1) \: \: \: \: \: \: \: ( \sqrt{3} = 1.732) \\ \\ h = 20 \times 0.732 \\ \\ h = 14.64m$

Thus,

The height of the tower is 14.64m

ʘ‿ʘ

Answer 2

here is your answer mate ✔✔