Question

the angle of elevation of the top a tower is observed to be 60 degree.At a point 40m directly above the first point of observation,the elevation is found to be 45 degree.find the height of the tower?

Answer 1

ANSWER:-

Given:

The angle of elevation of the top a tower is observed to be 60°. At a point 40m directly above the first of observation, the elevation to be 45°.

To find:

Find the height of the tower.

Solution:

Let the AB be the tower of height h m.

Let the two points be C & D such that, CD=40m,

⚫Angle ADE= 45°

⚫Angle ACB= 60°

Therefore,

In ∆ADE,

$= > tan45 \degree = \frac{AE}{DE} \\ \\ = > 1 = \frac{AE}{DE} \\ \\ = > DE = AE$

In ∆ACB,

$= > tan60 \degree = \frac{AB}{BC} \\ \\ = > \sqrt{3} = \frac{AB}{BC} \\ \\ = > AE + 40 = \sqrt{3} BC \: \: \: \: \: \: \: \: \: [AE = DE = BC] \\ \\ = > BC = \frac{40}{ \sqrt{3} - 1 } \\ \\ = > BC = \frac{40}{1.732 - 1} = \frac{40}{0.732} \\ \\ = > BC = 54.64m$

AB= 40m + 54.64m

AB= 94.64m

Hence,

The height of the tower is 94.64m.

Hope it helps ☺️