Question

The angle of elevation of the top of a tower as observed form a point in a horizontal plane through the foot of the tower is 32°. When the observer moves towards the tower a distance of 100 m, he finds the angle of elevation of the top to be 63°. Find the height of the tower and the distance of the first position from the tower. [Take tan 32° = 0.6248 and tan 63° = 1.9626]

Answer 1

Step-by-step explanation:

which is the required ans.

Answer 2

$\huge\bigstar\mathfrak\green{\underline{\underline{SOLUTION:}}}$

Let AB be the tower of height h metres.

Let P be the first point of observation and Q be the second point of observation.

Let ∠AQB= 32° & ∠APB=63°

Now,

PQ= 100m, Let BP=x m

In ∆ABP,

$tan63 \degree = \frac{</strong><strong>AB</strong><strong>}{</strong><strong>BP</strong><strong>} \\ \\ = > 1.9626 = \frac{h}{x} \\ \\ = > h = 1.9626x .............(1)$

In ∆ABQ,

$tan32 \degree = \frac{</strong><strong>AB</strong><strong>}{</strong><strong>BQ</strong><strong>} \\ \\ = > 0.6248 = \frac{h}{x + 100} \\ \\ = > h = 0.248x + 62.48...........(2)$

Putting the value of h from(1) in(2), we get

=) 1.9626x= 0.6248x+62.48

=) 1.3378x= 62.48

=)x= 62.48/1.3378

=) x= 46.7

So, distance of the first point from the tower = 46.7m.

Now, from (1), we get

h= (1.9626 × 46.7)m

=) h= 91.7m

So,height of tower= 91.7m

hope it helps ☺️