Question

The angle of elevation of the top of a tower as observed from 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:

• Let the Height of tower beABin△ABCtan6 =BCAB

⇒ 1.96AB =BC...........(1)

In △ABD

tan32 = BDAB

⇒BD= 0.62AB ............(2)

and it is given that CD=100 m=BD−BC

From (1) and (2)

⇒ 0.62AB − 1.96AB

=100

⇒2.12AB

=100

⇒AB= 2.12100

=47.16 m

and

BC= 1.96AB

= 1.9647.16

=24.06 m