Question

12. In the adjoining figure, the angle of elevation from a point
P of the top of a tower QR, 50 m high is 60° and that of
the tower PT from a point Q is 30°. Find the height of
the tower PT, correct to the nearest metre. (2018)​

Answer 1

Given:

The height of the tower QR is 50 m.

To find:

Height of the tower PT.

Solution:

1) As the figure is not given in the question so it is given in the attached picture. Refer that.

2) In ΔPQR

tan60°= RQ/QP

• tan60°= 50/x
• √3 = 50/x
• x = 50/√3

3) In ΔTPQ

tan30°=PT/PQ

• 1/√3 = y/x
• 1/√3 = y/50/√3
• y = 50/3 = 16.66

The height of the tower PT is 16.66 m (Approx. 17m)

Answer 2

Answer:

tan 60 = 50/x

√3 = 50/x

50/√3= x

now

tan 30 = y/x

1/√3=y/50/√3

50 / √3 × 1/ √3

== 50 / 3

= 16.67 or 17 m