Question

an observer observes the top of a tower from a point P and found its angle of elevation 30 degree the tower is 200 metre high distance of point P is 200✓3If the observer moves towards the tower and from point q he observes the top of tower the angle of elevation 45 degree then the distance of tower is?​

Answer 1

Answer:

ANSWER

Let BD be the height of the tower and AE be the distance of the eye level of the observer from the ground level.

Draw EC parallel to AB such that AB=EC.

Given AB=EC=30

3

m and AE=BC=1.5m

In right angled △DEC,

tan30

∘

=

EC

CD

⇒CD=ECtan30

∘

=

3

30

3

∴CD=30m

Thus, the height of the tower, BD=BC+CD

=1.5+30=31.5m

Step-by-step explanation:

hope it helps