Question

III LIIS point.
The angles of elevation of the top of a tower from two points at a distance of 4 m and
9 m from the base of the tower and in the same straight line with it are complementary,
Prove that the height of the tower is 6 m.​

Answer 1

Step-by-step explanation:

Let the height of the tower be x.

If angle from the point 4m away from the base is θ then angle at 9m away will be (90 - θ) since its complementary.

Thus,

Tan (90 - θ) = height of tower/distance of point from base.

Tan (90 - θ) = x/9; (i)

Now,

Tan (90 - θ) = cot θ; (ii)

Thus,

cot θ = distance of point from base / height of tower = 1/tan θ

cot θ = 4/x (since, θ is 4m away from the base) (iii)

From eq. (i), (ii) and (iii);

x/ 9 = 4/x

x^2 = 36;

x = 6m;

Height of tower will be same if we exchange the positon of θ and (90 - θ).

That's all.