The angles of elevation of the top of a tower from two points X and Y (on opposite sides of the tower) at distances b² m and a² m, respectively, from the base and in the same straight line with it are complementary. The height (in m) of the tower is ________.

Correct option is D

Given:

Distances from points to tower base:

XB = $$b^2$$​, YB = $$a^2$$​​

Angles of elevation from points X and Y to top of tower A:

$$\angle$$​ AXB = $$\theta$$​, $$\angle$$​AYB = $$90^\circ - \theta$$​ (complementary)

Need to find: Height of the tower h = AB

Concept Used:

Also, if $$\theta$$​ and $$90^\circ - \theta$$​ are complementary, then:

$$\tan \theta \cdot \tan(90^\circ - \theta)$$​ = 1

​$$\tan(90^\circ - \theta) = \cot \theta$$​​

Solution: 

From both triangles formed (right-angled):

From point X:

​$$\tan \theta = \frac{h}{b^2}$$​​

From point Y:

​$$\tan(90^\circ - \theta) = \cot \theta = \frac{h}{a^2}$$​​

Multiply both:

​$$\frac{h}{b^2} \cdot \frac{h}{a^2} = 1 \\ \ \\\Rightarrow \frac{h^2}{a^2 b^2} = 1\\ \ \\\Rightarrow h^2 = a^2 b^2\\ \ \\\Rightarrow h = ab$$​​