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.

Let AQ is tower and C and D are the two point 4 m and 9 m away from the foot of tower.
The angle of elevations are complementary. So if one angle is θ, then the other will be 90- θ.
In ΔAQR,
AQ/QR = tan θ
⇒ AQ/4 = tan θ …(1)
In ΔAQS,
AQ/QS = tan (90 – θ)
⇒ AQ/9 = cot θ …(2)
Multiplying the equations (1) and (2), we have
(AQ/4)(AQ/9) = tan θ. cot θ
⇒ AQ²/36 = 1
⇒ AQ² = 36
⇒ AQ = ±6
Height can’t be negative. Hence, the height of the tower is 6 m.