The angles of elevation of the top of a tower from two points on the ground at distance 16 m and 25 m from its base are complementary to each other. The height of the tower is?

20 m

BD = 25 m

tan (90 - Θ) = \(\frac{AB}{BC}\) ⇒ AB = 16 cot Θ ..... (i)

tan Θ = \(\frac{AB}{BD}\) ⇒ AB = 25 tan Θ ..... (ii)

multiply (i) by (ii)

AB² = 16 cot Θ × 25 tan Θ

AB = \(\sqrt {16 × 25}\) = 4 × 5 = 20 m       (tanΘ cotΘ  = 1)

height of tower is 20 m