Question

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

Solution :-

Let AB be the tower. C and D be the two points with distance 4 m and 9 m from the base respectively. As per question,

In right ΔABC,

⇒tan x = AB/BC

⇒tan x = AB/4

⇒AB = 4 tan x … (i)

Again, from right ΔABD,

⇒tan (90°-x) = AB/BD

⇒cot x = AB/9

⇒AB = 9 cot x … (ii)

Multiplying equation (i) and (ii)

AB² = 9 cot x * 4 tan x

⇒ AB² = 36 (because cot x = 1/tan x)

⇒ AB = ± 6

Since height cannot be negative. Therefore, the height of the tower is 6 m.

Hence Proved !!