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.

Solution:
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Answer 1

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.

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Answer 2

Answer:

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.

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