SSC CGL 1 Sep 2016 Paper

Question : 71 of 100

Marks: +1, -0

The angle of elevation of the top of a tower from two horizontal points (in opposite sides) at distances of 25 metre and 64 metre from the base of tower are x and 90° – x respectively. The height of the tower will be

39 metre
89 metre
1.6 metre
40 metre

Validate

Solution:

Let AB = Height of tower = h metre
BC = 25 metre
BD = 64 metre
∠ACB = x° and ÐADB = (90 – x)
In ΔABC,
tan x =

{AB}/{BC}

⇒ tan x =

h/{25}

In ΔABD
tan (90° - x) =

{AB}/{BD}

⇒ cot x =

h/{64}

∴ tan x . cot x =

h/{25} × h/{64}

⇒

h^2

= 25 × 64
∴ h =

√{25×64}

= 5 × 8
= 40 metre

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