RRB NTPC Mains or Stage 2 Model Paper 3 with solutions

Question : 1 of 120

Marks: +1, -0

P and Q are two points observed from the top of a building

10√ 3

m high. I f the angles of depression of the points are complementary and PQ = 20 m, then the distance of P from the building is

Solution:

tan θ=10√3/x+20

tan (90 – θ) ={10√ 3}/x

cot θ ={10 √3}/x

So,

{10 √3}/{x + 20}=x/{10 √3}

x^2 + 20x – 300 = 0

x = 10 or x = – 3 (x can not be negative)
∴ x = 10 meter.

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