SSC CGL 19 Oct 2014 Shift 2 Paper

Question : 130 of 200

Marks: +1, -0

The angle of elevation of the top of a vertical tower situated perpendicularly on a plane is observed as 60° from a point P on the same plane. From another point Q, 10m vertically above the point P, the angle of depression of the foot of the tower is 30°. The height of the tower is

15 m
30 m
20 m
25 m

Validate

Solution:

AB = Tower = h metre
PQ = 10 metre
∠ APB = 60°,
∠ CQB = ∠ QBP = 30°
In Δ PBQ,
tan 30° =

{PQ}/{PB}

⇒

1/√3

{10}/{PB}

⇒ PB = 10

√3

metre
In Δ APB
tan 60° =

{AB}/{PB}

⇒

√3

h/{10√3}

⇒ h =

√3 × 10√3

= 30 metre

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