RRB Junior Engineer Stage 1 Model Papers 10

Question : 4 of 98

Marks: +1, -0

The angles of elevation of the top of a tower standing on a horizontal plane from two points on a line passing through the foot of the tower at a distance 9 ft and 16 ft respectively are complementary angles. Then the height of the tower is

9 ft
12 ft
16 ft
144 ft

Validate

Solution:

In ∆ABC

\tan α=h/9

....(1)
In ∆ABD

\tan β=h/16

α+β=90°

(given)

β=90-α

Since

\tan β=h/16

\tan(90-α)=h/16⇒\cot α=h/16 \or \tanα=16/h

....(2)

From eqn.(1) and (2)

h/9=16/h⇒h^2=16×9⇒h=12 feet

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