SSC CPO SI and ASI Model Paper 9

Question : 149 of 200

Marks: +1, -0

An insect at point P makes an angle of elevation of 30° to the top of tower on the same plane. It moves towards the tower until the angle of elevation becomes 45. If the height of the tower is

100/3 m

, then through how much distance did the insect move?

$100(3 -√3)m$
${100(3-√ 3)}/ 3 m$
$100( √3 -3 )m$
${100 (√3 -3)}/3 m$

Validate

Solution:

(2) Let the distance that the insectmoved be x m.

Now,

tan30°={AB}/{PB }⇒1/√3

=100/√{3 PB} ⇒PB=100m

Also,
tan45°={AB}/{QB }=

100 √3 /{QB}⇒ QB100/√3 m

Distance moved by insect =

=X=PB-QB=100- 100/√3

=100×{√3 -1}/√3= {100 (3-√ 3)}/3 100 m

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