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Question

The angle of elevation of a jet plane from a point on the ground is60°. After a flight of 20sec at the speed of 432km/hr, the angle of elevation changes to 30°. If the jet plane is flying at a constant height, then its height is:



A

12003m

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B

18003m

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C

36003m

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D

24003m

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Solution

The correct option is A

12003m


Given

The initial angle of elevation =60°

The changed angle of elevation =30°

Step 1 - Distance traveled PQ

Speed (v)=432km/hr

v=432×518m/sec

v=120m/sec

we know that,

distance traveled PQ=speed ×time

=120×20=2400m

Step 2- the jet plane is flying at a constant height

InPACtan60°=hAD3=hADAD=h3m

InAQDtan30°=hAC13=hACAC=3hm

Now,

AD=AC+CD3h=h3+2400(CD=PQ=2400m)3h-h3=24002h3=2400h=2400×32mh=12003m

Therefore, the jet plane is flying at a constant height of 12003m

Hence, Option A is correct.


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