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+3 votes
123k views
in Mathematics by (33.0k points)
closed by

The angle of elevation of an aeroplane from a point on the ground is 45° after flying for 15seconds, the elevation changes to 30° . If the aeroplane is flying at a height of 2500 meters, find the speed of the areoplane.

2 Answers

+1 vote
by (17.0k points)
selected by
 
Best answer

AA' and BB' denote the height of the plane from the ground.

∴ AA' = BB' = 2500 m

From △OAA'

tan 45° = \(\frac{AA'}{OA'}\) 

\(1=\frac{AA'}{OA'}\)

OA' = AA' = 2500 m

From △OBB'

tan 30° = \(\frac{BB'}{OB'}\) 

\(\frac 1{\sqrt 3}=\frac{BB'}{OB'}\)

OB' = √3 BB' = 2500√3 m

Distance covered by the plane from A to B is AB or A'B'

⇒ A'B' = OB' – OA'

= (2500√3 − 2500)

= 2500(√3 − 1)

=2500(1.73 − 1)

= 2500 × 0.73

= 1825 m

Time taken by the plane in moving from A to B = 15 seconds

Speed of the plane = \(\frac{1825}{15}\)

= 121.66 ms−1

+4 votes
by (60.2k points)

Let the height of flying of the aeroplane be PQ = BC and point A be the point of observation. 

We have,

So, the speed of the aeroplane is 122 m/s or 439.2 km / s .

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