Question

Sujana observes an aeroplane flying at a, height of 1.5 km at an angle of elevation of 30°. After 20 seconds, the plane moves away from her and makes an angle of elevation of 15° from the same height. The speed of the plane (in kmph) is 500 540 490 510​

Answer 1

Answer:

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Step-by-step explanation:

Let P and Q be the two position of the plane.

Given angles of elevation of the plane in two position P and Q as ∠PAB=60

0

and ∠QAC=30

0

PB=QC= 15 km

Now in right angled △ABP

tan60

0

=

AB

BP

⇒

3

=

AB

1.5

⇒AB=

3

1.5

⇒(0.5)

3

km

Again in right angled triangle △ACQ,

tan30

0

=

AC

QC

⇒

3

1

=

AB

1.5

⇒AC=(1.5)

3

km

PQ=BC=AC−AB

=(1.5)

3

−(0.5)

3

=(1.5−0.5)

3

=

3

km

The plane travels PQ=

3

km in 15 secs

∴ Speed of the aeroplane =

time

distance

⇒

3600

15

3

⇒240

3

km/h