THE ANGLE OFELEVATION OF AN AEROPLANE FROM A POINT ON THE GROUND IS 60.AFTER FLYING 15SECONDS THE ANGLE OF ELEVATION CHANGES TO 30.IF THE AEROPLANE IS FLYING AT A CONSTANT HEIGHT OF 1500âˆš3 ,FIND THE SPEED OF THE AEROPLANE

From triangle ACD

$\\\\tan30=\frac{CD}{AD}\\\frac{1}{\sqrt{3}}=\frac{1500\sqrt{3}}{AD}\\AD=4500m$

From triangle ABE

$\\\\tan60=\frac{EB}{AE}\\\sqrt{3}=\frac{1500\sqrt{3}}{AE}\\AE=1500m$

The distance travelled in 15sec BC=ED=AD-AE=4500-1500=3000m

Speed=distance travelled/time=3000/15=200m/sec

Find the answer to a similar type of question in the following link

https://learn.careers360.com/school/question-the-angle-of-elevation-of-an-aeroplane-from-a-point-a-on-the-ground-is-after-a-flight-of-seconds-the-angle-of-elevation-changes-to-if-the-plane-is-flying-at-a-constant-height-of-metres-find-the-speed-of-the-aeroplane-103411/

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