An aero plane, when 3000m high, passes vertically above another plane at an instant when the angles of elevation of the two aero planes from the same point on the ground are 600 and 450 respectively. Find the vertical distance between the two aero planes.

Let P and Q be the position of two Aeroplane. When Q is vertically below P and OP = 5000 m

 

Given: the angle of elevation of P and Q at a point on ground (say A) are 60° and 45°

∴∠PAO = 60° and QAO = 45°

 

Now in ∆AOP and AOQ we have

 

Vertical distance between P and Q = PQ = OP – OQ

You can solve further by putting the value of according to the question.

  • 10

tan 60=3000/x

root 3=3000/x

x=3000root3/3 =1000root3

now tan 45=height of 2 aeorpkllane/base

    which wioll be equalso the distance between them will be 3000-1000root 3

hope this to be right

  • 0
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