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.
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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
tan 60=OP/OA. = root3=3000/OA
tan45= OQ/OA.
OA= 3000/ ROOT3
∴ Vertical distance between P and Q = PQ = OP – OQ
OQ= 3000/root3
= 1000 root3
therefore,,, 3000-1000root 3
=1000root3 ( root3-1) 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
tan 60=OP/OA. = root3=3000/OA
tan45= OQ/OA.
OA= 3000/ ROOT3
∴ Vertical distance between P and Q = PQ = OP – OQ
OQ= 3000/root3
= 1000 root3
therefore,,, 3000-1000root 3
=1000root3 ( root3-1) m
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