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Question

The angle of elevation of a jet plane from a point A on the ground is 60∘. After a flight of 30 seconds, the angle of elevation changes to 30∘. If the jet plane is flying at a constant height of 3600√3 metres, find the speed of the jet plane.

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Solution

Let A be the point of observation and let AX be a horizontal line through A and QC⊥AX. Let P and Q be the two positions of the plane. Let PB⊥AX. Then, PB=QC=3600√3 metres, ∠BAP=60∘ and ∠BAQ=30∘ From right ΔABP, we have ABBP=cot 60∘=1√3⇒AB3600√3=1√3 ⇒AB=3600√3×1√3m=3600 m ...... (i) Let BC=PQ=x metres Then, AC=AB+BC=(x+3600) m [using (i)] From right ΔACQ, we have ACCQ=cot 30∘=√3⇒x+36003600√3=√3 ⇒x+3600=(3600×3)=10800 ⇒x=10800−3600=7200 Thus, PQ = 7200 metres Now, 7200 m is covered in 30 seconds. ∴ speed of the jet plane =(720030×60×601000)km/hr =864 km/hr. Hence, the speed of the jet plane is 864 km/hr.

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