Question
Download Solution PDFThe angle of elevation from a point on a ground to the airplane is 60°. After 15 seconds angle of elevation of the plane become 30°. If the plane is flying at the height of 1500√3 meters then the speed of the plane is?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFAB = DE = 1500√3
IN Δ ABC
\(tan\;60^\circ = \frac{{AB}}{{BC}}\)
\(BC = \frac{{1500\sqrt 3 }}{{\sqrt 3 }} = 1500\)
In Δ DEC
\(tan\;30^\circ = \frac{{DE}}{{CE}}\)
\(\frac{1}{{\sqrt 3 }} = \frac{{1500\sqrt 3 }}{{{\rm{CE}}}}\)
CE = 1500 × 3 = 4500
BE = CE – BC
BE = 4500 – 1500
BE = 3000
Speed \( = \frac{{{\rm{distance}}}}{{{\rm{time}}}} = \frac{{3000}}{{15}} = 200\)
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