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Height and Distance
Height and Distance
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Question

If the angles of elevation of the top of a tower from three collinear points A,BA, B and C,C, on a line leading to the foot of the tower, are 30∘{30^ \circ }, 45∘{45^ \circ } and 60∘{60^ \circ } respectively, then the ratio, AB:BC,AB:BC, is :

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Solution

As PBPB bisects ∠APC,\angle APC, therefore ABAB :: BCBC =PA:PC=PA:PC Also in ΔAPQ,sin⁡30∘=hPA⇒PA=2h\Delta APQ,\sin {30^ \circ } = {h \over {PA}} \Rightarrow PA = 2h and in ΔCPQ,\Delta CPQ, sin⁡60∘=hPC⇒PC=2h3\sin {60^ \circ } = {h \over {PC}} \Rightarrow PC = {{2h} \over {\sqrt 3 }} ∴\therefore AB:BC=2h:2h3=3:1AB:BC = 2h:{{2h} \over {\sqrt 3 }} = \sqrt 3 :1

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