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Height and Distance
Height and Distance
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Question

ABAB is a vertical pole with BB at the ground level and AA at the top. AA man finds that the angle of elevation of the point AA from a certain point CC on the ground is 60∘{60^ \circ }. He moves away from the pole along the line BCBC to a point DD such that CD=7CD=7 m. From DD the angle of elevation of the point AA is 45∘{45^ \circ }. Then the height of the pole is :

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Solution

In ΔABC\Delta ABC hx=tan⁡60∘=3{h \over x} = \tan {60^ \circ } = \sqrt 3 ⇒x=h3 \Rightarrow x = {h \over {\sqrt 3 }} In ΔABDhx+7\Delta ABD{h \over {x + 7}} =tan⁡45∘=1 = \tan {45^ \circ } = 1 ⇒h=x+7⇒h−h3=7 \Rightarrow h = x + 7 \Rightarrow h - {h \over {\sqrt 3 }} = 7 ⇒h=733−1×3+13+1 \Rightarrow h = {{7\sqrt 3 } \over {\sqrt 3 - 1}} \times {{\sqrt 3 + 1} \over {\sqrt 3 + 1}} ⇒h=732(3+1 m) \Rightarrow h = {{7\sqrt 3 } \over 2}\left( {\sqrt 3 + 1\,m} \right)

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