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Height and Distance
Height and Distance
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Question

Let a vertical tower ABA B of height 2h2 h stands on a horizontal ground. Let from a point PP% on the ground a man can see upto height hh of the tower with an angle of elevation 2α2 \alpha. When from PP, he moves a distance dd in the direction of AP→\overrightarrow{A P}, he can see the top BB of the tower with an angle of elevation α\alpha. If d=7hd=\sqrt{7} h, then tan⁡α\tan \alpha is equal to

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Solution

ΔAPM\Delta{APM} gives tan⁡2α=hx\tan 2\alpha = {h \over x} ..... (i) ΔAQB\Delta{AQB} gives tan⁡2α=2hx+d=2hx+h7\tan 2\alpha = {{2h} \over {x + d}} = {{2h} \over {x + h\sqrt 7 }} ...... (ii) From (i) and (ii) tan⁡2α=2 . tan⁡2α1+7 . tan⁡2α\tan 2\alpha = {{2\,.\,\tan 2\alpha } \over {1 + \sqrt 7 \,.\,\tan 2\alpha }} Let t=tan⁡αt = \tan \alpha ⇒t=22t1−t21+7 . 2t1−t2 \Rightarrow t = {{2{{2t} \over {1 - {t^2}}}} \over {1 + \sqrt 7 \,.\,{{2t} \over {1 - {t^2}}}}} ⇒t2−27t+3=0 \Rightarrow {t^2} - 2\sqrt 7 t + 3 = 0 t=7−2t = \sqrt 7 - 2

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