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Height and Distance
Height and Distance
Medium

Question

A man on the top of a vertical tower observes a car moving at a uniform speed towards the tower on a horizontal road. If it takes 18 min. for the angle of depression of the car to change from 30 o to 45 o ; then after this, the time taken (in min.) by the car to reach the foot of the tower, is :

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Solution

Assume height of tower = AB = h From Δ\Delta ABD, tan45 o = ABAD{{AB} \over {AD}} \Rightarrow $$$$\,\,\, ABAD=1{{AB} \over {AD}} = 1 [ as tan45 o = 1] ⇒\Rightarrow    \,\,\, AB = AD ∴   \therefore\,\,\, AD = h From Δ\Delta BAC, tan30 o = ABAC{{AB} \over {AC}} ⇒\Rightarrow    \,\,\, hAC=13{h \over {AC}} = {1 \over {\sqrt 3 }} ⇒\Rightarrow    \,\,\, AC = h 3\sqrt 3 As, AC = AD + DC \Rightarrow $$$$\,\,\, DC = AC −- AD = 3h\sqrt 3 h −- h Given that, time taken to reach from point C to D = 18 min. ∴   \therefore\,\,\, Car speed = distancetime{{distance} \over {time}} = CD18{{CD} \over {18}} = (3−1)h18{{(\sqrt 3 - 1)h} \over {18}} ∴   \therefore\,\,\, Time taken to move from D to A = Distance  of  DAspeed{{Distan ce\,\,of\,\,DA} \over {speed}} = h(3−1)h18{h \over {{{\left( {\sqrt 3 - 1} \right)h} \over {18}}}} = 18(3−1){{18} \over {\left( {\sqrt 3 - 1} \right)}} = 18(3+1)3−1{{18\left( {\sqrt 3 + 1} \right)} \over {3 - 1}} = 9 (3+1)\left( {\sqrt 3 + 1} \right) min.

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