A man on the top of a vertical tower observes a car moving at a uniform speed towards the …

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 30o to 45o ; then after this, the time taken (in min.) by the car to reach the foot of the tower, is :

Accepted Answer

Assume height of tower = AB = h

From $$\Delta $$ABD,

tan45o = $${{AB} \over {AD}}$$

$$ \Rightarrow $$ $\,\,\,$${{AB} \over {AD}} = 1$$  [ as    tan45o = 1]

$$ \Rightarrow $$\,\,\,$AB = AD$\therefore\,\,\,$AD = h

From$\Delta $BAC,

tan30o =${{AB} \over {AC}}$$ \Rightarrow $$\,\,\,$${h \over {AC}} = {1 \over {\sqrt 3 }}$$ \Rightarrow $$\,\,\,$AC = h$\sqrt 3 $As,    AC = AD + DC$ \Rightarrow $$ $\,\,\,$$ DC = AC - AD = $$\sqrt 3 h$$ - h

Given that, time taken to reach from point C to D = 18 min.

$$\therefore\,\,\,$$ Car speed = $${{distance} \over {time}}$$ = $${{CD} \over {18}}$$ = $${{(\sqrt 3  - 1)h} \over {18}}$$

$$\therefore\,\,\,$$ Time taken to move from D to A

= $${{Distan ce\,\,of\,\,DA} \over {speed}}$$

= $${h \over {{{\left( {\sqrt 3  - 1} \right)h} \over {18}}}}$$

= $${{18} \over {\left( {\sqrt 3  - 1} \right)}}$$

= $${{18\left( {\sqrt 3  + 1} \right)} \over {3 - 1}}$$

= 9 $$\left( {\sqrt 3  + 1} \right)$$ min.

Solutions · 1