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# A man on the top of a vertical observation tower observes a car moving at a uniform speed coming directly towards it. If it takes 12 minutes for the angle of depression to change from 30° to 45°, how soon after this will the car reach the observation tower ?

A. 14 min. 35 sec.

B. 15 min. 49 sec.

C. 16 min. 23 sec.

D. 18 min. 5 sec.

### Solution(By Examveda Team)

Let AB be the tower and C and D be the two positions of the car.
Then, $\angle ACB = {45^ \circ },$    $\angle ADB = {30^ \circ }$
Let, AB = h, CD = x and AC = y
\eqalign{ & \frac{{AB}}{{AC}} = \tan {45^ \circ } = 1 \cr & \Rightarrow \frac{h}{y} = 1 \cr & \Rightarrow y = h \cr & \frac{{AB}}{{AD}} = \tan {30^ \circ } = \frac{1}{{\sqrt 3 }} \cr & \Rightarrow \frac{h}{{x + y}} = \frac{1}{{\sqrt 3 }} \cr & \Rightarrow x + y = \sqrt 3 h \cr & \therefore x = \left( {x + y - y} \right) \cr & \,\,\,\,\,\,\, = \sqrt 3 h - h \cr & \,\,\,\,\,\,\, = h\left( {\sqrt 3 - 1} \right) \cr}
Now, $$h\left( {\sqrt 3 - 1} \right)$$   is covered in 12 min.
So, h will be covered in→
\eqalign{ & \left[ {\frac{{12}}{{h\left( {\sqrt 3 - 1} \right)}} \times h} \right] \cr & = \frac{{12}}{{\left( {\sqrt 3 - 1} \right)}}\min \cr & = \left( {\frac{{1200}}{{73}}} \right)\min \cr & = 16\min ,\,\,23\sec \cr}

This Question Belongs to Arithmetic Ability >> Height And Distance

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