Examveda

A man on the top of a vertical observation tower observers a car moving at a uniform speed coming directly towards it. If it takes 8 minutes for the angle of depression to change from 30° to 45°, how soon after this will the car reach the observation tower?

A. 8 min 17 second

B. 10 min 57 second

C. 14 min 34 second

D. 12 min 23 second

Solution(By Examveda Team)

Consider the diagram shown above. Let AB be the tower. Let D and C be the positions of the car.
Then, ∠ ADC = 30° , ∠ ACB = 45°
Let AB = h, BC = x, CD = y
\eqalign{ & \tan {45^ \circ } = \frac{{AB}}{{BC}} = \frac{h}{x} \cr & \Rightarrow 1 = \frac{h}{x} \cr & \Rightarrow h = x\,......\left( 1 \right) \cr & \tan {30^ \circ } = \frac{{AB}}{{BD}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{AB}}{{\left( {BC + CD} \right)}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{h}{{x + y}} \cr & \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{h}{{x + y}} \cr & \Rightarrow x + y = \sqrt 3 \,h \cr & \Rightarrow y = \sqrt 3 \,h - x \cr}
$$\Rightarrow y = \sqrt 3 \,h - h$$     (∵ Substituted the value of x from equation 1 )
$$\Rightarrow y = h\left( {\sqrt 3 - 1} \right)$$
Given that distance y is covered in 8 minutes.
i.e, distance $$h\left( {\sqrt 3 - 1} \right)$$   is covered in 8 minutes.
Time to travel distance x
= Time to travel distance h (∵ Since x = h as per equation 1).
Let distance h is covered in t minutes.
since distance is proportional to the time when the speed is constant, we have
\eqalign{ & h\left( {\sqrt 3 - 1} \right) \propto 8\,......\left( {\text{A}} \right) \cr & h \propto t\,..............\left( {\text{B}} \right) \cr & \frac{{\left( {\text{A}} \right)}}{{\left( {\text{B}} \right)}} \Rightarrow \frac{{h\left( {\sqrt 3 - 1} \right)}}{h} = \frac{8}{t} \cr & \Rightarrow \left( {\sqrt 3 - 1} \right) = \frac{8}{t} \cr & \Rightarrow t = \frac{8}{{\left( {\sqrt 3 - 1} \right)}} \cr & \,\,\,\,\,\,\,\,\,\,\, = \frac{8}{{\left( {1.73 - 1} \right)}} \cr & \,\,\,\,\,\,\,\,\,\,\, = \frac{8}{{.73}} \cr & \,\,\,\,\,\,\,\,\,\,\, = \frac{{800}}{{73}}\,{\text{minutes}} \cr & \,\,\,\,\,\,\,\,\,\,\, = 10\frac{{70}}{{73}}\,{\text{minutes}} \cr & \approx {\text{10}}\,{\text{minutes}}\,{\text{57}}\,{\text{seconds}} \cr}

This Question Belongs to Arithmetic Ability >> Height And Distance

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