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 ?

Answer: Option C

Solution

Answer: Option C

Solution:

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
$$\begin{array}{rl}& \frac{AB}{AC}=\tan 45^{\circ }=1\\ & \Rightarrow \frac{h}{y}=1\\ & \Rightarrow y=h\\ & \frac{AB}{AD}=\tan 30^{\circ }=\frac{1}{\sqrt{3}}\\ & \Rightarrow \frac{h}{x+y}=\frac{1}{\sqrt{3}}\\ & \Rightarrow x+y=\sqrt{3}h\\ & \therefore x=\left(x+y-y\right)\\ & =\sqrt{3}h-h\\ & =h\left(\sqrt{3}-1\right)\end{array}$$
Now, $$h\left(\sqrt{3}-1\right)$$   is covered in 12 min.
So, h will be covered in→
$$\begin{array}{rl}& \left[\frac{12}{h\left(\sqrt{3}-1\right)}\times h\right]\\ & =\frac{12}{\left(\sqrt{3}-1\right)}\min \\ & =\left(\frac{1200}{73}\right)\min \\ & =16\min ,23\sec \end{array}$$