question_answer

A man observes a car from the top of a tower, which is moving towards the tower with a uniform speed. If the angle of depression of the car changes from \[30{}^\circ \] to \[45{}^\circ \] in 12 minutes, find the time taken by the car now to reach the tower.

Answer:

Let AB is a tower, car is at point D at \[30{}^\circ \] and goes to C at \[45{}^\circ \] in 12 minutes.

In \[\Delta \,ABC\],

\[\frac{AB}{BC}=\tan \,\,45{}^\circ \]

\[\Rightarrow \]               \[\frac{h}{x}=1\Rightarrow h=x\]                                                            ?(i)

In \[\Delta \,ABD,\]

\[\frac{AB}{BD}=\tan \,\,30{}^\circ \]

\[\Rightarrow \]               \[\frac{h}{x+y}=\frac{1}{\sqrt{3}}\Rightarrow h=\frac{x+y}{\sqrt{3}}\]                                      ?(ii)

Comparing eq. (i) & (ii), we get

\[x=\frac{x+y}{\sqrt{3}}\Rightarrow \sqrt{3}x=x+y\]

\[\Rightarrow \]               \[\left( \sqrt{3}-1 \right)x=y\]

Car covers the distance y in time \[\text{=12 min}\text{.}\]

So \[\left( \sqrt{3}-1 \right)x\] distance covers in 12 min

Distance x covers in time \[=\frac{12}{\sqrt{3}-1}\times \frac{\sqrt{3}+1}{\sqrt{3}+1}\]

\[=\frac{12\left( \sqrt{3}+1 \right)}{3-1=2}\]

\[=6\left( \sqrt{3}+1 \right)\,\min \]

\[=6\times 2.732=16.39\]

Now car reaches to tower in 16.39 minutes.