Question

A police inspector spots a thief standing 7 km away from him on a straight road that runs East-West. The inspector is standing on the eastern side while the thief is on the western side of the road. On spotting the inspector, the thief takes his bicycle and tries to cut across the field next to the road, riding away at a uniform speed of \(9\sqrt{2}\) km/hour in a direction making an angle of \(45^{\circ}\) with the road towards North-East. The inspector starts on his scooter at the same instant, moving at a uniform speed of \(15\) km/hour, and catches the thief.

Time taken by the inspector to catch the thief is:

• A. 12 minutes
• B. 15 minutes
• C. 18 minutes
• D. 20 minutes

Hint:

Split the thief's velocity into East and North parts, then find the time at which the straight-line distance from the inspector's start point equals the distance the inspector himself covers.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Instead of coordinates, picture a triangle formed by the inspector's start point $I$, the thief's start point $P$, and the meeting point $M$. We know one full side of this triangle, $IP=7$ km, and one angle, so the Sine Rule can give us the rest.

Step 2: Key Formula or Approach:
Since the thief moves at $45^{\circ}$ to the road (the road being the line $IP$), the angle of the triangle at $P$, between $PI$ and the thief's path $PM$, is exactly $45^{\circ}$. The Sine Rule gives $\dfrac{IM}{\sin P}=\dfrac{PM}{\sin I}=\dfrac{IP}{\sin M}$.

Step 3: Detailed Explanation:
In time $T$, the thief covers $PM=9\sqrt{2}T$ and the inspector covers $IM=15T$.
From $\dfrac{PM}{\sin I}=\dfrac{IM}{\sin P}$:
\[ \sin I = \dfrac{PM\sin P}{IM}=\dfrac{9\sqrt{2}T\times \sin 45^{\circ}}{15T}=\dfrac{9\sqrt{2}\times \tfrac{1}{\sqrt2}}{15}=\dfrac{9}{15}=0.6 \]
So angle $I$ is the angle of a $3$-$4$-$5$ right triangle: $\sin I=0.6$, $\cos I=0.8$.
The third angle is $M=180^{\circ}-45^{\circ}-I$, so $\sin M=\sin(45^{\circ}+I)=\sin45^{\circ}\cos I+\cos45^{\circ}\sin I=\dfrac{1}{\sqrt2}(0.8+0.6)=\dfrac{1.4}{\sqrt2}\approx0.99$.
Now using $\dfrac{IP}{\sin M}=\dfrac{IM}{\sin P}$:
\[ 15T = \dfrac{IP\times\sin P}{\sin M} = \dfrac{7\times0.7071}{0.99}\approx5 \]
So $T=\dfrac{5}{15}=\dfrac{1}{3}$ hour.

Step 4: Final Answer:
$\dfrac{1}{3}$ hour is $20$ minutes, the same answer as before, confirming option D.