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.

The distance the inspector has to travel is:

• A. 3 km
• B. 3.75 km
• C. 5 km
• D. 6 km

Hint:

Once the time taken to catch the thief is known, the distance is simply the inspector's speed multiplied by that time.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Picture the triangle formed by the inspector's start point $I$, the thief's start point $P$, and the meeting point $M$. Side $IP=7$ km is known, and the angle at $P$ between $PI$ and the thief's path $PM$ is $45^{\circ}$, since the thief moves at $45^{\circ}$ to the road. The Sine Rule can give the side $IM$, which is exactly the distance the inspector travels.

Step 2: Key Formula or Approach:
$\dfrac{IM}{\sin P}=\dfrac{PM}{\sin I}=\dfrac{IP}{\sin M}$, where $PM=9\sqrt2\,T$ and $IM=15T$ for the meeting time $T$.

Step 3: Detailed Explanation:
From $\dfrac{PM}{\sin I}=\dfrac{IM}{\sin P}$: $\sin I=\dfrac{9\sqrt2 T\sin45^{\circ}}{15T}=\dfrac{9}{15}=0.6$, so $\cos I=0.8$ (a $3$-$4$-$5$ triangle).
Angle $M=180^{\circ}-45^{\circ}-I$, so $\sin M=\sin(45^{\circ}+I)=\sin45^{\circ}\cos I+\cos45^{\circ}\sin I=\dfrac{1}{\sqrt2}(0.8+0.6)\approx0.99$.
Using $\dfrac{IP}{\sin M}=\dfrac{IM}{\sin P}$:
\[ IM = \dfrac{IP\times \sin P}{\sin M} = \dfrac{7\times0.7071}{0.99}\approx5 \text{ km} \]

Step 4: Final Answer:
The inspector travels about $5$ km to catch the thief, the same answer as before, confirming option C.