UPSC CDS I 2019 Math Paper

Concept:The plane's horizontal distance from point $P$ determines the angle of elevation. Equal angles occur at symmetric positions on the circular path.Explanation:Let the circle have center $C$ (above the airport) and radius $R$. Let $A$ be the point on the circle directly above $P$. Let $O$ be the point diametrically opposite to $A$. The plane completes one round in 3 minutes (180 seconds), so its angular speed is $\omega = \frac{360^{\circ}}{180\,\text{s}} = 2^{\circ}/\text{s}$. The angle of elevation from $P$ is equal at times $t$ and $t+30$. This implies the horizontal distances from $P$ to the plane are equal. Such equal distances occur when the plane's two positions are symmetric about either the radius through $A$ or the radius through $O$. Since the plane later flies vertically above $P$ at $A$, the symmetric pair must be about $O$ (otherwise the time to $A$ would be 15 seconds, which is not an option). Let the positions at $t$ and $t+30$ be $B$ and $C$, symmetric about $O$. The central angle between $B$ and $C$ is $\omega \times 30 = 60^{\circ}$. The midpoint $O$ is reached in $\frac{30}{2} = 15$ seconds from $B$. From $O$ to $A$, the plane covers a half-circle ($180^{\circ}$) in $\frac{180}{2} = 90$ seconds. Thus, total time from $B$ (at time $t$) to $A$ (at time $t+x$) is $15 + 90 = 105$ seconds. Therefore, $x = 105$ seconds.Answer:$x = 105$ seconds (Option C).