UPSC CDS 1 2021 Math Paper

Concept:The condition that the two halves subtend equal angles at the point on the ground implies that the line joining the midpoint to the point is the angle bisector of the angle formed by the ends of the pole at that point.Explanation:Let the pole be $AB$ with $A$ at the top and $B$ at the base.Let $D$ be the midpoint of $AB$ so that $AD = DB$.Let $P$ be the point on the ground at a distance $x$ from $B$.The two halves $AD$ and $DB$ subtend equal angles at $P$, i.e., $\angle APD = \angle DPB$.Therefore, $PD$ is the internal bisector of $\angle APB$ in $\triangle APB$.By the Angle Bisector Theorem, $\frac{AD}{DB} = \frac{PA}{PB}$.Since $AD = DB$, we have $PA = PB$.Thus $PA = PB = x$ (because $PB = x$).Now, the pole leans at $60^{\circ}$ with the vertical, so it makes $30^{\circ}$ with the horizontal ground.Take $B$ at $(0,0)$ and the pole lying in the vertical plane. Then $A$ is at $(L\cos30^{\circ},\; L\sin30^{\circ}) = \left(\frac{L\sqrt{3}}{2},\; \frac{L}{2}\right)$, where $L$ is the length of the pole.Point $P$ is at $(x,0)$.The condition $PA = x$ gives:$\sqrt{ \left(\frac{L\sqrt{3}}{2} - x\right)^2 + \left(\frac{L}{2}\right)^2 } = x$.Squaring and simplifying:$\left(\frac{L\sqrt{3}}{2} - x\right)^2 + \frac{L^2}{4} = x^2$.Expanding: $\frac{3L^2}{4} - L\sqrt{3}\,x + x^2 + \frac{L^2}{4} = x^2$.Cancel $x^2$ and combine $L^2$ terms: $L^2 - L\sqrt{3}\,x = 0$.Factor: $L(L - \sqrt{3}\,x) = 0$, so $L = \sqrt{3}\,x$ (since $L>0$).Hence the length of the pole is $\sqrt{3}\,x$ metres.Answer:$\sqrt{3}\,x$ metre (Option B).