UPSC CDS 1 2023 Math Paper

Concept:Using tangent ratios in two right triangles formed by the line of sight to the top and base of the tower.Explanation:Let the woman’s eye be at point $A$, $h$ metres above water level.Let $D$ be the base of the tower at water level, and $C$ the top of the tower.Draw a horizontal line from $A$ to meet the tower at point $E$. So $AE$ is the horizontal distance to the tower.In triangle $AED$, angle of depression $=30^{\circ}$, opposite side $ED = h$, adjacent side $AE$.$\tan 30^{\circ} = \frac{h}{AE} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{AE} \Rightarrow AE = h\sqrt{3}$.In triangle $AEC$, angle of elevation $=60^{\circ}$, opposite side $CE$ (tower height above eye level), adjacent $AE = h\sqrt{3}$.$\tan 60^{\circ} = \frac{CE}{AE} \Rightarrow \sqrt{3} = \frac{CE}{h\sqrt{3}} \Rightarrow CE = 3h$.Total height of tower $= CE + ED = 3h + h = 4h$.Answer:$4h$