IPMAT Indore 2024 · Quantitative Ability · Question 16

Step 1 — set up. Let the pole be PQ with foot Q and height h. Let B be the nearer point, A the farther one, with AB=20.

\tan{30}^{∘}=\frac{h}{AQ}\Rightarrow AQ=h\sqrt{3}, \tan{45}^{∘}=\frac{h}{BQ}\Rightarrow BQ=h

Step 2 — use the 20 m walk.

AQ-BQ=20 \Rightarrow h\sqrt{3}-h=20 \Rightarrow h\left( \sqrt{3}-1 \right)=20

Step 3 — rationalise.

h=\frac{20}{\sqrt{3}-1}\times \frac{\sqrt{3}+1}{\sqrt{3}+1}=\frac{20\left( \sqrt{3}+1 \right)}{3-1}=10\left( \sqrt{3}+1 \right)

Numerically h\approx 10\left( 2.732 \right)\approx 27.3 m — comfortably more than the 20 m walked, as it must be.

Answer: Option 4 — 10\left( \sqrt{3}+1 \right)

Shortcut. The two-angle formula. When the elevation goes from {\theta}_{1} to {\theta}_{2} over a walk of d,

h=\frac{d}{\cot{\theta}_{1}-\cot{\theta}_{2}}

Here \frac{20}{\sqrt{3}-1}=10\left( \sqrt{3}+1 \right). Even faster: a {45}^{∘} elevation means the remaining distance equals h, so the picture reduces to one equation on sight. Option (c) is the trap for anyone who forgets to rationalise and halve.