CAT 2022 Slot 2 · Quant · Question 1

Both altitudes can be measured against the same side AB, which is what makes them comparable.

The altitude AD runs from A perpendicular to BC, so triangle ABD has a right angle at D.

In that right triangle the angle at B is \angle ABC=\theta, and AB is the hypotenuse.

So AD=AB\sin \theta.

The altitude BE runs from B perpendicular to AC, so triangle ABE has a right angle at E.

In that right triangle the angle at A is \angle BAC={45}^{∘}, and AB is again the hypotenuse.

So BE=AB\sin{45}^{∘}=\frac{AB}{\sqrt{2}}.

Dividing, \frac{AD}{BE}=\frac{AB\sin \theta}{\frac{AB}{\sqrt{2}}}.

The common factor AB cancels, leaving \sqrt{2}\sin \theta — the first option.

Shortcut. Each altitude equals AB times the sine of the angle it faces, so the ratio is just \frac{\sin \theta}{\sin{45}^{∘}} and the side length never matters. Option 4 is also ruled out at a glance, since the ratio must vary with \theta.