CAT 2023 Slot 3 · Quant · Question 17

Let the equal base angles at B and C be \beta, so the apex angle at A is {180}^{∘}-2\beta.

Since the triangle is isosceles, the altitude AD also bisects the apex angle, so \angle OAB={90}^{∘}-\beta.

In the right triangle ABE, the angle at E is {90}^{∘} and the angle at A is the apex angle, so \angle ABE={90}^{∘}-\left({180}^{∘}-2\beta \right)=2\beta -{90}^{∘}.

In triangle AOB the three angles sum to {180}^{∘}, so \angle AOB={180}^{∘}-\left({90}^{∘}-\beta \right)-\left( 2\beta -{90}^{∘}\right).

Simplifying, \angle AOB={180}^{∘}-\beta.

Setting this equal to {105}^{∘} gives \beta ={75}^{∘}, so the apex angle is {180}^{∘}-{150}^{∘}={30}^{∘}.

Now take AB=AC=1 to compare the two altitudes.

In the right triangle ABD, AD=AB\sin \beta =\sin{75}^{∘}.

In the right triangle ABE, BE=AB\sin A=\sin{30}^{∘}=\frac{1}{2}.

So \frac{AD}{BE}=\frac{\sin{75}^{∘}}{\frac{1}{2}}=2\sin{75}^{∘}.

Since \sin{75}^{∘}=\cos{15}^{∘}, this is 2\cos{15}^{∘} — the third option.

Shortcut. Both altitudes drop from the same side AB, so each is AB times the sine of the angle it faces. The ratio is therefore just \frac{\sin \beta}{\sin A}, and the angle chase reduces to the single fact \angle AOB={180}^{∘}-\beta.