CAT 2022 Slot 2 QA Question & Solution

GeometryMedium

Question

In triangle ABC, altitudes AD and BE are drawn to the corresponding bases. If $\angle BAC = 45^\circ$ and $\angle ABC = \theta$, then $\frac{AD}{BE}$ equals

Options

$\sqrt{2} \cos \theta$

$\frac{(\sin \theta + \cos \theta)}{\sqrt{2}}$

$\sqrt{2} \sin \theta$

Solution

It is given, angle BAE = 45 degrees

This implies AE = BE

Let AE = BE = x

In right-angled triangle ABD, it is given $\angle ABC = \theta$

$\sin\theta = \frac{AD}{AB}$

$\sin\theta = \frac{AD}{x\sqrt{2}}$

$\sqrt{2}\,\sin\theta = \frac{AD}{BE}$

The answer is option D.