cat 2022 Question 1

Question

In triangle ABC, altitudes AD and BE are drawn to the corresponding bases. If $\angle B A C=45^{\circ}$ and ∠ABC=θ , then $\frac{A D}{B E}$ equals

Accepted Answer

It is given, Angle BAE = 45 degrees

This implies AE = BE

Let AE = BE = x

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

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

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

$\sqrt{2}\sin heta = \frac{AD}{BE}$

Question 1.