CAT 2022 Slot 2 QA Question 22: In triangle ABC, altitudes AD and BE are drawn to the corresponding bases. If $\angle BAC = 45^\circ$ and $\angle ABC = heta$, then $\dfrac{AD}{BE}$ equals

Question

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

Accepted Answer

Using the area of triangle $ABC$ in two ways:

$\dfrac{1}{2} 	imes BC 	imes AD = \dfrac{1}{2} 	imes AC 	imes BE$
$\dfrac{AD}{BE} = \dfrac{AC}{BC}$

By the sine rule:

$\dfrac{AC}{BC} = \dfrac{\sin \angle ABC}{\sin \angle BAC}$
$\dfrac{AC}{BC} = \dfrac{\sin 	heta}{\sin 45^\circ}$
$= \dfrac{\sin

In triangle ABC, altitudes AD and BE are drawn to the corresponding bases. If ∠BAC=45∘\angle BAC = 45^\circ∠BAC=45∘ and ∠ABC=θ\angle ABC = \theta∠ABC=θ, then ADBE\dfrac{AD}{BE}BEAD​ equals

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