In triangle \ABC, altitudes \AD and \BE are drawn to the corresponding bases. If \angle\BAC=45^\circ and \angle\ABC=0, then \frac{\AD}{\BE} equals

(\sin\\theta +\cos \\theta)/\sqrt2

CAT 2022 Slot 2 Paper

Question : 49 of 66

Marks: +1, -0

In triangle $\ABC$ , altitudes $\AD$ and $\BE$ are drawn to the corresponding bases. If $∠\BAC=45°$ and $∠\ABC=0$ , then ${\AD}/{\BE}$ equals

Solution:

It is given, Angle

{\BAE}=45

degrees
This implies

\A \E=\B \E

Let

{\AE}= {\BE}= x

In right-angled triangle

{\ABD}

, it is given

∠ \ABC=\θ

\sin \θ={\AD}/{\AB}

\sin \θ={\AD}/{x {√{2}}}

{√{2}} \sin \θ={\AD}/{\BE}

The answer is option D.

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