In the trapezium ABCD the sides AB and CD are parallel.The value of \frac{\sin \angleBAC}{\sin \angleBAD} is

Correct Answer is : AD–––AC–––

\stackrel{ –––}{ \AC}/ {\AD} ^{ –––}

\stackrel{ –––}{ \BC}/ {\AD} ^{ –––}

\stackrel{ –––}{ \AC}/ {\CD} ^{ –––}

\stackrel{ –––}{ \AB}/ {\CD} ^{ –––}

\stackrel{ –––}{ \AD}/ {\AC} ^{ –––}

Xavier Aptitude Test 2019 Paper

Question : 59 of 98

Marks: +1, -0

In the trapezium ABCD the sides AB and CD are parallel.The value of

{\sin ∠BAC}/{\sin ∠BAD}

Solution:

In right ΔAEC,

\sin \EAC=\CE/\AC

⇒\sin \BAC=\CE/\AC

In right ΔAFD,

\sin \FAD =\DF/\AD

⇒\sin \BAD =\DF/\AD

Now,

{\sin \BAC}/{\sin \BAD}={\CE/\AC}/{\DF/\AD}=\AD/\AC

[∵ \CE=\DF]

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