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ABCD is a trapezium such that AB and CD are parallel and BC\perpCD.

If \angleADB =0,BC= p and CD = q,then AB is equal to:

  • Option 1)

    \frac{\left ( p^{2} +q^{2}\right )\sin \Theta }{\left ( p\cos \Theta +q\sin \Theta \right )^{2}}

  • Option 2)

    \frac{\left ( p^{2} +q^{2}\right )\sin \Theta }{ p\cos \Theta +q\sin \Theta }

  • Option 3)

    \frac{ p^{2} +q^{2}\cos \Theta }{ p\cos \Theta +q\sin \Theta }

  • Option 4)

    \frac{ p^{2} +q^{2} }{ p^{2}\cos \Theta +q^{2}\sin \Theta }

 

Answers (1)

best_answer

As we learnt in 

Trigonometric Ratios of Functions -

\sin \Theta = \frac{Opp}{Hyp}

\cos \Theta = \frac{Base}{Hyp}

\tan \Theta = \frac{Opp}{Base}

- wherein

Trigonometric Ratios of Functions

 

In \bigtriangleup DAM

\tan \left ( \pi -\theta -\alpha \right ) =\frac{p}{x-q}

\tan (\theta + \alpha ) = \frac{p}{q-x}

x = q - p cot (\theta + \alpha )

= q - p \left [ \frac{\cot \theta \cot \alpha -1}{\cot 2 + \cot \theta } \right ]

=q-p\frac{(q\cos \theta - p\sin \theta )}{q\sin \theta+p\cos \theta }

Thus AB = \frac{\left ( p^{2}+q^{2} \right )\sin \theta }{p\cos \theta +q\sin \theta } 

 


Option 1)

\frac{\left ( p^{2} +q^{2}\right )\sin \Theta }{\left ( p\cos \Theta +q\sin \Theta \right )^{2}}

Correct

Option 2)

\frac{\left ( p^{2} +q^{2}\right )\sin \Theta }{ p\cos \Theta +q\sin \Theta }

Incorrect

Option 3)

\frac{ p^{2} +q^{2}\cos \Theta }{ p\cos \Theta +q\sin \Theta }

Incorrect

Option 4)

\frac{ p^{2} +q^{2} }{ p^{2}\cos \Theta +q^{2}\sin \Theta }

Incorrect

Posted by

divya.saini

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