Question

In the given figure, if $TS||QV,SV||TQ,\angle QRS={115}^{\circ }$, then $\angle TQR$ is equal to

• A. ${65}^{\circ }$
• B. ${115}^{\circ }$
• C. ${45}^{\circ }$
• D. ${25}^{\circ }$

Answer 1

The correct option is D ${25}^{\circ }$

Given, A parallelogram $QTSV,\angle QRS={115}^{\circ }$

From parallelogram

$\angle S+\angle V={180}^{\circ }$

Because $TS\parallel QV$

So, ${90}^{\circ }+\angle V=180$

$\angle V={90}^{\circ }$

similarity $\angle TQV+\angle V={180}^{\circ }$

so, $\angle TQV={90}^{\circ }$

$\angle TRQ=180-\angle QRS=180-115={65}^{\circ }$

Now we know that alternate angle will be equal as $RQ$ is transversal to $TS\parallel QV$

so, $\angle TRQ=\angle RQV={65}^{\circ }$

But $\angle TQV=90\Rightarrow \angle RQV+\angle TQR={90}^{\circ }$

$\angle TQR=90-65$

$\angle TQR={25}^{\circ }$ option$-$D