Question

In the given figure, line RT is drawn parallel to SQ. If $\angle QPS={100}^o,\angle PQS={40}^o,\angle PSR={85}^o$ and $\angle QRS={70}^o$, then $\angle QRT$ is:

• A. ${45}^o$
• B. ${65}^o$
• C. ${85}^o$
• D. ${90}^o$

Answer 1

Answer: (B)

${65}^o$

From $\Delta$PSQ
$\angle PSQ={180}^o-({100}^o+{40}^o)={40}^o$
$\therefore \Delta QSR={85}^o-\angle PSQ$
$={85}^o-{40}^o={45}^o$
Since RT || SQ, therefore
$\angle TRX={45}^o$
Now, ${70}^o+x^o+{45}^o={180}^o$
$\therefore x^o={180}^o-{115}^o={65}^o$