Question

Given $\Delta PQR$, in Figure, $mP={70}^o,\overline{QS}$ bisects $\angle Q$, and $\overline{RS}$ bisects $\angle R,m\angle S=$

• A. ${110}^o$
• B. ${115}^o$
• C. ${120}^o$
• D. ${125}^o$
• E. ${130}^o$

Answer 1

The correct option is D ${125}^o$

In triangle $PQR$,

$\angle$$P+\angle Q+\angle R={180}^o$

$\Rightarrow \angle Q+\angle R={180}^o-{70}^o={110}^o$
In triangle $QRS$,

$\angle$ $QSR+SQR+QRS={180}^o$

$\Rightarrow QSR={180}^o-(SQR+SRQ)$

$={180}^o-\frac{(\angle Q+\angle R)}{2}$

$=180-\frac{110}{2}$

$=180-55$

$={125}^o$