Question
In the given figure, $\triangle\text{PQR}$ is an isosceles triangle with $PQ = PR$ and $\text{m}\angle\text{PQR}=35^\circ.$ Find $\text{m}\angle\text{QSR}$ and $\text{m}\angle\text{QTR}.$ 
✓
Answer
We have, $\angle\text{PQR}=35^\circ$ Since, $\triangle\text{PQR}$ is an isosceles triangle with $PQ = PR$.
Then, $\angle\text{PQR}=\angle\text{PRQ}=35^\circ$ In
$\triangle\text{PQR},$ by angle sum property $\angle\text{p}+\angle\text{PQR}+\angle\text{PRQ}=180^\circ$
$\Rightarrow\angle\text{p}+35^\circ+35^\circ=180^\circ$
$\Rightarrow\angle\text{p}=180^\circ-35^\circ-35^\circ=110^\circ$
$\therefore\angle\text{QSR}=\angle\text{p}=110^\circ$ [Angles in same segment]
Now, $\angle\text{QSR}+\angle\text{QTR}=180^\circ$ [Opposite angles of cyclic quad.]
$\Rightarrow110^\circ+\angle\text{QTR}=180^\circ$
$\Rightarrow\angle\text{QTR}=180^\circ-110^\circ=70^\circ$
Then, $\angle\text{PQR}=\angle\text{PRQ}=35^\circ$ In
$\triangle\text{PQR},$ by angle sum property $\angle\text{p}+\angle\text{PQR}+\angle\text{PRQ}=180^\circ$
$\Rightarrow\angle\text{p}+35^\circ+35^\circ=180^\circ$
$\Rightarrow\angle\text{p}=180^\circ-35^\circ-35^\circ=110^\circ$
$\therefore\angle\text{QSR}=\angle\text{p}=110^\circ$ [Angles in same segment]
Now, $\angle\text{QSR}+\angle\text{QTR}=180^\circ$ [Opposite angles of cyclic quad.]
$\Rightarrow110^\circ+\angle\text{QTR}=180^\circ$
$\Rightarrow\angle\text{QTR}=180^\circ-110^\circ=70^\circ$
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