In $$\triangle$$PQR, the side QR is extended to S such that RS = PR. If $$\angle$$QPS = 110$$^\circ$$ and $$\angle$$PRQ = 70$$^\circ$$, then the value of $$\angle$$PQR is

Let  $$\angle$$PSR = $$x$$

From the figure,

$$\angle$$PRQ + $$\angle$$PRS = 180$$^\circ$$

$$=$$>  70$$^\circ$$ + $$\angle$$PRS = 180$$^\circ$$

$$=$$>  $$\angle$$PRS = 110$$^\circ$$

In $$\triangle\ $$PRS, RS = PR

Angles opposite to equal sides are are equal

$$=$$>  $$\angle$$RPS = $$\angle$$PSR

$$=$$>  $$\angle$$RPS = $$x$$

$$\angle$$PRS + $$\angle$$RPS + $$\angle$$PSR = 180$$^\circ$$

$$=$$>  110$$^\circ$$ + $$x$$ + $$x$$ = 180$$^\circ$$

$$=$$>   2$$x$$ = 70$$^\circ$$

$$=$$>   $$x$$ = 35$$^\circ$$

$$=$$>  $$\angle$$RPS = 35$$^\circ$$

Given,  $$\angle$$QPS = 110$$^\circ$$

$$=$$>  $$\angle$$QPR+ $$\angle$$RPS = 110$$^\circ$$

$$=$$>  $$\angle$$QPR + 35$$^\circ$$ = 110$$^\circ$$

$$=$$>  $$\angle$$QPR = 75$$^\circ$$

In $$\triangle\ $$PQR,

$$\angle$$PQR + $$\angle$$PRQ + $$\angle$$QPR = 180$$^\circ$$

$$=$$>  $$\angle$$PQR + 70$$^\circ$$ + 75$$^\circ$$ = 180$$^\circ$$

$$=$$>  $$\angle$$PQR + 145$$^\circ$$ = 180$$^\circ$$

$$=$$>  $$\angle$$PQR = 35$$^\circ$$

Hence, the correct answer is Option D