Internal bisectors of ∠Q and ∠R of ΔPQR intersect at O. If ∠ROQ = 96° then the value of ∠RPQ is

To find : $$\angle$$RPQ = $$\theta$$ = ?

Solution : Let $$\angle$$PQR = $$2x$$ and $$\angle$$PRQ = $$2y$$

=> $$\angle$$OQR = $$x$$ and $$\angle$$ORQ = $$y$$ [SInce, QO & RO are angle bisectors]

In $$\triangle$$PQR

=> $$\theta$$ + $$\angle$$PQR + $$\angle$$PRQ = 180°

=> $$\theta$$ = 180° -2$$(x+y)$$ ---------Eqn(1)

In $$\triangle$$QOR

=> x + y + 96° = 180°

=> x + y = 84°

Putting value of (x+y) in eqn (1)

=> $$\theta$$ = 180 - 2*84 = 180-168 = 12°