In a $$\triangle$$ABC, the sides AB and AC are extended to P and Q, respectively. The bisectors of $$\angle$$PBC and $$\angle$$QCB intersect at a point R If $$\angle$$R = $$66^\circ$$, then the measure of $$\angle$$A is:

As per the given question,

Let $$\angle PBR=\angle RBC=x$$ and $$\angle QCR=\angle RCB=y$$

But $$\angle PBR +\angle RBC +\angle ABC=180^\circ$$

$$\angle ABC=180^\circ-2x$$--------(i) (sum of angle on the straight line)

And  $$\angle ACB=180^\circ-2y$$--------(ii) (sum of angle on the straight line)

Now in $$\triangle RBC$$

$$\angle RBC+\angle BCR +\angle CRB=180^\circ$$

$$x+y +66=180^\circ$$

$$x+y =180^\circ-66^\circ=114^\circ ---------------(iii)$$

Now in $$\triangle ABC$$

$$\angle A+\angle ACB +\angle ACB=180^\circ$$

$$\angle A=180^\circ-(180^\circ-2x)-(180^\circ-2y)$$

$$\angle A=+2x-180^\circ+2y$$

$$\angle A=+2x-180^\circ+2y$$

$$\angle A=2(x+y)-180^\circ ------------(iv)$$

Now from the equation (iii) and (iv)

$$\angle A=2(114)-180^\circ$$

$$\angle A=228^\circ-180^\circ=48^\circ$$