Off Center Angle

Since $\angle APB + \angle BPC = 180^{\circ}$ , then $\angle BPC = \angle BAP + \angle ABP = \angle BAP + 49^{\circ}$ .

Thus, we are attempting to find $\angle BAP$ for a solution, and we can do so by extending the $OC$ line to meet point $D$ on the circumference before joining it with point $B$ , as shown below:

According to Inscribed Angles Theorem , $\angle BAP = \angle BDO$ , and because $\triangle BDO$ is an isosceles triangle, $\angle DBO = \angle BDO$ .

Furthermore, from Inscribed Angles Theorem , $\angle ABD = \angle ACD = 17^{\circ}$ .

Thus, $\angle ABP = 49^{\circ} = \angle ABD + \angle DBO = 17^{\circ} + \angle BAP$ .

Hence, $\angle BAP = 49^{\circ} -17^{\circ} = 32^{\circ}$ .

Finally, $\angle BPC = 49^{\circ} + 32^{\circ} = 81^{\circ}$ .

Joining OA, We see that both the blue angles are equal and so are the 2 yellow angles. (By isoceles triangle theorem since radii are equal)

So , Using exterior angle theorem in triangle $PAB$ , We get $Red=2Yellow-Blue$

Thus $Red = 2*49-17=98-17=81$

Upper image for diagram

Lower image for solution.. easy to understand

1) OBC Is an isosceles triangle.

2) 2(Angle BAC) = Angle BOC.

Let us take Angle BAC=θ.

So angle BOC =2θ.

Since BAP and BOC cut the same arc, BAP = 1/2 * BOC

BPC = BAP + 49 = BOC + 17 because an exterior angle = sum of the two non adjacent angles

Substituting 1/2*BOC + 49 = BOC + 17

Solving BOC = 64

64 + 17 = 81 degrees