∵ CP and CQ are tangents to a circle with centre O.
∴ CP = CQ = 11 cm
Similarly, BR and BQ are tangents to the same circle
∴ BR = BQ
But    BQ = CQ – BC
= 11 cm – 7 cm. = 4 cm.
Hence, BR = 4 cm.

We know that, A line drawn through the centre to the point of contact is perpendicular on it.
∠OTP = 90°
Now, in right ΔOPT
OP² = OT² + PT² (10)² = OT² + (8)²
⇒    100 = OT² + 64
⇒    OT² = 100 – 64
⇒    OT² = 30
⇒    OT = 6 cm
Hence, Radius of the circle is 6 cm.          

Since the tangents from an external point to a circle are equal in length, therefore
PA = PB, CA = CR and DB = DR
Now,    PA = PB
⇒ PC + CA = PD + DB
⇒    PC + CR = PD + DR.

∠AOP + ∠POB = 180° [Linear pair]
⇒ ∠AOP + 115 = 180
⇒ ∠AOP = 180 – 115
= 65° ∠AOP = 90
[Line drawn through the centre to point of contact]
Now, In ΔAPO,
∠OAP + ∠AOP + ∠APO = 180
⇒    90 + 65 + ∠APO = 180
⇒    ∠APO = 180 – 155
= 25°.