兩條 only a_a 兩條 only a_a

Q.1
Let O be centre of circle. Angle AOB = 2 x Angle ACB = 2 48 = 96. (Angle at centre is two times angle at circumference).
Angle PBO = Angle PAO = 90. (Property of tangent to circle).
Therefore, for quadrilateral BOAP, Angle BPA + 90 + 90 + 96 = 360 (sum of angles of quadrilateral). Therefore, Angle BPA = 180 - 96 = 84.
Q.2
BP = BQ (Property of tangent to circle). Therefore, triangle PBQ is an isos. triangle, therefore, angle PQB = (180- 72)/2 = 54.
Therefore, angle PRQ = angle PQB = 54 (angles in alternate segment).