Question

(1)
ABC.
44 In the given figure, O is the centre of the
circle and arc ABC subtends an angle of
130° at the centre. If AB is extended to P,
find ZPBC.
5/In the given figure, ABCD is a cyclic
which AE is drawn​

Answer 1

Consider a point D on the arc CA and join DC and AD

Consider a point D on the arc CA and join DC and ADWe know that the angle subtended by an arc is twice the angle subtended by it on the circumference in the alternate segment

Consider a point D on the arc CA and join DC and ADWe know that the angle subtended by an arc is twice the angle subtended by it on the circumference in the alternate segmentSo we get

Consider a point D on the arc CA and join DC and ADWe know that the angle subtended by an arc is twice the angle subtended by it on the circumference in the alternate segmentSo we get∠2=2∠1

Consider a point D on the arc CA and join DC and ADWe know that the angle subtended by an arc is twice the angle subtended by it on the circumference in the alternate segmentSo we get∠2=2∠1By substituting the values

Consider a point D on the arc CA and join DC and ADWe know that the angle subtended by an arc is twice the angle subtended by it on the circumference in the alternate segmentSo we get∠2=2∠1By substituting the values130

Consider a point D on the arc CA and join DC and ADWe know that the angle subtended by an arc is twice the angle subtended by it on the circumference in the alternate segmentSo we get∠2=2∠1By substituting the values130 =2∠1

Consider a point D on the arc CA and join DC and ADWe know that the angle subtended by an arc is twice the angle subtended by it on the circumference in the alternate segmentSo we get∠2=2∠1By substituting the values130 =2∠1So we get

Consider a point D on the arc CA and join DC and ADWe know that the angle subtended by an arc is twice the angle subtended by it on the circumference in the alternate segmentSo we get∠2=2∠1By substituting the values130 =2∠1So we get∠1=65

Consider a point D on the arc CA and join DC and ADWe know that the angle subtended by an arc is twice the angle subtended by it on the circumference in the alternate segmentSo we get∠2=2∠1By substituting the values130 =2∠1So we get∠1=65 From the figure we know that the exterior angle of a cyclic quadrilateral = interior opposite angle

Consider a point D on the arc CA and join DC and ADWe know that the angle subtended by an arc is twice the angle subtended by it on the circumference in the alternate segmentSo we get∠2=2∠1By substituting the values130 =2∠1So we get∠1=65 From the figure we know that the exterior angle of a cyclic quadrilateral = interior opposite angle∠PBC=∠1

Consider a point D on the arc CA and join DC and ADWe know that the angle subtended by an arc is twice the angle subtended by it on the circumference in the alternate segmentSo we get∠2=2∠1By substituting the values130 =2∠1So we get∠1=65 From the figure we know that the exterior angle of a cyclic quadrilateral = interior opposite angle∠PBC=∠1So we get ∠PBC=65

Consider a point D on the arc CA and join DC and ADWe know that the angle subtended by an arc is twice the angle subtended by it on the circumference in the alternate segmentSo we get∠2=2∠1By substituting the values130 =2∠1So we get∠1=65 From the figure we know that the exterior angle of a cyclic quadrilateral = interior opposite angle∠PBC=∠1So we get ∠PBC=65 Therefore, ∠PBC=65