Question

In figure, AC is the diameter of circle and angle ADB =20° then find angle BPC.

Answer 1

Answer:

110°

Step-by-step explanation:

Given that AC is the diameter of the circle,

Also, Given that ∠ADB = 20°,

Now, if we consider AB as the chord of the circle and the major segment

ADCB, we have

∠ACB = ∠ADB ( Angles in the same segment)

∠ACB =  20°.

Also,

∠ABC = 90°( Angle subtnded by the diameter of a circle)

Now, in triangle ABC,

∠ABC + ∠BCA +∠CAB = 180°

=> ∠CAB = 180° - 90° - 20°

=>∠CAB = 70°.

Now, consider the quadrilateral ACPB, we can observe that all the vertices

of the quadrilateral lie on the same circle, hence ACPB is a cyclic

quadrilateral.

We know that, in a cyclic quadrilateral opposite angles are supplementary

Hence, ∠CAB + ∠BPC = 180°

=>∠BPC = 180° - 70°

= 110°

Answer 2

Answer:

110°

Step-by-step explanation:

Given that AC is the diameter of the circle,

Also, Given that ∠ADB = 20°,

Now, if we consider AB as the chord of the circle and the major segment

ADCB, we have

∠ACB = ∠ADB ( Angles in the same segment)

∠ACB =  20°.

Also,

∠ABC = 90°( Angle subtnded by the diameter of a circle)

Now, in triangle ABC,

∠ABC + ∠BCA +∠CAB = 180°

=> ∠CAB = 180° - 90° - 20°

=>∠CAB = 70°.

Now, consider the quadrilateral ACPB, we can observe that all the vertices

of the quadrilateral lie on the same circle, hence ACPB is a cyclic

quadrilateral.

We know that, in a cyclic quadrilateral opposite angles are supplementary

Hence, ∠CAB + ∠BPC = 180°

=>∠BPC = 180° - 70°

= 110°