Question

In the figure given below, PC is a tangent to the free from the point P and B is a point on the
circle such that PB = CB. Find DCP, if DPC = 20°

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Answer 1

Given :- In the figure given below, PC is a tangent to the free from the point P and B is a point on the circle such that PB = CB. ∠DPC = 20°

To Find :- ∠DCP = ?

Answer :-

given that,

→ PB = CB

so,

→ ∠BCP = ∠BPC = 20° (Angle opposite to equal sides are equal.)

then,

→ ∠CDP = ∠BCP (By alternate segment theorem.)

now, in ∆DCP,

→ ∠DCP + ∠CDP + ∠DPC = 180° (Angle sum property.)

→ ∠DCP + 20° + 20° = 180°

→ ∠DCP = 180° - 40°

→ ∠DCP = 140° (Ans.)

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