Given:
O is the center of circle, PG is tangent to the circle,
G is the point of contact and PG=DG
To find: The value of angle ∠DPG
Here,
- angle ∠DGP = 90° [ Diameter drawn from the point of contact to a circle is perpendicular to the tangent of the circle. ]
- angle ∠PDG = angle ∠DPG [ being the base of isosceles triangle ΔDGP ]
- ∠DGP + ∠PDG + ∠DPG = 180° [ being sum of the angles of triangle ΔDGP ]
Or, 90° + ∠DPG + ∠DPG = 180°
Or, 2 ∠DPG = 90°
Or, ∠DPG = 90° / 2 = 45°
∴ the value of ∠DPG is 45°, required answer.