Answer:
70
Step-by-step explanation:
Given
∠PDB = 110°
PDis the tangent to the circle whose centre is E
Construction
Join points D and E.
∠DPE= 90° [Since angle in a semi circle is 90°]
Also ∠PDE = 90° [Since radius ⊥ tangent]
From the figure we have,
∠PDB=∠PDE + ∠DBE
i.e. 110° = 90° + ∠DBE
Therefore, ∠DBE =20°
Now in ΔBED, BE = ED [Radii]
So, ∠EDB = ∠EBD =20°
In ΔBPD, we have
∠DPE = 90° & ∠DBE = 20°
Therefore, ∠DPB = 70°
