Given: t is tangent at B is diameter

If m = 110°, then m =

70
80
110

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°

a is the right answer