Answer:
a) 60°
b) 80°
c) 100°
d) 50°
e) 30°
Step-by-step explanation:
Draw radius lines from center O to points A, B, and D.
AB is parallel to EC.  So ∠EOA = ∠OAB, and ∠COB = ∠OBA.  Since ΔOAB is an isosceles triangle, ∠OAB = ∠OBA.  So by substitution, ∠EOA = ∠COB.  And since we know ∠COB = ∠AOB, and ∠EOA + ∠COB + ∠AOB = 180°, we can show they all equal 60°.
Using inscribed angle theorem, ∠ABD is half of arc AED.  And arc AED = AE + ED = 60° + (180° − 80°) = 160°.  So ∠ABD = 80°.
∠OBD = ∠ABD − ∠ABO = 80° − 60° = 20°.  Angles of a triangle add up to 180°, so ∠OFB = 180° − (20° + 60°) = 100°.  And vertical angles are congruent, so ∠DFC = 100°.
Tangent lines are perpendicular to the radius line, so ∠OAP = 90°.  Which means ∠PAB = 30°.  ∠ABP is supplementary to ∠ABD, so ∠ABP = 100°.  And since angles of a triangle add up to 180°, ∠P = 50°
As found earlier, ∠PAB = 30°.
