Guys I’m serious, last time I posted this no one helped me, I’ve tried so many times but I can’t figure it out and I’ve reached the point where I’m crying on my floor from all the stress I’m dealing with. So please can someone please please help a girl out before I have another anxiety attack. Even if you help with finding some parts I just need something to work off of.

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°.

3/5

step-by-step explanation:

double number line is not a standard concept, so it must have been defined in your textbook. assume top line is numerator.