Can someone help with the work shown to find them

m(ar AB) = 60°, m(ar BAC) = 240°, m(ar BC) = 120°,

m(ar EA) = 92°, m(ar ECA) = 268°

Solution:

Let us take O be the center of the circle.

Given m∠AOB = 60° and m∠BOE = 32°

Sum of the adjacent angles = 180°

m∠EOC + 32° + 60° = 180°

m∠EOC + 92° = 180°

m∠EOC = 88°

Angle in the diameter is 180°.

∠AOC = 180°

The angle measure of the central angle is congruent to the measure of the intercepted arc.

m∠AOB = m(ar AB) = 60°

Central angle = intercepted arc

m∠AOC = m(ar AOC) = 180°

m(ar BAC) = m(ar AOC) + m(ar AOB)

                 = 180° + 60°

m(ar BAC) = 240°

m(ar BC) = m(ar BE) + m(ar EC)

              = 32° + 88°

m(ar BC) = 120°

m(ar EA) = m(ar EB) + m(ar AB)

              = 32° + 60°

m(ar EA) = 92°

m(ar ECA) = m(ar EC) + m(ar AOC)

                 = 88° + 180°

m(ar ECA) = 268°

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