Given: AB ∥ DC , m CB =62°, m∠DAB=104° Find: m∠DEA, m∠ADB m∠DEA = _, m∠ADB =_

The values of the angles are;

m∠DEA = 62°, m∠ADB = 45°

Step-by-step explanation:

Specify an arc or an angle three letters  

Angle opposite an arc on the circumference

m DA ≅ m CB = 62°  (Arc between parallel lines are congruent)

∠CAB = 1/2 × m CB = 1/2 × 62° = 31° (Angle at the center = 2 × Angle st the circumference)

∠DBA = 31° (Angle at the center m DA = 2 × Angle st the circumference)

m∠DAB = 104° (Given)

∠ADB = 180° - m∠DAB - ∠DBA = 180° - 104° - 31° = 45° (Interior angles of triangle ΔADB

m∠ADB = 45°

∠AEB = 180 - ∠CAB - ∠DBA = 180° - 31° - 31° = 118°

∠AEB ≅ ∠COD (Vertically opposite angles)

∠DEA ≅ ∠CEB (Vertically opposite angles)

∠AEB + ∠COD + ∠DEA + ∠CEB = 360° (Sum of angles at a point)

118° + 118° + ∠DEA + ∠CEB = 360°

∠DEA + ∠CEB = 360° - 118° - 118° = 124°

Given that ∠DEA = ∠CEB we have;

2 × ∠DEA = 124°

∠DEA = 124°/2 = 62°

m∠DEA = 62°.

step-by-step explanation:

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