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

Given: AB ∥ DC , m CB =62°, m∠DAB=104° Find: m∠DEA, m∠ADB-example-1

1 Answer

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Answer:

m∠DEA = 62°

m∠ADB (arc) = 194°

Angle ∠ADB = 21°.

Explanation:

The given information are;


AB\left | \right |DC, m CB (arc) = 62°, m∠DAB (arc) = 104°

arc∠BCD = 360° - 104° = 256°

m DC (arc) = arc∠BCD - arc CB = 256° - 62° (Sum of angles)

Therefore DC (arc) = 194°

m DA ≅ m CB = 62° (Parallel lines congruent arc theorem. Arc between two parallel lines)

m∠DEA = 1/2×(arc DA + arc CB) = 1/2×(62° + 62°) =62°

m∠DEA = 62°

Arc AB = m∠DAB (arc) - m DA = 104° - 62° = 42°

m∠ADB (arc) = 360 - m∠DAB (arc) - m CB (arc) (Sum of angles around a circle or point)

∴ m∠ADB (arc) = 360 - 104 - 62 = 194°

m∠ADB (arc) = 194°

Angle ADB = subtended by arc AB = ∴1/2×arc AB

Angle ∠ADB = 42/2 = 21°.

Angle = 21°

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