In the given figure, BAD || EF, ∠AEF = 55° and ∠ACB = 25°, find ∠ABC.

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Solution

In the given figure, EF || BD and CE is the transversal.

∴ ∠CAD = ∠AEF (Pair of corresponding angles)

⇒ ∠CAD = 55°

In ∆ABC,

∠CAD = ∠ABC + ∠ACB (Exterior angle of a triangle is equal to the sum of the two interior opposite angles)

⇒ 55° = ∠ABC + 25°

⇒ ∠ABC = 55° − 25° = 30°

Thus, the measure of ∠ABC is 30°.

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$In the given figure, B A D | | E F, ∠ A E F = 55^{\circ} a n d ∠ A C B = 25^{\circ}, f i n d ∠ A B C .$

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∠ A C B = 31^{\circ}

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