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In ∆ABC, side BC has been produced to D. If ∠ACD = 132° and ∠A = 54°, then ∠B = ?

(a) 48°
(b) 78°
(c) 68°
(d) 58°

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Solution

$(b) 78 ° \begin{array}{l} ∠ ACD= ∠ ABC+ ∠ BAC (exterior angle property) \\ => ∠ ABC=132° - 54 ° = 78 ° \end{array}$

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Mark (✓) against the correct answer In ∆ABC, side BC has been produced to D. If ∠ACD = 132° and ∠A = 54°, then ∠B = ? (a) 48° (b) 78° (c) 68° (d) 58°

(b) 78° ∠ACD=∠ABC+∠BAC (exterior angle property) =>∠ABC=132°−54°=78°

Mark (✓) against the correct answer
In ∆ABC, side BC has been produced to D. If ∠ACD = 132° and ∠A = 54°, then ∠B = ?

(a) 48°
(b) 78°
(c) 68°
(d) 58°

$(b) 78 ° \begin{array}{l} ∠ ACD= ∠ ABC+ ∠ BAC (exterior angle property) \\ => ∠ ABC=132° - 54 ° = 78 ° \end{array}$