In ∆ABC, side BC is produced to D. If ∠ABC = 40° and ∠ACD = 120°, then ∠A = ?
(a) 60°
(b) 40°
(c) 80°
(d) 50°

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Solution

(c) 80°

$∴ ∠ A + ∠ B = ∠ A C D \Rightarrow ∠ A + 40 ° = 120 ° \Rightarrow ∠ A = 80 °$

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In ∆ABC, side BC is produced to D. If ∠ABC = 40° and ∠ACD = 120°, then ∠A = ? (a) 60° (b) 40° (c) 80° (d) 50°

(c) 80° ∴∠A+∠B=∠ACD⇒∠A+40°=120°⇒∠A=80°

In ∆ABC, side BC is produced to D. If ∠ABC = 40° and ∠ACD = 120°, then ∠A = ?
(a) 60°
(b) 40°
(c) 80°
(d) 50°

(c) 80°

$∴ ∠ A + ∠ B = ∠ A C D \Rightarrow ∠ A + 40 ° = 120 ° \Rightarrow ∠ A = 80 °$