In ∆ABC, side BC is produced to D. If ∠ABC = 50° and ∠ACD = 110° then ∠A = ?
(a) 160°
(b) 60°
(c) 80°
(d) 30°

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Solution

$∴ ∠ A + ∠ B = ∠ A C D \Rightarrow ∠ A + 50 ° = 110 ° \Rightarrow ∠ A = 60 °$
Hence, the correct answer is option (b).

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$I n a Δ A B C, s i d e B C i s p r o d u c e d t o D . I f ∠ A B C = 50^{\circ} a n d ∠ A C D = 110^{\circ} t h e n ∠ A = ? (a) 160^{\circ} (b) 60^{\circ} (c) 80^{\circ} (d) 30^{\circ}$

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In ∆ABC, side BC has been produced to D. If ∠BAC = 45° and ∠ABC = 55°, then ∠ACD = ?

(a) 80°
(b) 90°
(c) 100°
(d) 110°

Q. In a ΔABC, if AB = AC and BC is produced to D such that ∠ACD = 100°, then ∠A =

(a) 20°

(b) 40°

(c) 60°

(d) 80°

Q. ABC is a triangle in which AB = AC. The base BC is produced to D. If

∠

ACD =

130^{\circ}

, then

∠

A = ___.

ABC is a triangle in which AB = AC. The base BC is produced to D and $∠ A C D = 130^{\circ}$ . Then $∠ A$ equals

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In ∆ABC, side BC is produced to D. If ∠ABC = 50° and ∠ACD = 110° then ∠A = ? (a) 160° (b) 60° (c) 80° (d) 30°

∴∠A+∠B=∠ACD⇒∠A+50°=110°⇒∠A=60° Hence, the correct answer is option (b).

In ∆ABC, side BC is produced to D. If ∠ABC = 50° and ∠ACD = 110° then ∠A = ?
(a) 160°
(b) 60°
(c) 80°
(d) 30°

$∴ ∠ A + ∠ B = ∠ A C D \Rightarrow ∠ A + 50 ° = 110 ° \Rightarrow ∠ A = 60 °$
Hence, the correct answer is option (b).