Question : In a triangle, ABC, BC is produced to D so that CD = AC. If $\angle BAD=111^{\circ}$ and $\angle ACB=80^{\circ}$, then the measure of $\angle ABC$ is:
Option 1: 31°
Option 2: 33°
Option 3: 35°
Option 4: 29°
Correct Answer: 29°
Solution :
Given: AC = CD, $\angle$BAD = 111°, and $\angle$ACB = 80°
So, $\angle$ACD = 180° – 80° = 100°
In isosceles $\triangle$ACD,
$\angle$ACD + $\angle$CAD + $\angle$ADC = 180°
⇒ 2$\angle$CAD = 180° – 100°
⇒ $\angle$CAD = 40°
⇒ $\angle$CAB = 111° – 40° = 71°
⇒ $\angle$ABC = 180° – 71° – 80° = 29°
Hence, the correct answer is 29°.
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