Question : In $\triangle $ABC, D is a point on side BC such that $\angle$ADC = 2$\angle$BAD. If $\angle$A = 80° and $\angle$C = 38°, then what is the measure of $\angle$ADB?
Option 1: 58°
Option 2: 62°
Option 3: 52°
Option 4: 56°
Correct Answer: 56°
Solution :
Let $\angle$BAD = z then $\angle$ADC = 2z
In $\triangle $ABC,
⇒ $\angle$A + $\angle$B + $\angle$C = 180º
⇒ 80° + $\angle$B + 38° = 180º
⇒ $\angle$B + 118° = 180º
⇒ $\angle$B = 62º
In $\triangle $ABD,
⇒ $\angle$A + $\angle$B = $\angle$ADC (The sum of two opposite angles is equal to the external angle)
⇒ z + 62º = 2z
⇒ z = 62º
And,
⇒ $\angle$ABD + $\angle$BAD + $\angle$ADB = 180º
⇒ 62º + 62º + $\angle$ADB = 180º
⇒ $\angle$ADB = 180º – 124º = 56º
Hence, the correct answer is 56º.
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