Question : In $\triangle A B C, \angle B=78^{\circ}, A D$ is a bisector of $\angle A$ meeting BC at D, and $A E \perp B C$ at $E$. If $\angle D A E=24^{\circ}$, then the measure of $\angle A C B$ is:
Option 1: 30$^\circ$
Option 2: 38$^\circ$
Option 3: 32$^\circ$
Option 4: 42$^\circ$
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Correct Answer: 30$^\circ$
Solution :
$\angle$B = 78$^\circ$
AD is the bisector of $\angle$A meeting BC at D.
AE$\perp$BC at E
$\angle$ DAE = 24$^\circ$
If AD is the angle bisector and AE is the perpendicular on base BC, then
$\angle$DAE = $\frac{1}{2}((\angle \text{B} - \angle \text{C}))$
Let the value of $\angle$ACB be x.
Now, According to the concept used $\angle$ DAE = $\frac{1}{2}(\angle \text{B} - \angle \text{C})$
⇒ 24$^\circ$ = $\frac{1}{2}(78^\circ - \text{x})$
⇒ 48$^\circ$ = 78$^\circ$ – x
$\therefore$ x = 30$^\circ$
Hence, the correct answer is 30$^\circ$.
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