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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Question : In a triangle ${ABC}, {D}$ is a point on ${BC}$ such that $\frac{A B}{A C}=\frac{B D}{D C}$. If $\angle B=68^{\circ}$ and $\angle C=52^{\circ}$, then measure of $\angle B A D$ is equal to:
Question : In $\triangle ABC$, M and N are the points on side BC such that AM $\perp$ BC, AN is the bisector of $\angle A$, and M lies between B and N. If $\angle B=68^{\circ}$, and $\angle \\{C}=26^{\circ}$, then the measure of $\angle MAN$ is:
Question : $D$ is a point on the side $BC$ of a triangle $ABC$ such that $AD\perp BC$. $E$ is a point on $AD$ for which $AE:ED=5:1$. If $\angle BAD=30^{\circ}$ and $\tan \left ( \angle ACB \right )=6\tan \left ( \angle DBE \right )$, then $\angle ACB =$
Question : In $\triangle {ABC}$, D is a point on BC such that $\angle {ADB}=2 \angle {DAC}, \angle {BAC}=70^{\circ}$ and $\angle {B}=56^{\circ}$. What is the measure of $\angle A D C$?
Question : In an isosceles triangle $ABC$, $AB = AC$ and $\angle A = 80^\circ$. The bisector of $\angle B$ and $\angle C$ meet at $D$. The $\angle BDC$ is equal to:
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