Question : In $\triangle A B C, O$ is the point of intersection of the bisectors of $\angle B$ and $\angle A$. If $\angle B O C=108^{\circ}$, then $\angle B A O=$?
Option 1: 27°
Option 2: 40°
Option 3: 18°
Option 4: 36°
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Correct Answer: 18°
Solution : In $\triangle$ABC, O is the point of intersection of the bisectors of $\angle$B and $\angle$A $\angle$BOC = 108° If the bisector of $\angle$ABC and $\angle$ACB of a triangle ABC meet at a point O then, $\angle$BOC = $90^\circ + \frac{1}{2}$$\angle$A ⇒ $108^\circ = 90^\circ + \frac{1}{2}$$\angle$A ⇒ $\frac{1}{2} \angle \text{A} = 108^\circ - 90^\circ$ ⇒ $\frac{1}{2} \angle\text{ A} = 18^\circ$ ⇒ $\angle$A = 36$^\circ$ We know that an angle bisector of an angle of a triangle divides the opposite side into two parts that are proportional to the other two sides of the triangle. ⇒ $\angle$ BAO = $\frac{\angle \text{A}}{2}$ ⇒ $\angle$ BAO = $\frac{36}{2}$ ⇒ $\angle$ BAO = 18$^\circ$ $\therefore$ The value of $\angle$BAO = 18$^\circ$ Hence, the correct answer is 18$^\circ$.
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Question : In $\triangle A B C, \angle A=66^{\circ}$ and $\angle B=50^{\circ}$. If the bisectors of $\angle B$ and $\angle C$ meet at P, then, $\angle B P C-\angle P C A=$?
Question : AB is a chord in the minor segment of a circle with centre O. C is a point on the minor arc (between A and B ). The tangents to the circle at A and B meet at a point P. If $\angle \mathrm{ACB}=108^{\circ}$, then $\angle \mathrm{APB}$ is equal to:
Question : In a $\triangle A B C, \angle B+\angle C=110^{\circ}$, then find the measure of $\angle A$.
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 : A, B, and C are three angles of a triangle. If $A-B=45^{\circ}$ and $B-C=15^{\circ}$ then $\angle A=$?
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