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:
Option 1: 21º
Option 2: 28º
Option 3: 24º
Option 4: 22º
Latest: SSC CGL preparation tips to crack the exam
Don't Miss: SSC CGL complete guide
New: Unlock 10% OFF on PTE Academic. Use Code: 'C360SPL10'
Correct Answer: 21º
Solution : $\triangle$ABC, M and N are the side on BC such that AM $\perp$ BC, AN is the bisector $\angle$A, and M lies between B and N $\angle$B = 68$^\circ$, and $\angle$C = 26$^\circ$ Concept used: $\triangle$ABC, M and N are the sides on BC such that AM ⊥ BC, AN is the bisector $\angle$A, and M lies between B and N. $\angle$MAN = $\frac{(\angle \text{B} - \angle \text{C})}{2}$ Solution: ⇒ $\angle$MAN = $\frac{(\angle \text{B} - \angle \text{C})}{2}$ ⇒ $\angle$MAN = $\frac{(68^\circ - 26^\circ)}{2}$ = 21$^\circ$ Hence, the correct answer is 21$^\circ$.
Candidates can download this ebook to know all about SSC CGL.
Answer Key | Eligibility | Application | Selection Process | Preparation Tips | Result | Admit Card
Question : In a circle, O is the centre of the circle. Chords AB and CD intersect at P. If $\angle AOD=32^{\circ}$ and $\angle CO B=26^{\circ}$, then the measure of $\angle APD$ lies between:
Question : In $\triangle {ABC}$, O is the incentre and $\angle {BOC}=135^{\circ}$. The measure of $\angle {BAC}$ is:
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:
Question : If the measure of one angle of a right triangle is 30º more than the measure of the smallest angle, then the measure of the smallest angle 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 $\angle ACB =$
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile