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
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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$^\circ$.
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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:
Option 1: 18º and 22º
Option 2: 26º and 30º
Option 3: 30º and 34º
Option 4: 22º and 26º
Question : In $\triangle {ABC}$, O is the incentre and $\angle {BOC}=135^{\circ}$. The measure of $\angle {BAC}$ is:
Option 1: 90º
Option 2: 55º
Option 3: 80º
Option 4: 45º
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$^\circ$
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 $\angle A D C$?
Option 1: $72^{\circ}$
Option 2: $54^{\circ}$
Option 3: $74^{\circ}$
Option 4: $81^{\circ}$
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 $\angle ACB =$
Option 1: $30^{\circ}$
Option 2: $45^{\circ}$
Option 3: $60^{\circ}$
Option 4: $15^{\circ}$
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