MCQ
In the given figure, $\text{AM }\bot\text{ BC}$ and AN is the bisector of $\angle\text{A}.$ If $\angle\text{ABC}=70^\circ$ and $\angle\text{ACB}=20^\circ,$ then $MAN =?$
- A$20^\circ$
- B$30^\circ$
- C$15^\circ$
- ✓$25^\circ$
✓
Answer
Correct option: D.
$25^\circ$
In $\triangle\text{ABC}$
$\angle\text{BAC}+\angle\text{ABC}+\angle\text{BCA}=180^\circ$ (Angle sum property)
$\angle\text{BAC}=180^\circ-70^\circ-20^\circ$
$\angle\text{BAC}=90^\circ$
In $\triangle\text{ANC}$
$\angle\text{ANC}+\angle\text{NAC}+\angle\text{ACN}=180^\circ$ (Angle sum property)
$\angle\text{ANC}+45^\circ+20^\circ=180^\circ$ (AN is angle bisector of ∠A)
$\angle\text{ANC}=115^\circ$
In $\triangle\text{AMN}$
$\angle\text{AMN}+\angle\text{MAN}=\angle\text{ANC}$ (Measure of exterior angle is equla to the sum of two opposite interior angles)
$90^\circ+\angle\text{MAN}=115^\circ$
$\angle\text{MAN}=25^\circ.$
$\angle\text{BAC}+\angle\text{ABC}+\angle\text{BCA}=180^\circ$ (Angle sum property)
$\angle\text{BAC}=180^\circ-70^\circ-20^\circ$
$\angle\text{BAC}=90^\circ$
In $\triangle\text{ANC}$
$\angle\text{ANC}+\angle\text{NAC}+\angle\text{ACN}=180^\circ$ (Angle sum property)
$\angle\text{ANC}+45^\circ+20^\circ=180^\circ$ (AN is angle bisector of ∠A)
$\angle\text{ANC}=115^\circ$
In $\triangle\text{AMN}$
$\angle\text{AMN}+\angle\text{MAN}=\angle\text{ANC}$ (Measure of exterior angle is equla to the sum of two opposite interior angles)
$90^\circ+\angle\text{MAN}=115^\circ$
$\angle\text{MAN}=25^\circ.$
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