MCQ
- A95º
- B73º
- C72º
- D74º
✓
Answer
- 72º
Solution:
Given that $\triangle\text{ABC}$
BE is bisector of $\angle\text{B}$ and AD is bisector of $\angle\text{BAC}$
$\angle\text{B} = 2\angle\text{C}$
By exterior angle theorem in triangle ADC
$\angle\text{ADB} = \angle\text{DAC} + \angle\text{C} ...\ \text{(i)}$
In $\triangle\text{ADB},$
$\angle\text{ABD} + \angle\text{BAD} + \angle\text{ADB} = 180^\circ$
$2\angle\text{C} + \angle\text{BAD} + \angle\text{DAC} + \angle\text{C} = 180^\circ$ [From (i)]
$3\angle\text{C} + \angle\text{BAC} = 180^\circ$
$\angle\text{BAC} = 180^\circ - 3\angle\text{C} ...\ \text{(ii)}$
Therefore,
$\text{AB = CD}$
$\angle\text{C} = \angle\text{DAC}$
$\angle\text{C} = \frac{1}{2}\angle\text{BAC}\ ...\ \text{(iii)}$
Putting value of Angle C in (ii), we get
$\angle\text{BAC} = 180^\circ - \frac{1}{2} \angle\text{BAC}$
$\angle\text{BAC} +\frac{3}{2} \angle\text{BAC} =180^\circ$
$\frac{5}{2} \angle\text{BAC} =180^\circ$
$\angle\text{BAC} = \frac{180\times2}{2}$
$=72^\circ$
$\angle\text{BAC} = 72^\circ$
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