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Triangles — Maths STD 9 — Question

Gujarat BoardEnglish MediumSTD 9MathsTriangles1 Mark
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Answer

  1. 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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