MCQ
If in a triangle $ABC , b =\sqrt{3}, c =1$ and $B-C=90^{\circ}$, then $\angle A$ is
  • $30^{\circ}$
  • B
    $45^{\circ}$
  • C
    $75^{\circ}$
  • D
    $15^{\circ}$

Answer

Correct option: A.
$30^{\circ}$
(A) $\tan \left(\frac{ B - C }{2}\right)=\frac{ b - c }{ b + c } \cot \frac{ A }{2}$
$\Rightarrow \tan \left(\frac{90^{\circ}}{2}\right)=\frac{\sqrt{3}-1}{\sqrt{3}+1} \cot \frac{A}{2}$
$\Rightarrow \tan \left(\frac{ A }{2}\right)=\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{3+1-2 \sqrt{3}}{2}=2-\sqrt{3}$
$\Rightarrow \frac{ A }{2}=15^{\circ} \Rightarrow A =30^{\circ}$

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