MCQ
In a $\triangle\text{ABC},$ if $a = 2, \angle\text{B}=60^{\circ}$ and $\angle\text{C}=75^{\circ},$ then $b =$
  • A
    $\sqrt{3}$
  • $\sqrt{6}$
  • C
    $\sqrt{9}$
  • D
    $1+\sqrt{2}$

Answer

Correct option: B.
$\sqrt{6}$
It is given that $\text{a}=2,\angle\text{B}=60^{\circ}$ and $\angle\text{C}=75^{\circ}.$
In $\triangle\text{ABC},$
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^{\circ} ($Angle sum property$)$
$\Rightarrow\angle\text{A}+60^{\circ}+75^{\circ}=180^{\circ}$
$\Rightarrow\angle\text{A}=180^{\circ}-135^{\circ}=45^{\circ}$
Using sine rule, we get
$\frac{2}{\sin45^{\circ}}=\frac{\text{b}}{\sin60^{\circ}}$ $\Big(\frac{\text{a}}{\sin\text{A}}=\frac{\text{b}}{\sin\text{B}}=\frac{\text{c}}{\sin\text{C}}\Big)$
$\Rightarrow\text{b}=\frac{2\times\frac{\sqrt{3}}{2}}{\frac{1}{\sqrt{2}}}=\sqrt{6}$
Hence, the correct answer is option $(b).$

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