In a triangle $A B C, a=(1+\sqrt{3}) \mathrm{cm}, b=2 \mathrm{~cm}$ and angle $C=60^{\circ}$. Then the other two angles are

• Correct answer
  A

  $45^{\circ}$ and $75^{\circ}$
• B

  $30^{\circ}$ and $90^{\circ}$
• C

  $105^{\circ}$ and $15^{\circ}$
• D

  $100^{\circ}$ and $20^{\circ}$

Solution

Now as a > b $\therefore \quad \angle \mathrm{A} \gt \angle \mathrm{B}$ Now from Sine Rule, $\begin{aligned} & \frac{\sin A}{a}=\frac{\sin B}{b} \\ & \frac{\sin A}{1+\sqrt{3}}=\frac{\sin B}{2} \end{aligned}$ From option (A), $\begin{aligned} & \frac{\sin 75}{1+\sqrt{3}}=\frac{\sin 45}{2} \\ & \frac{\sqrt{6}+\sqrt{2}}{4(1+\sqrt{3})}=\frac{1}{2 \sqrt{2}} \\ & 2 \sqrt{12}+4=4+4 \sqrt{3} \\ & 4+4 \sqrt{3}=4+4 \sqrt{3} \end{aligned}$ $\therefore \quad$ Option (A) is correct.

Hence, the correct answer is option: 'A'.