Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively.

If $a=2,b=\sqrt{6}$ and $c=\sqrt{3}-1$ then the angles of the triangle are:

• A. $A={45}^0,B={120}^0,C={15}^0$
• B. $A={15}^0,B={120}^0,C={45}^0$
• C. $A={120}^0,B={15}^0,C={45}^0$
• D. $A={45}^0,B={15}^0,C={120}^0$

Answer 1

Answer: (A)

$A={45}^0,B={120}^0,C={15}^0$

$cosA=\frac{b^2+c^2-a^2}{2bc}$

$=\frac{6+3+1-2\sqrt{3}-4}{2\times (\sqrt{3}-1)\sqrt{6}}$

$=\frac{3-\sqrt{3}}{\sqrt{6}(\sqrt{3}-1)}$

$cosA=\frac{1}{\sqrt{2}}$

$A={45}^{\circ }$

$cosB=\frac{a^2+c^2-b^2}{2ac}$

$=\frac{4+3+1-2\sqrt{3}-6}{2\times 2\times (\sqrt{3}-1)}$

$=\frac{2-\sqrt{3}}{2(\sqrt{3}-1)}$

$cosB=\frac{-1}{2}$

$B={120}^{\circ }$

Now, we know that $A+B+C=180$

$\therefore C={15}^{\circ }$