Question

Let $a,b,c$ are sides of $△ABC.$ If $a=\sqrt{3}+1,b=2,c=\sqrt{2},$ then $\angle B$ is equal to

• A. ${15}^0$
• B. ${45}^0$
• C. ${60}^0$
• D. ${90}^0$

Answer 1

Answer: (B)

${45}^0$

By cosine formula, we have

$\cos B=\frac{c^2+a^2-b^2}{2ca}=\frac{{\left(\sqrt{2}\right)}^2+{(\sqrt{3}+1)}^2-2^2}{2\times \left(\sqrt{2}\right)\times (\sqrt{3}+1)}$

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

$=\frac{2(\sqrt{3}+1)}{2\sqrt{2}(\sqrt{3}+1)}=\frac{1}{\sqrt{2}}$ $=\cos {45}^0$

$\Rightarrow B={45}^0$