Question

Two sides of a triangle are $(\sqrt{3}+1)$ and $(\sqrt{3}-1)$ and they include an angle of ${60}^o$. Then the remaining of angles differ by.

• A. ${30}^o$
• B. ${45}^o$
• C. ${60}^o$
• D. ${90}^o$

Answer 1

The correct option is D ${90}^o$

Let $a=3+\sqrt{3},b=3-\sqrt{3}$ and $C={60}^{\circ }$

$\tan \frac{A-B}{2}=\frac{a-b}{a+b}\cot \frac{C}{2}\tan \frac{A-B}{2}=\frac{3+\sqrt{3}-3+\sqrt{3}}{3+\sqrt{3}+3-\sqrt{3}}\cot {30}^{\circ }\tan \frac{A-B}{2}=1\frac{A-B}{2}={45}^{\circ }A-B={90}^{\circ }$