Question

The smallest angle of the triangle whose sides are $6+\sqrt{12},\sqrt{48},\sqrt{24}$

• A. $\frac{\pi }{12}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{7}$

Answer 1

The correct option is A $\frac{\pi }{12}$

The smallest angle is opposite to the smallest side, we can use the law of cosines.

$a^2=b^2+c^2-2bc\cos A$-----------------------1

On simplification,

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

Substituting the values, we have

$\cos A=\frac{(6+2\sqrt{3}{)}^2+(4\sqrt{3}{)}^2-(2\sqrt{6}{)}^2}{2(6+2\sqrt{3})\left(4\sqrt{3}\right)}$

$=\frac{48+24\sqrt{3}+48-24}{2(6+2\sqrt{3})4\sqrt{3}}$

On taking numerator in terms of 2,

$\frac{2(24+12\sqrt{3}+24-12)}{2(6+2\sqrt{3})\left(4\sqrt{3}\right)}$

$=\frac{36+12\sqrt{3}}{(6+2\sqrt{3})4\sqrt{3}}$

On dividing by $(6+2\sqrt{3})$

$=\frac{6}{4\sqrt{3}}$

and finally,

$\cos A=\frac{\sqrt{3}}{2}$

$\therefore A={30}^{\circ }=\frac{\pi }{12}$ rad.