Question

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

• A. $\frac{\pi }{3}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{6}$
• D. $noneofthese$

Answer 1

Answer: (A)

$\frac{\pi }{3}$

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

$a^2=b^2+c^2-2bccosA$

On simplification:$cosA=\frac{b^2+c^2-a^2}{2cb}$

Substituting the values we have

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

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

On taking numerator in terms of 2,

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

On taking numerator in terms of 2,$=\frac{36+12\sqrt{3}}{(6+2\sqrt{3})\left(4\sqrt{3}\right)}$

And finally $cosA=\frac{3}{\sqrt{2}}=\cos {30}^{\circ }=\cos \frac{\pi }{12}$