Question

Sides of a triangle are $7,4\sqrt{3}$ and $\sqrt{13}$, then the smallest angle of the triangle is

• A. $\pi /2$
• B. $\pi /3$
• C. $\pi /4$
• D. $\pi /6$

Answer 1

The correct option is D $\pi /6$

Sides of triangle$=7,4\sqrt{3},\sqrt{3}$

The smallest side$=\sqrt{1}3$

So,$\cos \theta =\frac{7^2+{\left(4\sqrt{3}\right)}^2-{\left(\sqrt{13}\right)}^2}{\mathrm{2.7.4}\sqrt{3}}=\frac{49+48-13}{14.4\sqrt{3}}=\frac{84}{14.4\sqrt{3}}=\frac{\sqrt{3}}{2}\cos \theta =\frac{\sqrt{3}}{2}\theta ={\cos }^{-1}\frac{\sqrt{3}}{2}=\frac{\pi }{6}$

Answer $D$