Question

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

• A. $\frac{\mathrm{\pi }}{6}$
• B. $\frac{\mathrm{\pi }}{3}$
• C. $\frac{\mathrm{\pi }}{4}$
• D. $\frac{\mathrm{\pi }}{5}$

Answer 1

The correct option is A

$\frac{\mathrm{\pi }}{6}$

Explanation for the correct option:

Find the required value.

Given: sides of a triangle are respectively $7cm,4\sqrt{3}cm$ and $\sqrt{13}cm$.

Here smallest side is $\sqrt{13}$.

We know that $\cos C=\frac{(a^2+b^2–c^2)}{2ab}$

$$\begin{gathered} \cos C=\frac{\left(7^2+{\left(4\sqrt{3}\right)}^2–{\left(\sqrt{13}\right)}^2\right)}{2\times 7\times 4\sqrt{3}} \\ \cos C=\frac{\left(49+48-13\right)}{56\sqrt{3}} \\ \cos C=\frac{84}{56\sqrt{3}} \\ \cos C=\frac{84}{56\sqrt{3}} \\ \cos C=\frac{\sqrt{3}}{2} \\ C={\cos }^{-1}\frac{\sqrt{3}}{2} \\ C=\frac{\mathrm{\pi }}{6} \end{gathered}$$

Therefore, the smallest angle of given triangle is $\frac{\mathrm{\pi }}{6}$.

Hence, option $A$ is the correct answer.