Question

If the sides of the triangle are $p,q,\sqrt{\left.(p^2+q^2+pq\right)}$, then the greatest angle is,

• A. $\frac{\mathrm{\pi }}{2}$
• B. $\frac{5\mathrm{\pi }}{4}$
• C. $\frac{2\mathrm{\pi }}{3}$
• D. $\frac{7\mathrm{\pi }}{4}$

Answer 1

The correct option is C

$\frac{2\mathrm{\pi }}{3}$

Explanation of correct option.

The formula to calculate the angle is,

$\cos B=\frac{a^2+c^2-b^2}{2ac}$ where $b$ is the greatest side.

Now, out of the three sides $\sqrt{(p^2+q^2+pq}$ is the greatest side,

Therefore,

Find $\cos \left(\theta \right)$ as,

$$\begin{gathered} \cos \left(\theta \right)=\frac{p^2+q^2-{\left(\sqrt{\left.(p^2+q^2+pq\right)}\right)}^2}{2pq} \\ =\frac{{\left(p^2+q^2-p^2-q^2-pq\right)}^{}}{2pq} \\ =-\frac{1}{2} \\ \Rightarrow \theta ={\cos }^{-1}(-1/2) \\ =\frac{2\mathrm{\pi }}{3} \end{gathered}$$

Hence, Option (C) is correct.