Question

If the sides of a triangle are $p,q$ and $\sqrt{p^2+pq+q^2}$, then its largest angle is

• A. $\pi /2$
• B. $2\pi /3$
• C. $5\pi /4$
• D. $7\pi /4$
• E. $5\pi /3$

Answer 1

Answer: (A)

$\pi /2$

$p,q,\sqrt{p^2+pq+q^2}$

$\cos \theta =\frac{p^2+q^2-(\sqrt{p^2+pq+q^2}{)}^2}{2pq}$

$=\frac{\text{/}{p}^2+\text{/}{q}^2-\text{/}{p}^2-pq-\text{/}{q}^2}{2pq}=\frac{-\text{/}pq}{2\text{/}pq}$

$=-\frac{1}{2}\Rightarrow \cos \theta =-\frac{1}{2}$

$\theta ={120}^o$

$\Rightarrow 2\frac{\pi }{3}$.