In a $\triangle PQR$, $\cos P + \cos Q + \cos R = 2 - \cos Q$, find $\frac{p + r}{q}$ where $p$, $q$, $r$ are side lengths

Question

In a $\triangle PQR$, $\cos P + \cos Q + \cos R = 2 - \cos Q,$
find $\dfrac{p + r}{q}$ (where $p$, $q$, $r$ are side lengths of $\triangle PQR)$.

For solving this question I was given 4 minutes. I tried using cosine law, but equation started becoming more difficult.

There must be a shorter way to find $\frac{p + r}{q}$, but I can't get it.

I also tried $$\cos P + \cos R = 2(1 - \cos Q)$$ $$\implies 2\cos(\frac{P + R}{2})\cos(\frac{P - R}{2}) = 4 \sin^2\frac{Q}{2}$$ $$\implies \cos(\frac{P - R}{2}) = 2\sin\frac{Q}{2}$$ but I have no idea how to get the required ratio.

Answer 1

Your second approach is on the right track. We can continue from there.

Multiply the last equation by $2\cos\frac Q2$.

$$2\cos\frac Q2\cos\left(\frac{P-R}2\right)=4\sin\frac Q2\cos\frac Q2$$ $$2\sin\left(\frac{P+R}2\right)\cos\left(\frac{P-R}2\right)=2\sin Q$$

Note that this is

$$\sin P+\sin R=2\sin Q$$ $$\frac{\sin P+\sin R}{\sin Q}=2$$

According to sin rule, we can write,

$$\frac{p+r}q=2$$ $\square$