Question

Consider a traingle $PQR$ having sides of lengths $p,q$ and $r$ opposite to the angles $P,Q$ and $R,$ respectively. Then which of the following statements is(are) TRUE ?

• A. $\cos P\ge 1-\frac{p^2}{2qr}$
• B. $\cos R\ge \left(\frac{q-r}{p+q}\right)\cos P+\left(\frac{p-r}{p+q}\right)\cos Q$
• C. $\frac{q+r}{p}<2\frac{\sqrt{\sin Q\sin R}}{\sin P}$
• D. If $p<q$ and $p<r,$ then $\cos Q>\frac{p}{r}$ and $\cos R>\frac{p}{q}$

Answer 1

Answer: (A), (B)

$\cos R\ge \left(\frac{q-r}{p+q}\right)\cos P+\left(\frac{p-r}{p+q}\right)\cos Q$



$\cos P=\frac{q^2+r^2-p^2}{2qr}$ and $\frac{q^2+r^2}{2}\ge \sqrt{q^2\cdot r^2}(\because \text{AM}\ge \text{GM})$
$\Rightarrow q^2+r^2\ge 2qr$
So, $\cos P\ge \frac{2qr-p^2}{2qr}$
$\Rightarrow \cos P\ge 1-\frac{p^2}{2qr}$


$\frac{(q-r)\cos P+(p-r)\cos Q}{p+q}=\frac{q\cos P+p\cos Q-r(\cos P+\cos Q)}{p+q}$
Applying projection formulae
$\left[\begin{array}{c}p=q\cos R+r\cos Q\\ q=r\cos P+p\cos R\\ r=q\cos P+p\cos Q\end{array}\right]$

$=\frac{r(1-\cos P-\cos Q)}{p+q}=\frac{r-q+p\cos R-(p-q\cos R)}{p+q}=\frac{(r-p-q)+(p+q)\cos R}{p+q}$

$=\cos R+\frac{r-q-p}{p+q}\le \cos R(\because r<p+q)$


Apply sine rule
$\frac{p}{\sin P}=\frac{q}{\sin Q}=\frac{r}{\sin R}$
$\Rightarrow \frac{q+r}{p}=\frac{\sin Q+\sin R}{\sin P}\ge \frac{2\sqrt{\sin Q\cdot \sin R}}{\sin P}$


If $p<q$ and $q<r$
So, $p$ is the smallest side, therefore one of $Q$ or $R$ can be obtuse
So, one of $\cos Q$ or $\cos R$ can be negative
Therefore $\cos Q>\frac{p}{r}$ and $\cos R>\frac{p}{q}$ cannot hold always.