Question

Consider a triangle $PQR$ 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)

(A) $\cos P=\frac{q^2+r^2-p^2}{2qr},q^2+r^2\ge 2qr$,
by A.M. $\ge$ G.M
$\Rightarrow \cos P\ge 1-\frac{p^2}{2qr}$
(B) By triangle inequality $q+p>r$
by projection formula
$r\cos P+p\cos R+q\cos R+r\cos Q>p\cos Q+q\cos P$
$\Rightarrow \cos R>\frac{(q-r)}{(p+q)}\cos P+\frac{(p-r)}{(p+q)}\cos Q$
So inequality is true (equality does not hold)
(C) By AM. $\ge$ GM.
$q+r\ge 2\sqrt{qr}$
$\frac{q+r}{p}\ge \frac{2\sqrt{qr}}{p}$
$\frac{\sin P}{p}=\frac{\sin Q}{q}=\frac{\sin R}{r}$ by sine rule
$\frac{q+r}{p}\ge \frac{2\sqrt{\sin Q\sin R}}{\sin P}$
(D) If $\cos Q>\frac{p}{r}\Rightarrow r\cos Q>p$ ... (i)
$\cos R>\frac{p}{q}\Rightarrow q\cos R>p\dots$ (ii)
Add (i) and (ii)
$r\cos Q+q\cos R>2P\Rightarrow p>2p$
$\Rightarrow p<0$, false