Question

Consider a triangle $P Q R$ 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. $\cos P \geq 1-\frac{ p ^{2}}{2 qr }$
• B. $\cos R \geq\left(\frac{q-r}{p+q}\right) \cos P+\left(\frac{p-r}{p+q}\right) \cos Q$
• 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) $\cos P =\frac{ q ^{2}+ r ^{2}- p ^{2}}{2 qr }, q ^{2}+ r ^{2} \geq 2 qr$,
by A.M. $\geq$ G.M
$\Rightarrow \cos P \geq 1-\frac{ p ^{2}}{2 qr }$
(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. $\geq$ GM.
$q+r \geq 2 \sqrt{q r} $
$\frac{q+r}{p} \geq \frac{2 \sqrt{q r}}{p}$
$\frac{\sin P}{p}=\frac{\sin Q}{q}=\frac{\sin R}{r} $ by sine rule
$\frac{q+r}{p} \geq \frac{2 \sqrt{\sin Q \sin R}}{\sin P}$
(D) If $\cos Q > \frac{p}{r} \Rightarrow r \cos Q>p$ ... (i)
$\cos R>\frac{p}{q} \Rightarrow q \cos R>p \ldots$ (ii)
Add (i) and (ii)
$r \cos Q+q \cos R>2 P \Rightarrow p>2 p$
$\Rightarrow p <0$, false