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

The Correct Option is A, B

(A) \(\cos P = \frac{{q^2 + r^2 - p^2}}{{2qr}}\)

And \(\frac{{q^2 + r^2}}{2} \geq \sqrt{q^2 \cdot r^2} \quad \text{(AM} \geq \text{GM)}\)

\(⇒\) \(q^2 + r^2 \geq 2qr\)

So \(\cos P \geq \frac{{2qr - p^2}}{{2qr}}\)

\(\cos P \geq 1 - \frac{{p^2}}{{2qr}}\)

(B) \(\frac{{(q-r) \cos P + (p-r) \cos Q}}{{p+q}} = \frac{{(q \cos P + p \cos Q) - r(\cos P + \cos Q)}}{{p+q}}\)

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

=\(\frac{{r(q-p \cos R) - (p-q \cos R)}}{{p+q}}\)

= \(\frac{{(r-p-q) + (p+q)\cos R}}{{p+q}}\)

= \(\cos R + \frac{{r-q-p}}{{p+q}} \leq \cos R\)

(C) \(\frac{{q+r}}{p} = \frac{{\sin Q + \sin R}}{{\sin P}} \geq 2\sqrt{\frac{{\sin Q \sin R}}{{\sin P}}}\)

(D) If \(p < q \quad \text{and} \quad q < r\)

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, \(\cos Q > \frac{p}{r} \quad \text{and} \quad \cos R > \frac{p}{q} \text{ cannot hold always.}\)