Question

If cos A + cos B+ cos C = $\sqrt{2}$, then $\frac{r}{R}$ (ratio of inradius (r)

and R is circumradius) is a root of the equation –

• A. $\left( \frac{r}{R} \right)^{2}$ – 2 $\frac{r}{R}$+ $\frac{1}{2}$= 0
• B. $\left( \frac{r}{R} \right)^{2}$ – 2 $\frac{r}{R}$– 1 = 0
• C. $\left( \frac{r}{R} \right)^{2}$ + 2 $\frac{r}{R}$– 1 = 0
• D. $\left( \frac{r}{R} \right)^{2}$ + 3 $\frac{r}{R}$– 1 = 0

Answer 1

Answer: (C)

$\left( \frac{r}{R} \right)^{2}$ + 2 $\frac{r}{R}$– 1 = 0

Answer 2

cos A + cos B + cos C = $\sqrt{2}$

⇒ 1 + 4 sinA/2 sin B/2 sinC/2 = $\sqrt{2}$

⇒ 1 + $\frac{r}{R}$ = $\sqrt{2}$⇒ $\frac{r}{R}$ = $\sqrt{2}$ – 1