Question

If the area of triangle $\triangle A B C$ is $a^{2}-(b-c)^{2}$ then its circumradius $R=$

• A. $a / 16 \operatorname{cosec}^{2}(A / 2)$
• B. $(C / 16) \sin ^{2}(C / 2)$
• C. $a / 6 \sin ^{2}(A / 2)$
• D. $b / 16 \sin ^{2}(B / 2)$

Answer 1

Answer: (A)

$\Delta=a^{2}-(b-c)^{2}=a^{2}-b^{2}-c^{2}+2 b c$
$\cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}$
$=\frac{2 b c-\Delta}{2 b c}=1-\frac{\Delta}{2 b c}$
$=1-\frac{a b c}{4 R \times 2 b c}=1-\frac{a}{8 R}$
$\Rightarrow 8 R=\frac{a}{1-\cos A}=\frac{a}{2 \sin ^{2}(A / 2)}$
$\Rightarrow R=\left(\frac{a}{16}\right) \text{cosec} 2\left(\frac{A}{2}\right)$