Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively. Let $R$ be the circumradius of the triangle. Then the value of $2R$ is equal to?

• A. $\frac{b^2-c^2}{a\sin (B-C)}$
• B. $\frac{c^2-a^2}{b\sin (C-A)}$
• C. $\frac{a^2-b^2}{c\sin (A-B)}$
• D. $\frac{a}{\sin A}$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $\frac{b^2-c^2}{a\sin (B-C)}$
B $\frac{c^2-a^2}{b\sin (C-A)}$
C $\frac{a^2-b^2}{c\sin (A-B)}$
D $\frac{a}{\sin A}$

We know that
$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$
Hence, option D is correct.
Now, consider option A,
$\frac{b^2-c^2}{asin(B-C)}$
$=\frac{4R^2({\sin }^{2}B-{\sin }^{2}C)}{2R\sin A\sin (B-C)}$
$=\frac{2R\sin (B+C)\sin (B-C)}{\sin (B+C)\sin (B-C)}$
$=2R$
Now, taking option B,
$\frac{c^2-a^2}{bsin(C-A)}$
$=\frac{4R^2({\sin }^{2}C-{\sin }^{2}A)}{2R\sin B\sin (C-A)}$
$=\frac{2R\sin (C+A)\sin (C-A)}{\sin (C+A)\sin (C-A)}$
$=2R$
Now, taking option C,
$\frac{a^2-b^2}{csin(A-B)}$
$=\frac{4R^2({\sin }^{2}A-{\sin }^{2}B)}{2R\sin C\sin (A-B)}$
$=\frac{2R\sin (A+B)\sin (A-B)}{\sin (A+B)\sin (A-B)}$
$=2R$