Question

In a $∆\mathrm{ABC}$, $2\mathrm{acsin}\left(\frac{\mathrm{A}-\mathrm{B}+\mathrm{C}}{2}\right)$ is equal to

• A. $a^2+b^2-c^2$
• B. $c^2+a^2-b^2$
• C. $b^2-c^2-a^2$
• D. $c^2-a^2-b^2$

Answer 1

The correct option is B

$c^2+a^2-b^2$

Explanation for the correct option:

Finding the value of $2\mathrm{ac}·\sin \left(\frac{\mathrm{A}-\mathrm{B}+\mathrm{C}}{2}\right)$:

In the triangle, the sum of all interior angles is $180°$. So,

$$\begin{gathered} A+B+C=180° \\ \Rightarrow A+C=180°-B \end{gathered}$$

Substitute the value of $A+C$ in $2\mathrm{ac}·\sin \left(\frac{\mathrm{A}-\mathrm{B}+\mathrm{C}}{2}\right)$, then

$\begin{array}{rcl}2\mathrm{ac}·\sin \left(\frac{\mathrm{A}-\mathrm{B}+\mathrm{C}}{2}\right)& =& 2\mathrm{ac}·\sin \left(\frac{180°-2\mathrm{B}}{2}\right)\\ & =& 2\mathrm{ac}·\sin \left(90°-\mathrm{B}\right)\\ & =& 2\mathrm{ac}·\mathrm{cosB}\left[\because \sin \left(90°-A\right)=\mathrm{cosA}\right]\\ & =& 2\mathrm{ac}\left(\frac{{\mathrm{a}}^2+{\mathrm{c}}^2-{\mathrm{b}}^2}{2\mathrm{ac}}\right)\left[\mathrm{by}\mathrm{cosine}\mathrm{rule}\right]\\ & =& {\mathrm{a}}^2+{\mathrm{c}}^2-{\mathrm{b}}^2\end{array}$

Hence, the correct option is B.