Question

If in a triangle $ABC$, $\theta$ is the angle determined by $cos\theta =(a-b)/c$,

then which of the following is correct?

• A. $\frac{(a+b)\sin \theta }{2\sqrt{ab}}=\cos \frac{A-B}{2}$
• B. $\frac{(a+b)\sin \theta }{2\sqrt{ab}}=\cos \frac{A+B}{2}$
• C. $\frac{c\sin \theta }{2\sqrt{ab}}=\cos \frac{A-B}{2}$
• D. $\frac{c\sin \theta }{2\sqrt{ab}}=\cos \frac{A+B}{2}$

Answer 1

Answer: (A), (D)

$\frac{c\sin \theta }{2\sqrt{ab}}=\cos \frac{A+B}{2}$
$\frac{(a+b)\sin \theta }{2\sqrt{ab}}=\cos \frac{A-B}{2}$

$\cos \theta =\left(\frac{a-b}{c}\right)\sin \theta =\sqrt{1-\frac{(a-b{)}^2}{c^2}}=\frac{1}{c}\sqrt{c^2-(a-b{)}^2}$

(A) $\frac{(a+b)\sin \theta }{2\sqrt{ab}}\Rightarrow \frac{(a+b)}{2\times \sqrt{ab}\times c}\sqrt{c^2-(a-b{)}^2}\Rightarrow \frac{(a+b)}{2\sqrt{ab}\times c}\sqrt{(a^2+b^2-c^2)+2ab}$

=$\frac{(a+b)}{2\sqrt{ab}\times c}\sqrt{2ab(1-\cos c)}\Rightarrow \frac{(a+b)}{\sqrt{2}\times c}\times \sqrt{2}\sin \left(\frac{c}{2}\right)$

=$\frac{\sin A+\sin B}{\sin C}\times \sin \left(\frac{c}{2}\right)$

=$\frac{2\sin \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)}{2\sin \left(\frac{c}{2}\right)\cos \left(\frac{c}{2}\right)}\sin \left(\frac{c}{2}\right)$

=$\cos \left(\frac{A-B}{2}\right)$

Hence, $\frac{(a+b)\sin \theta }{2\sqrt{ab}}$=$\cos \left(\frac{A-B}{2}\right)$ is correct.

(D) $\frac{c\sqrt{c^2-(a-b{)}^2}}{2\sqrt{ab}}$ $\Rightarrow \frac{c\times \sqrt{2ab(1-\cos c)}}{2\sqrt{ab}\times c}$ $\Rightarrow \sin \frac{c}{2}\Rightarrow \cos \frac{A+B}{2}$

Hence, $\frac{c\sin \theta }{2\sqrt{ab}}$ =$\cos \frac{A+B}{2}$ is correct.