Question

If in a triangle $ABC,\theta$ is the angle determined by $\cos \theta =\frac{a-b}{c},$ then

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

Answer 1

Answer: (B), (C)

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

$1-{\sin }^2\theta =\frac{a^2+b^2-2ab}{c^2}\Rightarrow \frac{c^2-{(a-b)}^2}{c^2}={\sin }^2\theta$
$\Rightarrow \frac{c\sin \theta }{2\sqrt{ab}}=\sin \frac{C}{2}$