Question

In a triangle ABC, AD is the median. And $\angle DAC=\theta$, then $\sin \theta =$

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

Answer 1

The correct option is A $\frac{a\sin C}{\sqrt{2b^2+2c^2-a^2}}$

$AD=\sqrt{2b^2+2c^2-a^2}$

Apply sine rule to $△ADC$

$\frac{AD}{\sin C}=\frac{\frac{a}{2}}{\sin \theta }$

$\sin \theta =\frac{a\sin C}{2\sqrt{2b^2+2c^2-a^2}}$