Answer
$$\sin105^\circ=\frac{\sqrt2+\sqrt6}{4}$$
2 should be matched with A.
Work Step by Step
$$\sin 105^\circ$$
$105^\circ$ can be rewritten as the sum of $45^\circ$ and $60^\circ$.
$$105^\circ=45^\circ+60^\circ$$
Therefore, $$\sin105^\circ=\sin(45^\circ+60^\circ)$$
Now we apply the sine difference identity, which states
$$\sin(A+B)=\sin A\cos B+\cos A\sin B$$
That makes
$$\sin105^\circ=\sin 45^\circ\cos60^\circ+\cos45^\circ\sin60^\circ$$
$$\sin105^\circ=\frac{\sqrt2}{2}\frac{1}{2}+\frac{\sqrt2}{2}\frac{\sqrt3}{2}$$
$$\sin105^\circ=\frac{\sqrt2}{4}+\frac{\sqrt6}{4}$$
$$\sin105^\circ=\frac{\sqrt2+\sqrt6}{4}$$
That means we match 2 with A.
