The side AB of a triangle ABC is c. The median BD is of length k. If $$\angle BDE = \theta$$ and $$\theta < 90^\circ$$, then the area of triangle ABC is

A

$$\frac{k^2 \sin \theta}{2} + k \sin \theta \sqrt{c^2 + k^2 \sin^2 \theta}$$

B

$$\frac{k^2 \sin 2 \theta}{2} + k \sin \theta \sqrt{c^2 - k^2 \sin^2 \theta}$$

C

$$\frac{k^2 \cos 2 \theta}{2} + k \sin \theta \sqrt{c^2 - k^2 \sin^2 \theta}$$

D

$$\frac{k^2 \cos \theta}{2} + k \sin \theta \sqrt{c^2 + k^2 \sin^2 \theta}$$