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

Based on the information given in the question, a diagram is constructed where BM is the median of AC. Since the question deals with areas, we construct BM as the altitude for the triangle. $$\angle\ BDA=\theta\ $$ (given) and $$\angle\ BAD=\alpha\ $$ is assumed. The following figure is obtained. 

BM is the height, and so $$\angle\ BMA=\angle\ BMD=90^{\circ\ }$$

So, in $$\triangle\ BAM$$,

 $$\sin\alpha\ =\frac{BM}{AB}=>\ BM=c\ \sin\alpha\ $$

$$\cos\ \alpha\ =\frac{AM}{AB}=>AM=c\cos\alpha\ $$

Again, in $$\triangle\ BDM$$,

$$\sin\theta\ =\frac{BM}{BD}=>\ BM=k\ \sin\theta\ $$

$$\cos\ \theta\ =\frac{DM}{BD}=>DM=k\cos\theta\ $$

Now, area of $$\triangle\ ABD$$ can be calculated as $$\dfrac{1}{2}\times\ AD\times\ BM=\frac{1}{2}\times\ \left(AM+BM\right)\times\ BM$$

Putting the values obtained we have, 

$$ar\triangle\ ABD=\frac{1}{2}\times\ \left(c\ \cos\alpha\ +k\ \cos\theta\ \right)\times\ k\ \sin\theta\ $$ ________(A)

Now, $$c\ \sin a=\ k\ \sin\theta\ =>\sin\ \alpha\ =\dfrac{k\sin\theta\ }{c}\ $$

So, $$\cos\ \alpha\ =\sqrt{\ 1-\left(\frac{k\sin\theta\ }{c}\ \right)^2}=\dfrac{\sqrt{\ c^2-k^2\sin^2\theta\ }}{c}$$

Substituting this in (A) we have 

$$ar\triangle\ ABD=\frac{1}{2}\times\ \left(\sqrt{\ c^2-k^2\sin^2\theta\ }\ +k\ \cos\theta\ \right)\times\ k\ \sin\theta\ $$

or, $$ar\triangle\ ABD=\frac{1}{2}\left(\ k\sin\theta\ \sqrt{c^2-k^2\sin^2\theta\ }\ +k^2\ \cos\theta\sin\theta\ \right)$$

or, $$ar\triangle\ ABD=\frac{1}{2}\left(\ k\sin\theta\ \sqrt{c^2-k^2\sin^2\theta\ }\ +\frac{k^2}{2}\times\ \ 2\cos\theta\sin\theta\ \right)$$

or, $$ar\triangle\ ABD=\dfrac{1}{2}\left(\ k\sin\theta\ \sqrt{c^2-k^2\sin^2\theta\ }\ +\dfrac{k^2}{2}\sin2\theta\ \right)$$

Now, since BD is the median, this means the area of $$ar\triangle\ ABC=2\ ar\triangle\ ABD$$

so , $$ar\triangle\ ABC=\ k\sin\theta\ \sqrt{c^2-k^2\sin^2\theta\ }\ +\dfrac{k^2}{2}\sin2\theta\ $$ , which is option B.