IN the trapezium ABCD, the diagonals intercept at M. Let AM= a, BM= b, Cm = c and DM = d, and let Angle AMB be $\theta$.
a=6
b=4
c=3
d=2
AB=8
DC=4
$\cos(\theta) = -\frac{1}{4}$ and $\sin (\theta) = \frac{(15^{0.5})}{4}$ Find the area of the trapezium in exact form.
It's obvious that $\angle AMB + \angle AMD = 180^{\circ}$, so we have: $\angle AMD = 180^{\circ} - \theta$. Now we want to find the cosine of $\angle AMD$.
$$\cos {\angle AMD} = \cos ({180^{\circ} - \theta}) = \cos {180^{\circ}} \cdot \cos {\theta} + \sin {180^{\circ}} \cdot \sin {\theta}$$
We know that $\sin {180^{\circ}} = 0$ and $\cos {180^{\circ}} = -1$. So we substitute:
$$\cos {\angle AMD} = (-1)\cdot(-\frac 14) + 0 \cdot \frac{\sqrt{15}}{4} = \frac 14$$
Note that also $\cos {\angle BMC} = \frac 14$
Now apply the law of cosine on $\triangle AMD$ and $\triangle BMC$.
$$AD^2 = AM^2 + DM^2 - 2\cdot AM \cdot DM \cdot \cos {\angle AMD}$$ $$AD^2 = 6^2 + 2^2 - 2\cdot 6\cdot 2 \cdot \frac 14$$ $$AD^2 = 36 + 4 - 6 = 34 \implies AD = \sqrt{34}$$
$$BC^2 = BM^2 + CM^2 - 2\cdot BM \cdot CM \cdot \cos {\angle BMC}$$ $$BC^2 = 4^2 + 3^2 - 2 \cdot 4 \cdot 3 \cdot \frac 14$$ $$BC^2 = 16 + 9 - 6 = 19 \implies BC =\sqrt{19}$$
Now just apply Heron's formula for area of the triangle on $\triangle ACD$ and $\triangle BAC$. The area of the trapezium will be the sum of the areas of these two triangles. You can do this by yourself.
The final answer should be $\approx 26.14$ or in exact form $\frac{27 \cdot \sqrt{15}}{4}$