I have no idea how to solve this, as it is kinda beyond what I have already learned. It goes like this:
$M$ is the diagonal's point of intersection (or center I guess) in the parallelogram $ABCD$. We know the angles $BMC = 80°$ and $MBC = 40°$, we also know that $BC = 10$. We need to find $AB$, the area, the diagonals and the angles of the parallelogram.
This is probably a rather simple problem but I'd really appreciate your help, thanks!
I am not quite sure this is the ideal way to go, but by the sine rule: $$\frac{MC}{\sin 40^{\circ}}=\frac{MB}{\sin 60^{\circ}}=\frac{10}{\sin 80^{\circ}}$$
This gives, $$AC=2MC=\frac{\sin 40^{\circ}\cdot10 \cdot 2}{\sin 80^{\circ}}=\frac{10}{\cos 40^{\circ}}$$ And similarly, $$BD=2BM=\frac{\sin 60^{\circ}\cdot10 \cdot 2}{\sin 80^{\circ}}=\frac{10\sqrt3}{\sin 80^{\circ}}$$
Now we have the diagonals, area follows,
$$\text{Area}=\frac12\cdot AC\cdot BD\cdot\sin80^{\circ}=\frac{50\sqrt3}{\cos 40^{\circ}}$$
Taking $\angle MBA=\alpha$, we have:
$$\frac{MB^2}{\text{Area}}=\cot100^{\circ}+\cot\alpha$$
You can solve it to get the angles. And then the sides follow quite easily. Or alternately you can solve:
$$\frac12 \sin(\alpha+40)\cdot AB \cdot BC= \sin\alpha\cdot AB \cdot MB$$ $$5 \sin(\alpha+40)= \sin\alpha\cdot\frac{10\sqrt3}{\sin 80^{\circ}}$$