Here's an interesting fact:
If $A$, $B$, $C$ are angles with $A+B+C=180^\circ$, then $\sin A$, $\sin B$, $\sin C$ are the lengths of the sides of a (not the!) triangle with angles $A$, $B$, $C$.
Specifically, the "$\sin A$-$\sin B$-$\sin C$" triangle is the one inscribed in a circle of diameter $1$; you might consider it a fundamental representative of the family of triangles with angles $A$, $B$, $C$, but it is not the only triangle with those angles. You can get from the representative to any other member of the family by magnifying the side-lengths (and the circumdiameter) by some factor, $m$; conversely, any member of the family has side-lengths that are multiples of the side-lengths of the representative (since the family members are all similar).
A triangle has angles $A$, $B$, $C$ if and only if its side-lengths are $m\sin A$, $m\sin B$, $m \sin C$ for some (positive) $m$.
This is, in fact, exactly what the Law of Sines tells you:
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = m$$
where $m$ is that scale factor ... and also the diameter of the triangle's circumcircle.
In the case of a right triangle with hypotenuse $1$, recall Thales' Theorem:
A right angle is inscribed in a semi-circle.
Thus, an hypotenuse-$1$ right triangle is inscribed in a circle whose diameter matches that hypotenuse: in other words,
A unit-hypotenuse right triangle is the fundamental representative of the "$A$-$B$-$90^\circ$" family.
Such a triangle's sides are indeed $\sin A$, $\sin B$ (which we also call "$\cos A$", since $A$ and $B$ are complementary), and $\sin 90^\circ$ (which, of course, is $1$). Perhaps this is what you were getting-at in your comment to your question.
Introduce the second altitude as shown.
Use the areas of the same triangle but with different altitudes to calculate the ratio. (Or doing the above in the reverse order.) The proofing process will then be much simpler.
For example, red $= c \cos \theta$ and area $= \dfrac {red \cdot a}{2}$.
Best Answer
Clearly the greatest angle is opposite to the greatest side. Use Cosine Rule to get $$\begin{aligned}(1+\sin \alpha\cos\alpha)&=\sin^2\alpha+\cos^2\alpha-2\sin\alpha\cos\alpha\cos x\\ \dfrac{\sin\alpha\cos\alpha+1-1}{-2\sin\alpha\cos\alpha}&=\cos x\\ \cos x&=\dfrac{-1}{2}\implies x=\dfrac{2\pi}{3}=120^{\circ}\end{aligned}$$