<s>[INST] Right triangle inscribed in a square. Find the square area? I hope it's valid to ask for "a more neat solution" of a problem on this network, despite the fact that I don't have a strict definition of the word "neat".
Here is the square and the right triangle inscribed in it.

I did the following:
$$AC = Ah + hB$$
$$4\sin\theta = 4\cos\theta + 3\sin\theta$$
So
$$\tan\theta = 4$$
But
$$\sin\theta = \frac{\tan\theta}{\sqrt{1+\tan^2\theta}}$$
Therefore
$$AC = 4\sin\theta = \frac{16}{\sqrt {17}}$$
$$\text{Area} = \left(\frac{16}{\sqrt {17}}\right)^2$$
 [/INST] I think your way is good, but we don't need to find $\tan\theta$.
From $4\sin\theta=4\cos\theta+3\sin\theta$, we have
$$\sin\theta=4\cos\theta$$
Squaring the both sides gives
$$\sin^2\theta=16(1-\sin^2\theta)$$
from which we can have
$$\sin^2\theta=\frac{16}{17}\quad\Rightarrow\quad \text{(area)}=16\sin^2\theta=\frac{16^2}{17}$$
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