trigonometry: why this is not a triangle? 
As the photo shown, angle x doesn't nesscarry to be 90 degree.But when y=0, x=90 degree.
So here is my problem, what happen if I increase y, does x becomes larger (greater than 90) or less than 90 degree?

My attempt: 
My reasoning is that when you increase y (say set y=1). Then the base (12-y) becomes shorter, and the height (5+y) becomes taller. So that you can either swing (5+y) clockwise or counter clockwise. Therefore, angle x could be either greater than 90 or less than 90. But the solution says I am wrong. 
What are some intuitive way to explain this without using fancy equations?

Suppose that $y$ is a little more than $0$, so that the base is a little less than $12$. What height would be needed in order to have a right triangle with base $12-y$ and hypotenuse $13$? The Pythagorean theorem says that is would have to be
$$\sqrt{13^2-(12-y)^2}=\sqrt{25+24y-y^2}\;.$$
Is this more or less than $5+y$? If it’s more, $x$ must be more than $90°$: the $5+y$ side must tip over to the left as the $13$ side drops down. If it’s less, $x$ must be less than $90°$: the $5+y$ side must tip to the right as the $13$ side rises.
Now $(5+y)^2=25+10y+y^2$, so the question is whether $25+24y-y^2$ is more or less than $25+10y+y^2$, i.e., whether $14y-y^2$ is more or less than $y^2$. 


*

*Clearly $14y-y^2<y^2$ if and only if $14y<2y^2$, or $7y<y^2$. We need only consider $y$ such that $0<y<12$ (why?), so this occurs if $y>7$.

*It’s also clear that $14y-y^2>y^2$ if and only if $14y>2y^2$, or $7y>y^2$, which for $0<y<12$ occurs when $y<7$.
Thus, if we increase $y$ just a bit from $0$, we’ll be in the second case above, and $x$ will be more than $90°$. However, when $y=7$ the two sides will have changed places, and we’ll have a right triangle again. And as $y$ increases from $7$ towards (but not reaching) $12$, we’ll be in the first case above, and $x$ will be less than $90°$.
You can also use the cosine law, which says that
$$\begin{align*}
13^2&=(12-y)^2+(5+y)^2-2(12-y)(5+y)\cos x\\
&=169-14y+2y^2-2(60+7y-y^2)\cos x\;,
\end{align*}$$
i.e.,
$$(60+7y-y^2)\cos x=y^2-7y\;.$$
Moreover, the geometry makes it clear that $0\le y<12$, so $60+7y-y^2=(12-y)(5+y)\ne 0$. Thus,
$$\cos x=\frac{y^2-7y}{60+7y-y^2}=\frac{y(y-7)}{(12-y)(5+y)}\;.$$
You can now do a sign analysis. We know that $x=90°$ when $y=0$, so we’ll consider only $0<y<12$. On that interval $y>0$ and $12-y>0$, so the algebraic sign of $\cos x$ is the same as that of $\frac{y-7}{y+5}$. This is negative for $0<y<7$, $0$ at $y=7$, and positive for $7<y<12$. Thus, when you increase $y$ just a little from $0$, $\cos x$ becomes negative, and that can happen only if $x>90°$. When $y$ reaches $7$, however, $\cos x=0$, and we have the original right triangle, but with the base and height interchanged. And when $7<y<12$, $\cos x>0$, and we must have $x<90°$.
