{arccos{[(y-x)²+1600-(x+y)²]/[80(y-x)]}-arccos{[1600+(y-x)²-(y-x)²]/[80(y-x)]}=45°

首先，将角度转换为弧度：

$$45^\circ = \frac{\pi}{4}$$

然后，对方程两边进行化简：

$$\begin{aligned} & \arccos{\left(\frac{(y-x)^2+1600-(x+y)^2}{80(y-x)}\right)} - \arccos{\left(\frac{1600+(y-x)^2-(y-x)^2}{80(y-x)}\right)} = \frac{\pi}{4} \\ \Rightarrow & \arccos{\left(\frac{(y-x)^2+1600-x^2-y^2-2xy}{80(y-x)}\right)} - \arccos{\left(\frac{1600}{80(y-x)}\right)} = \frac{\pi}{4} \\ \Rightarrow & \arccos{\left(\frac{(y-x)^2-2xy}{80(y-x)}\right)} - \arccos{\left(\frac{20}{\sqrt{y-x}}\right)} = \frac{\pi}{4} \\ \Rightarrow & \cos{\left(\arccos{\left(\frac{(y-x)^2-2xy}{80(y-x)}\right)} - \arccos{\left(\frac{20}{\sqrt{y-x}}\right)}\right)} = \cos{\frac{\pi}{4}} \\ \Rightarrow & \cos{\left(\arccos{\left(\frac{(y-x)^2-2xy}{80(y-x)}\right)}\right)}\cos{\left(\arccos{\left(\frac{20}{\sqrt{y-x}}\right)}\right)} + \sin{\left(\arccos{\left(\frac{(y-x)^2-2xy}{80(y-x)}\right)}\right)}\sin{\left(\arccos{\left(\frac{20}{\sqrt{y-x}}\right)}\right)} = \frac{\sqrt{2}}{2} \\ \Rightarrow & \frac{(y-x)^2-2xy}{80(y-x)}\cdot\frac{20}{\sqrt{(y-x)^2-400}} + \sqrt{1-\frac{(y-x)^2-2xy}{6400}}\cdot\frac{y-x}{\sqrt{(y-x)^2-400}} = \frac{\sqrt{2}}{2} \\ \end{aligned}$$

这是一个关于 $y-x$ 的方程，可以通过数值计算或求解器求解。