it's the usual convention,

就表示矢量 A 的模长
if you're talking about the length of the vector A.

这是约定俗成的
First of all, it's clear from the Pythagoras' theorem

首先根据毕达哥拉斯定理
that a is the square root of (Ax)^2 + (Ay)^2.

显然有 A 等于 (Ax)^2 + (Ay)^2 的平方根
The angle θ that it makes with the x-axis

与 x 轴的夹角 θ
satisfies the condition tan is Ay over Ax..

满足其正切等于 Ay 除以 Ax
What this means--Now here's the main point.

这样的意义何在呢下面我要讲出重点
If you give me a pair of numbers, Ax and Ay,

如果给我一组数字 Ax 和 Ay
that's as good as giving me this arrow,

就相当于给了我这个箭头示意图
because I can find the length of the arrow by Pythagoras' theorem.

因为我可以利用毕达哥拉斯定理定理求出模长
And I can find the orientation of the arrow

我也可以求得矢量的方向
by saying the angle θ that satisfies tan is Ay over Ax.

通过求出满足正切等于 Ay 除以 Ax 的 θ角