Tangent of Angle in Cartesian Plane

Theorem
Let $P = \tuple {x, y}$ be a point in the cartesian plane whose origin is at $O$.

Let $\theta$ be the angle between the $x$-axis and the line $OP$.

Let $r$ be the length of $OP$.

Then:
 * $\tan \theta = \dfrac y x$

where $\tan$ denotes the tangent of $\theta$.

Proof

 * TangentCartesian.png

Let a unit circle $C$ be drawn with its center at the origin $O$.

Let a tangent line be drawn to $C$ parallel to $PS$ meeting $C$ at $R$.

Let $Q$ be the point on $OP$ which intersects this tangent line.

$\angle OSP = \angle ORQ$, as both are right angles.

Both $\triangle OSP$ and $\triangle ORQ$ share angle $\theta$.

By Triangles with Two Equal Angles are Similar it follows that $\triangle OSP$ and $\triangle ORQ$ are similar.

Thus:

Then:

When $\theta$ is obtuse, the same argument holds, but both $x$ and $\tan \theta$ are negative.

When $\theta = \dfrac \pi 2$ we have that $x = 0$.

Then $OP$ is parallel to the tangent line at $R$ which it therefore does not meet.

Thus when $\theta = \dfrac \pi 2$, it follows that $\tan \theta$ is not defined.

Likewise $\dfrac y x$ is not defined when $x = 0$.

Thus the relation holds for $\theta = \dfrac \pi 2$.

When $\pi < \theta < 2 \pi$ the diagram can be reflected in the $x$-axis.

In this case, $y$ is negative.

Thus the relation continues to hold.

When $\theta = 0$ and $\theta = \pi$ we have that $y = 0$ and $\tan \theta = 0 = \dfrac y x$.

Hence the result.