I am trying to help grand daughter express this is rectangular form:
thetha = pi/3
If $\displaystyle \displaystyle \begin{align*} \theta = \frac{\pi}{3} \end{align*}$ then you have a ray starting at the origin extending out infinitely long making this angle with the positive x-axis.
The Cartesian (rectangular) form for the equation of a line is $\displaystyle \displaystyle y = mx + c$ where m is the gradient and c is the y intercept. c is clearly 0. As for m, it can be calculated using $\displaystyle \displaystyle \begin{align*} m = \tan{(\theta)} = \tan{\left( \frac{\pi}{3} \right)} = \sqrt{3} \end{align*}$.
So the rectangular form of this equation is $\displaystyle \displaystyle \begin{align*} y = \sqrt{3}\,x \end{align*}$ where $\displaystyle \displaystyle \begin{align*} x \geq 0 \end{align*}$.
Hello, dgable!
$\displaystyle \text{Express in rectangular form: }\:\theta \,=\,\tfrac{\pi}{3}$
There are a few identities for converting polar to rectangular:
. . $\displaystyle \begin{array}{ccccccccc}r^2 &=&x^2+y^2 &\quad& r\cos\theta &=& x \\ \tan\theta &=&\dfrac{y}{x} && r\sin\theta &=& y \end{array}$
$\displaystyle \text{We are given: }\qquad\;\;\theta \:=\:\tfrac{\pi}{3}$
$\displaystyle \text{Take tangents: }\,\tan\theta \:=\:\tan\left(\tfrac{\pi}{3}\right)$
$\displaystyle \text{Substitute: }\qquad\quad\,\frac{y}{x} \:=\:\sqrt{3}$
Therefore: . . . . . $\displaystyle y \:=\:\sqrt{3}\,x$