Question

How do you convert the polar equation $\theta ={{30}^{\circ }}$ into rectangular form?

Answer 1

Hint: Consider the given polar equation and take the tangent function both the sides and use the value $\tan {{30}^{\circ }}=\dfrac{1}{\sqrt{3}}$ in the R.H.S to simplify. Now, consider the relation between the polar equation and Cartesian equation given by the relations: - \[y=r\sin \theta \] and $x=r\cos \theta $. Divide the first relation by the second relation, cancel the common terms and use the relation $\dfrac{\sin \theta }{\cos \theta }=\tan \theta $ to find the value of the tangent function in terms of x and y coordinate. Finally, find the relation between x and y to get the rectangular form of the equation.

Complete step by step answer:
Here, we have been provided with the polar equation $\theta ={{30}^{\circ }}$ and we are asked to convert it into the rectangular form, that means we have to obtain the relationship between x and y – coordinate.
Let us consider a point (P) with Cartesian coordinates P (x, y) and polar coordinates \[\left( r,\theta \right)\], where r is called the radius vector of the point P, then the relation between the two-coordinate system is given as:
$y=r\sin \theta ..................\left( i \right)$
$x=r\cos \theta .................\left( ii \right)$
Dividing equation (i) by equation (ii) and cancelling the common factors, we get,
\[\begin{align}
  & \Rightarrow \dfrac{y}{x}=\dfrac{r\sin \theta }{r\cos \theta } \\
 & \Rightarrow \dfrac{y}{x}=\dfrac{\sin \theta }{\cos \theta } \\
\end{align}\]
Using the conversion: $\dfrac{\sin \theta }{\cos \theta }=\tan \theta $, we get,
$\Rightarrow \dfrac{y}{x}=\tan \theta ..............\left( iii \right)$
Now, let us come to the polar equation provided to us. So, we have,
$\Rightarrow \theta ={{30}^{\circ }}$
Taking tangent function both the sides and using the value $\tan {{30}^{\circ }}=\dfrac{1}{\sqrt{3}}$, we get,
$\begin{align}
  & \Rightarrow \tan \theta =\tan {{30}^{\circ }} \\
 & \Rightarrow \tan \theta =\dfrac{1}{\sqrt{3}} \\
\end{align}$
Substituting the value of $\tan \theta $ from equation (iii) in the above relation, we get,
$\Rightarrow \dfrac{y}{x}=\dfrac{1}{\sqrt{3}}$
By cross-multiplication we get,
$\begin{align}
  & \Rightarrow \sqrt{3}y=x \\
 & \Rightarrow x=\sqrt{3}y \\
\end{align}$
Hence, the above relation represents the given equation in rectangular form.

Note:
One may note that if we are asked which type of curve the obtained equation represents then we can say that it represents a straight line with slope equal to $\sqrt{3}$ and y-intercept equal to 0. You must remember the relationship between the Cartesian coordinates and the polar coordinates to solve the above question. Also, remember the important relation: - \[{{x}^{2}}+{{y}^{2}}={{r}^{2}}\] as it will be used further in the coordinate geometry of circles.