How do I convert polar coordinates $\left( 6,60{}^\circ \right)$ to rectangular coordinates?

Hint: Like the rectangular coordinates, the polar coordinates are also a method for determining the position of a point in a plane. The rectangular coordinates are represented by the ordered pair $\left( x,y \right)$ whereas the polar coordinates are represented by the ordered pair $\left( r,\theta \right).$ These different coordinates are related by the relations, $x=r\cos \theta $ and $y=r\sin \theta .$

Complete step by step answer:
Let us consider the given data.
We are given with the polar coordinates $\left( 6,60{}^\circ \right).$
We know that like the rectangular coordinates, the polar coordinates are also a method of determining the position of a point in a plane.
And the general representation of the polar coordinate is $\left( r,\theta \right).$
In this ordered pair, $r$ is the distance of the point from the origin and $\theta $ is the angle made by the line joining the point and the origin with the $x-$axis.
Now, we can find the corresponding rectangular coordinates, represented by the ordered pair $\left( x,y \right)$ using the relations $x=r\cos \theta $ and $y=r\sin \theta .$
Here, in our case, $r=6$ and $\theta =60{}^\circ .$
Let us substitute these values in the relations written above to get the corresponding rectangular coordinates.
So, first let us find the coordinate $x$ using the relation $x=r\cos \theta .$
We will get $x=6\cos 60{}^\circ .$
We know that $\cos 60{}^\circ =\dfrac{1}{2}.$
So, $x=6\times \dfrac{1}{2}=3.$
Similarly, let us find the coordinate $y$ using the relation $y=r\sin \theta .$
We will get $y=6\sin 60{}^\circ .$
We have learnt that $\sin 60{}^\circ =\dfrac{\sqrt{3}}{2}.$
This will give us the value as $y=6\times \dfrac{\sqrt{3}}{2}=3\sqrt{3}.$

Hence the rectangular coordinates corresponding to the polar coordinates $\left( 6,60{}^\circ \right)$ are $\left( 3,3\sqrt{3} \right).$

Note: We know that $x=r\cos \theta $ and $y=r\sin \theta .$ Now, we will get ${{x}^{2}}+{{y}^{2}}={{r}^{2}}.$ This can be proved as ${{r}^{2}}{{\cos }^{2}}\theta +{{r}^{2}}{{\sin }^{2}}\theta ={{r}^{2}}\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right)={{r}^{2}}\times 1={{r}^{2}}.$ Also, we will take care of all types of calculation mistakes while solving the question.