Question

Prove that the polar coordinates $(0,0),(3,\frac{\pi }{2})$, and $(3,\frac{\pi }{6})$ form an equilateral triangle.

Answer 1

Let us convert polar co-ordinates into cartesian form,

(i) $(0,0)$

(ii) $r=3$ and $\theta =\frac{\pi }{2}$

$x=r\cos \theta$

$=3\cos \frac{\pi }{2}$

$=0$

And,

$y=r\sin \theta$

$=3\sin \frac{\pi }{2}$

$=3$

$(x,y)=(0,3)$

(iii) $r=3$ and $\theta =\frac{\pi }{6}$

$x=r\cos \theta$
$=3\cos \frac{\pi }{6}$

$=\frac{3\sqrt{3}}{2}$

$y=r\sin \theta$

$=3\sin \frac{\pi }{6}$

$=\frac{3}{2}$

$(x,y)=(\frac{3\sqrt{3}}{2},\frac{3}{2})$

Now, we have points $A(0,0),B(0,3)$ and $C(\frac{3\sqrt{3}}{2},\frac{3}{2})$

Now, we need to prove that $AB=BC=AC.$

$\begin{array}{cc}AB& =\sqrt{{\left(0-0\right)}^2+{\left(0-3\right)}^2}\\ & =\sqrt{0+9}\\ & =3\\ & \\ BC& =\sqrt{{\left(\frac{3\sqrt{3}}{2}-0\right)}^2+{\left(\frac{3}{2}-3\right)}^2}\\ & =\sqrt{{\left(\frac{3\sqrt{3}}{2}\right)}^2+{\left(-\frac{3}{2}\right)}^2}\\ & =\sqrt{\frac{27}{4}+\frac{9}{4}}\\ & =\sqrt{\frac{36}{4}}\\ & =\sqrt{9}\\ & =3\\ & \\ AC& =\sqrt{{\left(\frac{3\sqrt{3}}{2}-0\right)}^2+{\left(\frac{3}{2}-0\right)}^2}\\ & =\sqrt{{\left(\frac{3\sqrt{3}}{2}\right)}^2+{\left(\frac{3}{2}\right)}^2}\\ & =\sqrt{\frac{27}{4}+\frac{9}{4}}\\ & =\sqrt{\frac{36}{4}}\\ & =\sqrt{9}\\ & =3\end{array}$

Hence proved $AB=BC=CA$ which means that $△ABC$ is an equilateral triangle.