Question

Find the area of the triangle formed by the points with polar coordinates $(1,\frac{\pi }{6}),(2,\frac{\pi }{3}),(3,\frac{\pi }{2})$.

Answer 1

With polar coordinates $(a,\theta )$ the point is $(a\cos \theta ,a\sin \theta )$ in $X-Y$ coordinates

So the points $(1,\frac{\pi }{6}),(2,\frac{\pi }{3}),(3,\frac{\pi }{2})$

$=(1\cos 30°,1\sin 30°),(2\cos 60°,2\sin 60°),(3\cos 90°,3\sin 90°)$

$=(\frac{\sqrt{3}}{2},\frac{1}{2}),(1,\sqrt{3}),(0,3)$

Area of triangle of vertices $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ is

$\frac{1}{2}\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|$

$\frac{1}{2}\left|\begin{array}{cc}{x}_1-x_2& y_1-y_2\\ x_2-x_3& y_2-y_3\end{array}\right|$

$=\frac{1}{2}\left|\begin{array}{cc}\frac{\sqrt{3}}{2}-1& \frac{1}{2}-\sqrt{3}\\ 1-0& \sqrt{3}-3\end{array}\right|$

$=\frac{1}{2}|(\frac{\sqrt{3}}{2}-1)(\sqrt{3}-3)-1(\frac{1}{2}-\sqrt{3})|$

$=\frac{1}{2}(\frac{3}{2}-\sqrt{3}-\frac{3\sqrt{3}}{2}+3-\frac{1}{2}+\sqrt{3})$

$=\frac{1}{2}(4-\frac{3\sqrt{3}}{2})=(2-\frac{3\sqrt{3}}{4})$ sq. units