Converting Polar Coordinates to Cartesian
The polar coordinates are defined in terms of \(r\) and \(\theta\), where \(r\) is the distance of the point from the origin and \(\theta\) is the angle made with the positive \(x\)-axis.
Clearly, using trigonometry, if the Cartesian coordinates are \((x,y),\) then
\[\begin{array} &x = r \cos \theta, &y = r \sin \theta. \end{array}\]
If \(3\) is the distance of a point from the origin and \(\frac{\pi}{3}\) is the angle made with the positive \(x\)-axis, then what will be the Cartesian coordinates of the point?
The Cartesian coordinates can be obtained from the distance \(3\) of the point from the origin and the angle \(\frac{\pi}{3}\) made with the positive \(x\)-axis. Using trigonometry, we have
\[ \begin{align} x &= 3\cos\frac{\pi}{3} \\ &= \frac{3}{2} \\ y &= 3\sin\frac{\pi}{3} \\ &= \frac{3\sqrt{3}}{2}. \end{align} \]
Thus, the Cartesian coordinates of the point are \(\displaystyle \left ( \frac{3}{2}, \frac{3\sqrt{3}}{2}\right ).\) \(_\square\)
Write the Cartesian coordinates \( (2, 2\sqrt{3}) \) as polar coordinates.
Let \(a\) denote the distance between the origin and the point \( (2, 2\sqrt{3} ),\) then
\[ a = \sqrt{ 2^2 + (2\sqrt{3})^2} = 4. \qquad (1) \]
Now, let \( \theta\) denote the angle made with the positive \(x\)-axis, then we have
\[ \begin{align} \sin \theta &= \frac{2\sqrt{3}}{4} = \frac{\sqrt{3}}{2} \\ \cos \theta &= \frac{2}{4} = \frac{1}{2}, \end{align} \]
which both imply \( \theta=\frac{\pi}{3} .\qquad (2)\)
Thus from \( (1)\) and \((2),\) the polar coordinates of the Cartesian coordinates \( (2, 2\sqrt{3} )\) are \[ (a, \theta) = \left(4, \frac{\pi}{3}\right). \ _\square \]
What are the Cartesian coordinates of the point whose distance from origin is \(4\) and the angle made with the positive \(x\)-axis is \( \frac{\pi}{4} ?\)
Since the distance of the point from the origin is \(4\) and the angle made with the positive \(x\)-axis is \(\frac{\pi}{4} ,\) the Cartesian coordinates are
\[ \begin{align} x &= 4\cos\frac{\pi}{4} = 2\sqrt{2} \\ y &= 4\sin\frac{\pi}{4} = 2\sqrt{2}. \end{align} \]
Thus, the answer is \( \left(2\sqrt{2}, 2\sqrt{2} \right).\) \(_\square\)
Write the Cartesian coordinates \( \left(\sqrt{3}, 1\right) \) as polar coordinates.
Let \(a\) denote the distance between the origin and the point \( \left(\sqrt{3}, 1 \right),\) then
\[ a = \sqrt{ (\sqrt{3})^2 + 1^2} = 2. \qquad (1) \]
Now, let \( \theta\) denote the angle made with the positive \(x\)-axis, then we have
\[ \begin{align} \sin \theta &= \frac{1}{2} \\ \cos \theta &= \frac{\sqrt{3}}{2}, \end{align} \]
which both imply \( \theta=\frac{\pi}{6} . \qquad (2)\)
Thus, from \( (1)\) and \((2),\) the polar coordinates of the point \( \left(\sqrt{3}, 1\right)\) in Cartesian plane are \[ (a, \theta) = \left(2, \frac{\pi}{6} \right). \ _\square \]
What are the Cartesian coordinates of the point whose distance from origin is \(10\) and the angle made with the positive \(x\)-axis is \( \frac{\pi}{2} ?\)
Since the distance of point from the origin is \(10\) and the angle made with the positive \(x\)-axis is \(\frac{\pi}{2} ,\) the Cartesian coordinates are
\[ \begin{align} x &= 10\cos\frac{\pi}{2} = 0 , \\ y &= 10\sin\frac{\pi}{2} = 10. \end{align} \]
Thus, the answer is \( (0, 10).\) \(_\square\)
ParallelAngleBisector
The radius of the above circle is \(10\) and the Cartesian coordinates of \(A\) is \( (5, 5\sqrt{3} ).\) If \( \angle AOB\) is \(60^\circ ,\) what are the Cartesian coordinates of \( B?\)
From the cartesian coordinates of \(A\) \( (5,5\sqrt{3}) ,\) we can get the angle between \(\overline {OA}\) and the positive \(x\)-axis using the polar coordinates of \(A.\)
Let \( r\cos \theta \) and \( r\sin \theta \) be the polar coordinates of \( A,\) then since \(r\) is \(10,\) the angle \( \theta \) is established as follows:
\[ \begin{align} 10 \cos \theta &= 5 \\ \cos \theta &= \frac{1}{2}, \\ 10 \sin \theta &= 5\sqrt{3} \\ \sin \theta &= \frac{\sqrt{3}}{2}, \\ \Rightarrow \theta &= 60^\circ. \end{align} \]
Since the angle between \(\overline {OA} \) and the positive \(x\)-axis is \( 60^\circ ,\) the angle between \( \overline {OB}\) and the positive \(x\)-axis is \( 60^\circ + 60^\circ = 120^\circ.\)
Hence, the Cartesian coordinates of \(B\) are
\[ \begin{align} x&= 10 \cos 120^\circ = 10\left ( -\frac{1}{2} \right ) = -5, \\ y&= 10 \sin 120^\circ = 10\left( \frac{\sqrt{3}}{2} \right ) = 5\sqrt{3}. \ _\square \end{align} \]
