Question
Find two pairs of polar coordinates, one pair with $r>0$ and the other pair with $r<0$, for the point whose Cartesian coordinates are given. In each case, choose $\theta$ so that $0 \leq m^{\circ}(\theta)<360$.$(6,-2 \sqrt{3})$
Step 1
The point is \((6, -2\sqrt{3})\). Step 2: Calculate the radius \(r\) using the formula \(r = \sqrt{x^2 + y^2}\), where \(x = 6\) and \(y = -2\sqrt{3}\). \[ r = \sqrt{6^2 + (-2\sqrt{3})^2} = \sqrt{36 + 4 \cdot 3} = \sqrt{36 + 12} = \sqrt{48} = 4\sqrt{3} \] Step Show more…
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5-6 The Cartesian coordinates of a point are given. (i) Find polar coordinates $(r, \theta)$ of the point, where $r>0$ and $0 \leqslant \theta<2 \pi$. (ii) Find polar coordinates $(r, \theta)$ of the point, where $r<0$ and $0 \leqslant \theta<2 \pi$. (a) $(\sqrt{3},-1)$ (b) $(-6,0)$
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