2022-06-28

What is the polar form of $\left(-11,24\right)$ ?

Trey Ross

Step 1
The radius is given by the Pythagorean Theorem as
$\text{XXX}r=\sqrt{{\left(-11\right)}^{2}+{24}^{2}}=\sqrt{597}$
$\mathrm{tan}\left(\theta \right)=\frac{24}{-11}$
Unfortunately, $\mathrm{arctan}$ , the inverse of $\mathrm{tan}\left(\theta \right)$ is by definition an angle
$\in \left(-\frac{\pi }{2},+\frac{\pi }{2}\right)$
(i.e. in QI or Q IV)
In the diagram above, the angle returned by $\mathrm{arctan}\left(-\frac{24}{11}\right)$ is shown as ${\theta }_{2}$
Taking $\mathrm{arctan}\left(\frac{24}{-11}\right)=\mathrm{arctan}\left(\frac{-24}{11}\right)$
gives us an angle that is "pi" radians less than the true value of $\theta$
Therefore $\theta =\pi +\mathrm{arctan}\left(-\frac{24}{11}\right)$
$\left(r,\theta \right)=\left(\sqrt{597},\pi +\mathrm{arctan}\left(-\frac{24}{11}\right)\right)$
(using a calculator)
$\text{XXX}\approx \left(24.4,2.0\right)\approx \left(24.4,{114.6}^{\circ }\right)$

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