Addison Trujillo
2022-07-01
Answered

What is the polar form of $(11,55)$ ?

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Jaelynn Cuevas

Answered 2022-07-02
Author has **16** answers

Step 1

$\mathrm{tan}\left(\theta \right)=\frac{\text{opposite}}{\text{adjacent}}\to \theta =\text{arctan}\left(\frac{\text{opposite}}{\text{adjacent}}\right)$

$\theta =\text{arctan}\left(\frac{55}{11}\right)=\text{arctan}\left(5\right)\approx 1.3734\left(\text{}radians\right)\phantom{\rule{1ex}{0ex}}\text{or}\phantom{\rule{1ex}{0ex}}{78.69}^{\circ}$

By Pythagorean Theorem:

$r=\sqrt{{11}^{2}+{55}^{2}}\approx 56.09$

$\mathrm{tan}\left(\theta \right)=\frac{\text{opposite}}{\text{adjacent}}\to \theta =\text{arctan}\left(\frac{\text{opposite}}{\text{adjacent}}\right)$

$\theta =\text{arctan}\left(\frac{55}{11}\right)=\text{arctan}\left(5\right)\approx 1.3734\left(\text{}radians\right)\phantom{\rule{1ex}{0ex}}\text{or}\phantom{\rule{1ex}{0ex}}{78.69}^{\circ}$

By Pythagorean Theorem:

$r=\sqrt{{11}^{2}+{55}^{2}}\approx 56.09$

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