Trigonometry Examples

$(3, 60)$

Step 1

Convert from rectangular coordinates $(x, y)$ to polar coordinates $(r, θ)$ using the conversion formulas.

$r = \sqrt{x^{2} + y^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 2

Replace $x$ and $y$ with the actual values.

$r = \sqrt{{(3)}^{2} + {(60)}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3

Find the magnitude of the polar coordinate.

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Step 3.1

Raise $3$ to the power of $2$ .

$r = \sqrt{9 + {(60)}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2

Raise $60$ to the power of $2$ .

$r = \sqrt{9 + 3600}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.3

Add $9$ and $3600$ .

$r = \sqrt{3609}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.4

Rewrite $3609$ as $3^{2} \cdot 401$ .

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Step 3.4.1

Factor $9$ out of $3609$ .

$r = \sqrt{9 (401)}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.4.2

Rewrite $9$ as $3^{2}$ .

$r = \sqrt{3^{2} \cdot 401}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{3^{2} \cdot 401}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.5

Pull terms out from under the radical.

$r = 3 \sqrt{401}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = 3 \sqrt{401}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 4

Replace $x$ and $y$ with the actual values.

$r = 3 \sqrt{401}$

$θ = t a n^{- 1} (\frac{60}{3})$

Step 5

The inverse tangent of $20$ is $θ = 87.13759477 °$ .

$r = 3 \sqrt{401}$

$θ = 87.13759477 °$

Step 6

This is the result of the conversion to polar coordinates in $(r, θ)$ form.

$(3 \sqrt{401}, 87.13759477 °)$