Find the Cosecant Given the Point ( square root of 5,2)

$(\sqrt{5}, 2)$

To find the $\csc (θ)$ between the x-axis and the line between the points $(0, 0)$ and $(\sqrt{5}, 2)$ , draw the triangle between the three points $(0, 0)$ , $(\sqrt{5}, 0)$ , and $(\sqrt{5}, 2)$ .

Opposite : $2$

Adjacent : $\sqrt{5}$

Find the hypotenuse using Pythagorean theorem $c = \sqrt{a^{2} + b^{2}}$ .

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Rewrite ${\sqrt{5}}^{2}$ as $5$ .

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Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{5}$ as $5^{\frac{1}{2}}$ .

$\sqrt{{(5^{\frac{1}{2}})}^{2} + {(2)}^{2}}$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\sqrt{5^{\frac{1}{2} \cdot 2} + {(2)}^{2}}$

Combine $\frac{1}{2}$ and $2$ .

$\sqrt{5^{\frac{2}{2}} + {(2)}^{2}}$

Cancel the common factor of $2$ .

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Cancel the common factor.

$\sqrt{5^{\frac{2}{2}} + {(2)}^{2}}$

Rewrite the expression.

$\sqrt{5^{1} + {(2)}^{2}}$

Evaluate the exponent.

$\sqrt{5 + {(2)}^{2}}$

Simplify the expression.

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

$\sqrt{5 + 4}$

Add $5$ and $4$ .

$\sqrt{9}$

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

$\sqrt{3^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$3$

$\csc (θ) = \frac{Hypotenuse}{Opposite}$ therefore $\csc (θ) = \frac{3}{2}$ .

$\frac{3}{2}$

Approximate the result.

$\csc (θ) = \frac{3}{2} \approx 1.5$

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