Find the Cosecant Given the Point ( square root of 7,3)

$(\sqrt{7}, 3)$

To find the $\csc (θ)$ between the x-axis and the line between the points $(0, 0)$ and $(\sqrt{7}, 3)$ , draw the triangle between the three points $(0, 0)$ , $(\sqrt{7}, 0)$ , and $(\sqrt{7}, 3)$ .

Opposite : $3$

Adjacent : $\sqrt{7}$

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

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Rewrite ${\sqrt{7}}^{2}$ as $7$ .

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Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{7}$ as $7^{\frac{1}{2}}$ .

$\sqrt{{(7^{\frac{1}{2}})}^{2} + {(3)}^{2}}$

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

$\sqrt{7^{\frac{1}{2} \cdot 2} + {(3)}^{2}}$

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

$\sqrt{7^{\frac{2}{2}} + {(3)}^{2}}$

Cancel the common factor of $2$ .

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Cancel the common factor.

$\sqrt{7^{\frac{2}{2}} + {(3)}^{2}}$

Rewrite the expression.

$\sqrt{7^{1} + {(3)}^{2}}$

Evaluate the exponent.

$\sqrt{7 + {(3)}^{2}}$

Simplify the expression.

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

$\sqrt{7 + 9}$

Add $7$ and $9$ .

$\sqrt{16}$

Rewrite $16$ as $4^{2}$ .

$\sqrt{4^{2}}$

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

$4$

$\csc (θ) = \frac{Hypotenuse}{Opposite}$ therefore $\csc (θ) = \frac{4}{3}$ .

$\frac{4}{3}$

Approximate the result.

$\csc (θ) = \frac{4}{3} \approx 1.\overline{3}$

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