Answers:
[tex]\csc(\theta) = \frac{7\sqrt{33}}{33}\\\\\sin(\theta) = \frac{\sqrt{33}}{7}\\\\\cot(\theta) = \frac{4\sqrt{33}}{33}\\\\[/tex]
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Explanation:
Before we can find the trig ratios, we need to determine the length of the missing side. Apply the pythagorean theorem
[tex]a^2+b^2 = c^2\\\\4^2+b^2 = 7^2\\\\16+b^2 = 49\\\\b^2 = 49-16\\\\b^2 = 33\\\\b = \sqrt{33}[/tex]
The vertical side is [tex]\sqrt{33}[/tex] units long.
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The side we just found is the opposite side to angle [tex]\theta[/tex] (greek letter theta)
The hypotenuse is 7 units long. The hypotenuse is always opposite the right angle.
Divide the hypotenuse over opposite to get the cosecant.
[tex]\csc(\theta) = \frac{\text{hypotenuse}}{\text{opposite}}\\\\\csc(\theta) = \frac{7}{\sqrt{33}}\\\\\csc(\theta) = \frac{7\sqrt{33}}{\sqrt{33}*\sqrt{33}}\\\\\csc(\theta) = \frac{7\sqrt{33}}{\sqrt{33*33}}\\\\\csc(\theta) = \frac{7\sqrt{33}}{\sqrt{33^2}}\\\\\csc(\theta) = \frac{7\sqrt{33}}{33}\\\\[/tex]
Note: In the third step, I multiplied top and bottom by sqrt(33) to rationalize the denominator.
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Recall that sine is the reciprocal of cosecant, and vice versa.
So we'll use the same idea as before, but flip the original fraction.
[tex]\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}\\\\\sin(\theta) = \frac{\sqrt{33}}{7}\\\\[/tex]
The denominator is already a rational number, so there's not much else to do here.
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Cotangent is the ratio of adjacent over opposite
[tex]\cot(\theta) = \frac{\text{adjacent}}{\text{opposite}}\\\\\cot(\theta) = \frac{4}{\sqrt{33}}\\\\\cot(\theta) = \frac{4\sqrt{33}}{\sqrt{33}*\sqrt{33}}\\\\\cot(\theta) = \frac{4\sqrt{33}}{\sqrt{33*33}}\\\\\cot(\theta) = \frac{4\sqrt{33}}{\sqrt{33^2}}\\\\\cot(\theta) = \frac{4\sqrt{33}}{33}\\\\[/tex]
The steps are pretty much nearly identical as the section involving csc, except we're using the adjacent in place of the hypotenuse (so that means we use 4 in place of 7).
