Question

the terminal side of 0 passes through the point (8,-7) what is the exact value of cos 0 in simplified form​

Answer 1

Answer:  $\cos(\theta) = \frac{8\sqrt{113}}{113}$

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Work Shown:

x^2+y^2 = r^2

(8)^2+(-7)^2 = r^2

113 = r^2

r = sqrt(113)

The distance from (0,0) to (8,-7) is exactly sqrt(113) units.

This is the exact length of the hypotenuse of the right triangle.

Next, we do the following steps:

$\cos(\text{angle}) = \frac{\text{adjacent}}{\text{hypotenuse}}\\\\\cos(\theta) = \frac{x}{r}\\\\\cos(\theta) = \frac{8}{\sqrt{113}}\\\\\cos(\theta) = \frac{8\sqrt{113}}{\sqrt{113}*\sqrt{113}}\\\\\cos(\theta) = \frac{8\sqrt{113}}{\sqrt{113*113}}\\\\\cos(\theta) = \frac{8\sqrt{113}}{\sqrt{113^2}}\\\\\cos(\theta) = \frac{8\sqrt{113}}{113}$

Side note: cosine is positive in quadrant Q4.