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) = (8√(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) = (x)/(r)\\\\\cos(\theta) = (8)/(√(113))\\\\\cos(\theta) = (8√(113))/(√(113)*√(113))\\\\\cos(\theta) = (8√(113))/(√(113*113))\\\\\cos(\theta) = (8√(113))/(√(113^2))\\\\\cos(\theta) = (8√(113))/(113)\\\\`

Side note: cosine is positive in quadrant Q4.