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Mumz [18]
3 years ago
7

Point b has coordinates (-8,15) and lies on the circle whose equation is x^2+y^2=289. If an angles is drawn in standard position

with its terminal ray extending through point b, what is the cosine of the angle?
Mathematics
1 answer:
Alenkinab [10]3 years ago
8 0

Answer:

\cos \theta=-\dfrac{8}{17}

Step-by-step explanation:

Coordinates of Point b=(-8,15)

b lies on the circle whose equation is x^2+y^2=289

x^2+y^2=17^2

Comparing with the general form a circle with center at the origin: x^2+y^2=r^2

The radius of the circle =17 which is the length of the hypotenuse of the terminal ray through point b.

For an angle drawn in standard position through point b,

x=-8 which is negative

y=15 which is positive

Therefore, the angle is in Quadrant II.

\cos \theta=\dfrac{Adjacent}{Hypotenuse} \\$Adjacent=-8\\Hypotenuse=17\\\cos \theta=\dfrac{-8}{17} \\\cos \theta=-\dfrac{8}{17}

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