Question

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?

Answer 1

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.

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