Answer:
[tex]\cos \theta=-\dfrac{8}{17}[/tex]
Step-by-step explanation:
Coordinates of Point b[tex]=(-8,15)[/tex]
b lies on the circle whose equation is [tex]x^2+y^2=289[/tex]
[tex]x^2+y^2=17^2[/tex]
Comparing with the general form a circle with center at the origin: [tex]x^2+y^2=r^2[/tex]
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.
[tex]\cos \theta=\dfrac{Adjacent}{Hypotenuse} \\$Adjacent=-8\\Hypotenuse=17\\\cos \theta=\dfrac{-8}{17} \\\cos \theta=-\dfrac{8}{17}[/tex]
