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?

Respuesta :

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]