Helpppppppp????

Question

1) Let y=- 6/7 be the y coordinate of the point p(x,y) where the terminal side of the angle 0(standard position) meets the unit circle if p is in III what is the Cos(0)

the work includes drawing the triangle labeling the triangle and showing any algebra work circle your final answer.opp,adj and hyp

2) in the triangle the length of side b is 7 the M<A=60 find the exact length of b and c expect means that you cannot give me decimal ma you will need to give me either an integer a fraction or a radical.

Answer 1

y=-6/7, is in quadrant III or IV

where x is negative or positive

x= -/-sqr(1- (6/7)^2) = -+/sqr(49/49-36/49)

=+/-sqr(13/49)

=+/-sqr(13)/7

hypotenuse squared = 36/49 +13/49 = 1

cosT = adjacent side over hypotenuse = x/1 = x

=sqr(13)/7 = about .5151= about 1/2

Answer 2

If the point (x, y) lies on the unit circle, then x² + y² = 1.

Since y = -6/7, we have x² = 1 - 36/49 = 13/49. So, x = ±√13 / 7. Since we are in quadrant 3, x < 0.

So, x = - √13 / 7

Therefore p = (- √13 / 7, -6/7) = (cosθ, sinθ). Thus, cosθ = - √13 / 7.

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I will assume that the triangle is a right triangle, where c is the hypotenuse, a is the side opposite the 30° angle, and b is the side opposite the 60° angle.

If a = 7, then c = 2a = 14 and b = √3a = 7√3.