Θlies in quadrant ii . cosθ=- what is the exact value of sinθ in simplified form?

sin theta = sqrt(57) /11

Step-by-step explanation:

cos theta = -8/11

cos = opposite/ hypotenuse =x/ (sqrt(x^2+y^2))

sin = adjacent /hypotenuse= y /sqrt (x^2+y^2)

-8/11 = x/ (sqrt(x^2+y^2))

x = -8

sqrt((x^2+y^2)) =11

11^2 = x^2 + y^2

121 = 64+ y^2

57 = y^2

sqrt(57) = y

sin theta = sqrt(57) /11

If it lies in quadrant II

x is negative and y is positive

answer: yes i agree that you should not put it on/p>

step-by-step explanation: the reason for this is that it wastes other peoples time and also make them fail as well and that is never good.