The line with equation a + 2b = 0 coincides with the terminal side of an angle θ in standard position and cos θ<0.
What is the value of sinθ?
−2√5/5
√5
−1/2
√5/5
1a + 2b = 0
2b = -1a
b = (-1/2)a
The slope of this line is (-1/2)....this line lies in the II and IV quadrants...and since the cos < 0, then this must be a II quadrant angle so the sine must > 0
So.....find r = √ [ x^2 + y^2 ] = √ ] (-2)^2 + (1)^2 ] = √5
So....the sinθ = 1/ √5 = √5 / 5