+129852 7) Since we have a unit circle, the form is x^2 + y^2 = 1
So....we have that
x^2 + (0.6)^2 = 1
x^2 + .36 = 1
x^2 = 1 -.36
x^2 = .64
Since we are in Q2, x is negative so we need the negative value of the square root
x = -sqrt (.64) = -.0.8
The tangent = y / x = 0.6/ -0.8 = - 60/80 = -3/4
+129852 10 ) Using the same idea as in (7)....we have that
(sqrt(2)/3)^2 + y^2 = 1
(2/9) + y^2 = 1
y^2 = 1 - 2/9
y^2 = 7/9
In Q$....y is negative....so we need the negative square root
y = -sqrt(7)/3
So
tan θ = y/x = -sqrt(7)/ 3 -sqrt (7)
_________ = ________ rationalize the denominator and we get
sqrt(2) / 3 sqrt (2)
-sqrt (7) sqrt (2) -sqrt (14)
_____________ = _________
sqrt(2) sqrt (2) 2
