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A unit circle is centered at the origin.

 May 23, 2019
 #1
avatar+104069 
+1

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

 

cool cool cool

 May 23, 2019
 #2
avatar+104069 
+1

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

 

 

 

cool cool cool

 May 23, 2019

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