SOLUTION: I cant solve these problem's would like some help. 1.If cos 0= -4/5,find sec 0. 2.Find the value of sin 0 for angle 0 in standard position of the point (-2,-4)that lies on the

Question 173043: I cant solve these problem's would like some help.
1.If cos 0= -4/5,find sec 0.
2.Find the value of sin 0 for angle 0 in standard position of the point (-2,-4)that lies on the terminal side.
3.Suppose 0 is an angle in standard position whose terminal side lies in quadrant 2 and the tan 0=-1.Find the remaining trigonometric functions for the function 0.
4.Use the right triangle ABC to find side a if A=37 and b=9''.
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
I cant solve these problem's would like some help.
1.If cos 0= -4/5,find sec 0.
sec = 1/cos
sec = -5/4
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2.Find the value of sin 0 for angle 0 in standard position of the point (-2,-4)that lies on the terminal side.
The radius = sqrt(2^2 + 4^2)
r = 2sqrt(5)
sin = y/r = -4/(2sqrt(5))
sin = -2sqrt(5)/5
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3.Suppose 0 is an angle in standard position whose terminal side lies in quadrant 2 and the tan 0=-1.Find the remaining trigonometric functions for the function 0.
tan = sin/cos = y/x
In quadrant 2, y is + and x is negative and = absolute value
r = 1
x^2 + y^2 = 1 and x = -y
2x^2 = 1
x = -sqrt(2)/2 = cos
y = +sqrt(2)/2 = sin
The other 3 are the inverses.
4.Use the right triangle ABC to find side a if A=37 and b=9'
Angle B = 90-37 = 53�
Law of Sines:
a/sin(A) = b/sin(B)
a = 9*sin(37)/sin(53)
a =~ 6.782
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BTW, the plural of problem is problems, no apostrophe.

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