Friday

July 25, 2014

July 25, 2014

Posted by **Jon** on Sunday, February 10, 2008 at 5:13pm.

A)40

B)80

C)320

D)400

I chose A

(for the next one I dont know what its called but it looks like 0 with a horizontal line through the middle.)

2)Find the exact value of cos looks like 0 with a horizontal line through the middle if the terminal side of 0 with a horizontal line through the middle in standard postion contains the point (6,-8).

A)-4/5

B)3/5

C)4/5

D)-3/5

I chose B

- Trigonometric functions -
**Damon**, Sunday, February 10, 2008 at 5:26pmplus or minus all the way around or 360 changes nothing, so add 360 first

-400 + 360 = -40

well usually standard is counterclockwise from the x axis so

-40 = 360 -40 = 320

I would choose C

2,

That is called THETA but I will call it T

It is in the fourth quadrant between 270 and 360

the hypotenuse of this 3,4,5 triangle = 2*5 = 10

the cosine = adjacent/hypotenuse = +6/10==+3/5

B = agree

- Trigonometric functions -
**Alex**, Sunday, February 10, 2008 at 5:26pmC for the first one. All of them aside from 320 lie in Quadrant 1. 320 and -400 lie in Quadrant 4. Since 400 is negative it moves clockwise, when the other angles move counterclockwise.

I'm not sure what you're asking in the 2nd. The exact value of cos theta with a horizontal line through the middle of the angle?

- Trigonometric functions -
**Reiny**, Sunday, February 10, 2008 at 5:34pmRemember a negative angle is a clockwise rotation, so a -400º angle goes all the way around and down another 40º into the fourth quadrant.

Using the standard notation that would be an angle of 320º

for your second question you are probably looking at cosθ, or cos theta.

Theta is often used as the variable to represent an angle and is one of the Greek letters.

So draw a triangle by joining the origin to the point (6,-8), which is the fourth quadrant

cosθ = adjacent/hypotenuse

hypotenuse = √(6^2 + (-8)^2) = 10

adjacent = 6

so cosθ = 6/10 = 3/5

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