need help with this problem

[abel muroi]

I was given this information..

(theta) is in standard position, point (6, -7) is on the terminal side of angle. Find cos (theta)


ok so the first things i noticed (that i understand) is that the angle that i am looking for is in standard position, so the initial side of angle is on the positive side of the x axis and the terminal side of the angle is on the 4th quadrant. so i know how the angle looks like, but i don't know how to find the reference angle or the actual number of degrees that the angle is

can anyone help me?

 

[Ishuda]

I was given this information..

(theta) is in standard position, point (6, -7) is on the terminal side of angle. Find cos (theta)


ok so the first things i noticed (that i understand) is that the angle that i am looking for is in standard position, so the initial side of angle is on the positive side of the x axis and the terminal side of the angle is on the 4th quadrant. so i know how the angle looks like, but i don't know how to find the reference angle or the actual number of degrees that the angle is

can anyone help me?

Do you know the Pythagorean Theorem and the definition of the trig functions, i.e.
sin(theta) = opposite over hypotenuse
etc.?

 

[abel muroi]

Do you know the Pythagorean Theorem and the definition of the trig functions, i.e.
sin(theta) = opposite over hypotenuse
etc.?

i remember now. I was supposed to use the formula r = squareroot x² + squareroot y²

so here is my work

r = x² + y²
r = 6² + (-7)²
r = 36 + 49
r = 85
r = squareroot 85


so now that i have found the radius/hypotenuse of the triangle, i can find cos theta

cos theta = 6/(squareroot 85)

did i get the correct answer?

 

[pka]

i remember now. I was supposed to use the formula r = squareroot x² + squareroot y²
so here is my work
r = x² + y²
r = 6² + (-7)²
r = 36 + 49
r = 85
r = squareroot 85
cos theta = 6/(squareroot 85), did i get the correct answer?

That is correct: \(\displaystyle \large\cos(\theta)=\dfrac{x}{r}\)

 

[Ishuda]

i remember now. I was supposed to use the formula r = squareroot x² + squareroot y²

so here is my work

r = x² + y²
r = 6² + (-7)²
r = 36 + 49
r = 85
r = squareroot 85


so now that i have found the radius/hypotenuse of the triangle, i can find cos theta

cos theta = 6/(squareroot 85)

did i get the correct answer?

You got the correct answer but your intermediate steps had some errors/typos. For example you should have had
r = squareroot (x² + y²)
which is what you actually did. Also, you should have had
r² = x² + y²
r² = 6² + (-7)²
...
which, again, is what you did.

The fact that the intermediate steps were wrong is fine as long as it is just your work and you are the only one who is going to be looking at it since you know what you meant. However, if you are going to show it to someone else or, possibly, will need to refer to it later, you need to get it right. In any case, good job in getting the answer.

 

[Steven G]

i remember now. I was supposed to use the formula r = squareroot x² + squareroot y²

so here is my work

r = x² + y²
r = 6² + (-7)²
r = 36 + 49
r = 85
r = squareroot 85


so now that i have found the radius/hypotenuse of the triangle, i can find cos theta

cos theta = 6/(squareroot 85)

did i get the correct answer?

Abel, I am big on equal signs being valid. You write r=85 and the next line you write r=sqrt85. Well, which is it?? At the top you wrote r = squareroot x² + squareroot y²

then you wrote r = x² + y² . What is that all about?
You say that you wanted the reference angle (which is not needed). Well you can say that arctan(-7/6) = the reference angle.
You must be precise!