What is the angle that \vec{C} makes with the positive x axis?

[Paymemoney]

Homework Statement

Find the components of the vector $$\vec{C}$$ with length c = 1.00 and angle $$\phi = 30.0^{\circ}$$ as shown.

Homework Equations

$$opp=hypcos(\theta)$$
$$adj=hypcos(\theta)$$

The Attempt at a Solution

tried to use the above equations to answer the question, however when i looked in the answers it was incorrect.
Answer i got was x=0.866 and y=0.5

When i realized that it was between the vector and the y axis, i found that the angle didn't make contact with the x-axis, so my question is What is the angle that $$\vec{C}$$ makes with the positive x axis?

P.S

 

[tiny-tim]

Hi Paymemoney!

Homework Equations

$$opp=hypcos(\theta)$$
$$adj=hypcos(\theta)$$

erm … they can't both be right, can they?

When i realized that it was between the vector and the y axis, i found that the angle didn't make contact with the x-axis, so my question is What is the angle that $$\vec{C}$$ makes with the positive x axis?

Start at the positive x-axis, go round to the y-axis, then go round some more … that's your angle!

But the main thing you've done wrong is that the dotted line marked "x" on your diagram isn't x, is it?

 

[Paymemoney]

Hi Paymemoney!


erm … they can't both be right, can they?

oops, meant to be $$opp=hypsin(\theta)$$

Start at the positive x-axis, go round to the y-axis, then go round some more … that's your angle!

But the main thing you've done wrong is that the dotted line marked "x" on your diagram isn't x, is it?

ok so i got 600 degrees, what is the next step i should take to solve $$\vec{C}$$

 

[tiny-tim]

Hi Paymemoney!

(have a degree: º and a theta: θ and a phi: φ )

ok so i got 600 degrees …

hold it!

however did you get 600º ?

 

[Paymemoney]

well i went around the the axis 2 times o_o

 

[tiny-tim]

well i went around the the axis 2 times o_o

Still not 600º.

Anyway, just go from the x-axis to the y-axis anticlockwise, and then carry on to C.

 

[Paymemoney]

ok this is what i have done:

tell me if this is correct.

 

[tiny-tim]

Yes, that's exactly correct.

The "first quadrant" (top-right) is 0º to 90º, the "second quadrant" (top-left) is 90º to 180º, and so on.