#### Jomo

##### Elite Member
For example, if the red line is theta = 210o, then yes you are asked to find sin(210o), cos(210o) and tan(210o)

Please try this and see what you get. Using the blue reference will be helpful.

#### bushra1175

##### New member
I
For example, if the red line is theta = 210o, then yes you are asked to find sin(210o), cos(210o) and tan(210o)

Please try this and see what you get. Using the blue reference will be helpful.
Hi. I actually drew the graph myself so I can visualise the line segment, I just didn't know what they meant by 'angle made between the x axis and segment'. I was confused as to whether the angle starts from the first quadrant or third quadrant

#### Jomo

##### Elite Member
The angle starts on the positive x-axis. Just like the red angle you drew.

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#### bushra1175

##### New member
The angle starts on the positive x=axis. Just like the red angle you drew.
Okay I see. Thanks. That's that I wanted to know

#### Dr.Peterson

##### Elite Member
Here is the question:View attachment 21631

Does this mean find the trig ratios of the blue line or the red line?

View attachment 21630
The problem is poorly stated. The red is the angle "in standard position", and should be described as "between the positive x-axis and the ray". The blue is the reference angle, which is properly described as "between the ray and the x-axis". I think it's most likely, as others have said, that they mean the former, but it's possible they are asking about the reference angle.

#### pka

##### Elite Member
I must disagree that the problem is poorly stated. It is clearly stated: $$\theta$$ is the angle formed by the point $$(-3,-4)$$ and the positive $$x$$-axis.
Now the point $$(-3,-4)$$ determines a line segment of length $$r=\sqrt{(-3)^2+(-4)^2}=5$$ .
$$\sin(\theta)=\dfrac{x}{r}=\dfrac{-3}{5},~\cos(\theta)=\dfrac{y}{r}=\dfrac{-4}{5},~\&~\tan(\theta)=\dfrac{y}{x}=\dfrac{-4}{-3}$$ .,