#### bushra1175

##### New member

- Joined
- Jun 14, 2020

- Messages
- 44

- Thread starter bushra1175
- Start date

- Joined
- Jun 14, 2020

- Messages
- 44

- Joined
- Jun 14, 2020

- Messages
- 44

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 quadrantFor example, if the red line is theta = 210^{o}, then yes you are asked to find sin(210^{o}), cos(210^{o}) and tan(210^{o})

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

- Joined
- Jun 14, 2020

- Messages
- 44

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

- Joined
- Nov 12, 2017

- Messages
- 10,191

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.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

- Joined
- Jan 29, 2005

- Messages
- 10,731

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.Here is the question:View attachment 21631

Now the point \((-3,-4)\) determines a line segment of length \(r=\sqrt{(-3)^2+(-4)^2}=5\) .

\(\sin(\theta)=\dfrac{x}{r}=\dfrac{-3}{5},~\cos(\theta)=\dfrac{y}{r}=\dfrac{-4}{5},~\&~\tan(\theta)=\dfrac{y}{x}=\dfrac{-4}{-3}\) .,