# CAST Diagram

https://www.quora.com/profile/Bravewarrior/p-144899143
Here is the question and its solution. I need help with part b. If A is reflex wouldn't A be negative? And if B is acute then wouldn't that stay positive? So just plug in negative A and keep B positive in tan(A+B) and get 7/17? Then do tan inverse of that and get 22.38?
Original post by pigeonwarrior
https://www.quora.com/profile/Bravewarrior/p-144899143
Here is the question and its solution. I need help with part b. If A is reflex wouldn't A be negative? And if B is acute then wouldn't that stay positive? So just plug in negative A and keep B positive in tan(A+B) and get 7/17? Then do tan inverse of that and get 22.38?

If tan is positive, then the angle must lie in Q1 or Q3. If the angle is reflex so >180, then A could be either 180...270 or -90...-180 as theyre the same (Q3). It just depends on whether you think of cast diagram/angles as being 0..360 or -180..180.
(edited 6 months ago)
Original post by mqb2766
If tan is positive, then the angle must lie in Q1 or Q3. If the angle is reflex so >180, then A could be either 180...270 or -90...-180 as theyre the same (Q3). It just depends on whether you think of cast diagram/angles as being 0..360 or -180..180.

I guess I'm finally beginning to understand it more😅 . Thank you!
Original post by pigeonwarrior
I guess I'm finally beginning to understand it more😅 . Thank you!

One tip for understanding tan a bit more is that it, tan(theta), corresponds to the gradient of the line (hypotenuse of the unit right triangle). So obviously a line with positive gradient must lie in both Q1 and Q3 and the angles associated with the line are theta in Q1 and theta+180 in Q3. Its a straight line through the origin, so the two angles are 180 apart (tan repeats every 180). Similarly for angles/lines in Q2 and Q4 which have a negative gradient or tan(theta).
(edited 6 months ago)
Original post by undefined
One tip for understanding tan a bit more is that it, tan(theta), corresponds to the gradient of the line (hypotenuse of the unit right triangle). So obviously a line with positive gradient must lie in both Q1 and Q3 and the angles associated with the line are theta in Q1 and theta+180 in Q3. Its a straight line through the origin, so the two angles are 180 apart (tan repeats every 180). Similarly for angles/lines in Q2 and Q4 which have a negative gradient or tan(theta).

You are such a life saver, thank you so much! And would I just have to move in the clockwise direction to get negative solutions for tan? Thank you once again 🙂
Original post by pigeonwarrior
You are such a life saver, thank you so much! And would I just have to move in the clockwise direction to get negative solutions for tan? Thank you once again 🙂

Positive angles are anticlockwise from the positive x axis (as youd normally expect). Negative angles are clockwise from the positive x axis. Really it shouldnt matter which you work with / solve for as you can +/-360k (integer k) to get back to the domain you want.
Original post by mqb2766
Positive angles are anticlockwise from the positive x axis (as youd normally expect). Negative angles are clockwise from the positive x axis. Really it shouldnt matter which you work with / solve for as you can +/-360k (integer k) to get back to the domain you want.

Yes, that makes more sense now, thanks!