Now, I want to find 9, the angle. Normally, I would use cosine law with 1, 2, and 3 to find both 4 and 7. Then I would use the pythagorean theorum with 4 and 6 to find 5. Then I would use cosine law with 4, 5, 6 to find 8. Now I would add 8 to 7, and subtract that from 180 to find 9.
Is there a better way to do this, using less trig functions. I'm at 3 cos's and 1 square root function. If this is not clear enough, let me know. Thank you.
Triangle solution with less trig
Author
Here is a diagram:
Now, I want to find 9, the angle. Normally, I would use cosine law with 1, 2, and 3 to find both 4 and 7. Then I would use the pythagorean theorum with 4 and 6 to find 5. Then I would use cosine law with 4, 5, 6 to find 8. Now I would add 8 to 7, and subtract that from 180 to find 9.
Is there a better way to do this, using less trig functions. I'm at 3 cos's and 1 square root function. If this is not clear enough, let me know. Thank you.
Now, I want to find 9, the angle. Normally, I would use cosine law with 1, 2, and 3 to find both 4 and 7. Then I would use the pythagorean theorum with 4 and 6 to find 5. Then I would use cosine law with 4, 5, 6 to find 8. Now I would add 8 to 7, and subtract that from 180 to find 9.
Is there a better way to do this, using less trig functions. I'm at 3 cos's and 1 square root function. If this is not clear enough, let me know. Thank you.
(5) is known: have y1 and y2
(8) = sin (5)/(6); have opp and adj
(7) = 30deg (due to (1) being 120 , (2) and (3) being the same size.
(9) = 180 - (8) - (7)
Or are you looking for an algorithm
[edit]
RRRR y1 is unknown :( /me hides
(8) = sin (5)/(6); have opp and adj
(7) = 30deg (due to (1) being 120 , (2) and (3) being the same size.
(9) = 180 - (8) - (7)
Or are you looking for an algorithm
[edit]
RRRR y1 is unknown :( /me hides
Author
Dope, thank you for your second and third point. Can't believe I missed that. Anyone else have any further optimizations?
Angle9 = 150 - arctan(abs(y1-200) / 20);
// Not clear whether y1 is greater or less than y2 from your diagram.
// Not clear whether y1 is greater or less than y2 from your diagram.
Author
@Nypyren: y1 is less than y2. Could you please explain the equation you wrote. I am in Grade 11 ATM and have not learned either arctan or abs.
If tan(x) = y, then arctan (y) = x.
Also, abs(x) = abs(-x) = x. It's the absolute value, and basically strips the sign off a number.
Also, abs(x) = abs(-x) = x. It's the absolute value, and basically strips the sign off a number.
Quote:Original post by erissian
If tan(x) = y, then arctan (y) = x.
Sorry to be pedantic, but my tutor would have a fit if he saw that. Perhaps
If arctan(y) = x then tan(x) = y
would be a better statement. My problem is that you'd need a multifunctional definition of atan to avoid the counterexample:
tan(5pi/4) = 1
arctan(1) = pi/4
Your statement is fine so long as you restrict x to the range 0..2pi (or -pi..pi if your arctan works that way).
Regards
Admiral
No, don't feel bad for clarifying. He's here for help after all. I just didn't want to throw the ball too far over his head, as it were.
Quote:Original post by Nypyren
Angle9 = 150 - arctan(abs(y1-200) / 20);
// Not clear whether y1 is greater or less than y2 from your diagram.
My trig has become rusty. It looks like you are using inclination of a line to find the angle (slope = tan(theta)). I just don't see where the 150 is comming from. I would think: Ang9 + Ang7 = 180 - arctan(abs(y1-200) / 20)
Hint?
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