bugatti79
- 786
- 4
Find Lambda from this arbitrary triangle
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
4 replies · 2K views
- 1,687
- 344
I am still working on this one...
However, I feel like the secret will be in manipulating the equation you have.
Edited*
##\frac{\sin \psi }{ m - \cos \psi } = \frac{\sin( \pi - \psi) }{ a/p +\cos( \pi - \psi) } = \frac{p\sin( \pi - \psi) }{ a + p \cos ( \pi - \psi) }##
This should help you visualize the triangle you should be working with for ##\tan \lambda##.
However, I feel like the secret will be in manipulating the equation you have.
Edited*
##\frac{\sin \psi }{ m - \cos \psi } = \frac{\sin( \pi - \psi) }{ a/p +\cos( \pi - \psi) } = \frac{p\sin( \pi - \psi) }{ a + p \cos ( \pi - \psi) }##
This should help you visualize the triangle you should be working with for ##\tan \lambda##.
Last edited:
bugatti79
- 786
- 4
- 1,687
- 344
That looks good. You extension works to give you a solution for lambda. But it doesn't help get to that tan lambda formula you were looking for.
Try to extend the side length a to make a right triangle with hypotenuse of p. Use sin and cos to determine the side lengths.
Then you can directly get the formula for tangent using the opposite/adjacent of the appropriate right trangle which includes angle lambda.
Try to extend the side length a to make a right triangle with hypotenuse of p. Use sin and cos to determine the side lengths.
Then you can directly get the formula for tangent using the opposite/adjacent of the appropriate right trangle which includes angle lambda.
bugatti79
- 786
- 4
Hi RUber,
I have it. The extended segment of "a" which forms a right angled triangle with P is Pcos(pi-psi). The tan of lambda is obvious. Thank you very much.
I have learned a new way of tacking triangles!
I have it. The extended segment of "a" which forms a right angled triangle with P is Pcos(pi-psi). The tan of lambda is obvious. Thank you very much.
I have learned a new way of tacking triangles!