- #1
Click For Summary
The discussion revolves around deriving equation 3.2.31 from a schematic involving angles and magnitudes in a triangle. Participants suggest manipulating existing equations and extending side lengths to form right triangles for easier calculations. The use of sine and cosine is emphasized to determine side lengths and ultimately derive the tangent formula for angle lambda. One user successfully identifies the relationship involving the extended segment of "a" and expresses gratitude for the insights gained. The conversation highlights the importance of visualizing triangle relationships to solve complex problems.
- #2
RUber
Homework Helper
- 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:
- #3
- #4
RUber
Homework Helper
- 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.
- #5
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!
Similar threads
- · Replies 2 ·
- · Replies 9 ·
- · Replies 3 ·
- · Replies 1 ·
- · Replies 7 ·
- · Replies 1 ·
- · Replies 5 ·
- · Replies 18 ·
- · Replies 25 ·