- #1
Find Lambda from this arbitrary triangle
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Discussion Overview
The discussion revolves around deriving a specific equation related to angles and magnitudes in a triangle, particularly focusing on the calculation of the angle lambda (λ) from a given schematic. Participants are exploring methods to manipulate equations and visualize the triangle to find the desired relationships.
Discussion Character
- Exploratory, Technical explanation, Mathematical reasoning
Main Points Raised
- One participant expresses confusion about obtaining a specific equation from a schematic and seeks guidance on resources for learning about angles and magnitudes in complex schematics.
- Another participant suggests manipulating an equation involving sine and cosine to visualize the triangle relevant to calculating tan λ.
- A different participant shares their progress and an attachment, indicating their current understanding of the problem.
- One participant acknowledges the previous work but notes that it does not directly lead to the desired tan λ formula, recommending the construction of a right triangle to facilitate the calculation.
- A later reply claims to have found a solution by extending a segment to form a right triangle, asserting that the tangent of lambda is now clear.
Areas of Agreement / Disagreement
Participants appear to be working towards a solution with some shared understanding of the geometry involved, but there is no explicit consensus on the methods or final formula for tan λ.
Contextual Notes
Some participants' approaches depend on specific assumptions about the triangle's configuration and the relationships between the angles and sides, which may not be universally agreed upon.
- #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!
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