# Find Lambda from this arbitrary triangle

1. Nov 18, 2015

### bugatti79

Folks,

I am puzzled how one obtains equation 3.2.31 based on the schematic as attached! Can you help?
Is there an online source I can refer to to learn how to obtain angles and magnitudes of complex schematics?

Thanks

#### Attached Files:

• ###### Schematic.jpg
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2. Nov 18, 2015

### RUber

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 $\tan \lambda$.

Last edited: Nov 18, 2015
3. Nov 18, 2015

### bugatti79

Hi, thanks....
This is as far as I got....see attached.....

#### Attached Files:

• ###### Schematic 2.jpg
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4. Nov 18, 2015

### RUber

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.

5. Nov 19, 2015

### bugatti79

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!