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June 14th, 2016, 01:27 PM   #1
Ald
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Triangle inside another problem

Hi
I have a problem which I am sure is simple, but I can't get it right.
Given is beta, x, and y. Required is Phi.
Can you help?

Last edited by skipjack; June 14th, 2016 at 01:43 PM.
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June 14th, 2016, 02:55 PM   #2
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Notation: u=long hypotenuse, v=short hypotenuse, w=vertical side of triangles.
b=beta, p=phi

1)y/u=sinb or u=y/sinb.
2)w=ucosb.
3)(y-x)/w=tan(b-p).
4)p=b-arctan(b-p)=b-arctan((y-x)/w).

Carry out steps above in order.
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June 14th, 2016, 04:07 PM   #3
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Let $h$ be the height of the triangle. Then

$$h=\dfrac{y}{\tan\beta},\quad\phi=\beta-\tan^{-1}\left(\dfrac{y-x}{h}\right)$$
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June 14th, 2016, 04:25 PM   #4
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$\phi$ = tan$^{-1}$(sin(β)cos(β)/(y/x - sin²(β))) by mathman's method,
but greg1313's way of writing it is simpler and implies $\phi = \beta - \tan^{-1}((1 - x/y)\tan\beta)$.
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