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June 14th, 2016, 01:27 PM  #1 
Newbie Joined: Jun 2016 From: home Posts: 1 Thanks: 0  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. 
June 14th, 2016, 02:55 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,755 Thanks: 695 
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)(yx)/w=tan(bp). 4)p=barctan(bp)=barctan((yx)/w). Carry out steps above in order. 
June 14th, 2016, 04:07 PM  #3 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,934 Thanks: 1128 Math Focus: Elementary mathematics and beyond 
Let $h$ be the height of the triangle. Then $$h=\dfrac{y}{\tan\beta},\quad\phi=\beta\tan^{1}\left(\dfrac{yx}{h}\right)$$ 
June 14th, 2016, 04:25 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,622 Thanks: 2074 
$\phi$ = 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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