geometry problem

K

Kasa

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Feb 4, 2020
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View attachment 16478IMG_20200204_222808.jpghi, can anyone help me with this problem please? Need to solve alpha. I tried solving diferent equations and i wasnt able to solve them even wolframalpha couldnt solve them for alpha.
 
Please demonstrate your efforts. What do you know about Vertical Angles or the Sum of the interior angles of a triangle?

First, one might make an effort to write more clearly. Do we have [math]V_{\Lambda},\;V_{2},\;b, \;a,\;and\;\alpha\;twice?[/math]
Second, you may also improve the discussion by providing labels for at least some points.
 
To KASA, here is a simple question for you.
Why can you not post a clean, readable diagram?
What you posted is a mess not worthy of a reply.
 
Does the arc with a dot in it mean a right angle? Are there two other right angles you did not mark?

Are the given lengths called r1, r2, a, and b?

Please show us your work (by at least one method). Taking the diagram as I understand it, I get a trig equation that I can turn into a fourth-degree equation, which WA might be able to handle (numerically if not symbolically).
 
Hi, sorry for the mess and thank you for your replies. its r1, r2, a, b, and α twice, and yes, two arcs with dots mean a right angle. Its a timing belt problem so there has to be theroreticaly right angle and ri is the radius of pulleys. I cant write down anything right now but i can show you some equation from history in wolfram https://www.wolframalpha.com/input/?i=tg(x)=(a-cos(x)*r)/(b+sin(x)*r)+x=? Here the x means α, that was my best try (others was without any solution) it show some solution but i dont understand the plus 3.1416 n. and still it shows approximately.
 
Can you confirm that the two angles at the ends of segment "b" are both right angles, as they appear to be?
 
My equation, based on extending your short line to intersect that line, and then adding three segments that total a, is:

[MATH]r_1 \tan(\alpha) + \frac{b}{\sin(\alpha)} + r_2\tan(\alpha) = a[/MATH].​
 
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