Geometrical solution to static problem without friction

AI Thread Summary
The discussion revolves around proving the equation Δl = (d1*d2/l) * sin(α)Δα for a triangle with sides d1, d2, and l, and an angle α between d1 and d2. The user begins by applying the cosine addition formula and the law of cosines but struggles to progress further. It is noted that for small changes, sin(Δα) can be approximated as Δα and cos(Δα) as 1, simplifying the calculations. The conversation emphasizes expanding the left side of the equation and substituting the cosine terms to derive the desired result. Ultimately, the user expresses understanding after following the guidance provided.
maros522
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Homework Statement


We have triangle with sides d1,d2,l and angle \alpha between d1 and d2.
Assume small change \Delta\alpha of \alpha.


Homework Equations


Then we can write for \Deltal equation
\Deltal=(d1*d2)/(l) * sin\alpha\Delta\alpha.

How can I prove that?


The Attempt at a Solution


I start with function cos(\alpha+\Delta\alpha)=cos\alpha * cos\Delta\alpha - sin\alpha * sin\Delta\alpha
and laws of cossines (l+\Delta l)^2=d1^2+d2^2-2d1d2*cos(\alpha+\Delta\alpha)
But I don't know how to continue. :(
 
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maros522 said:

The Attempt at a Solution


I start with function cos(\alpha+\Delta\alpha)=cos\alpha * cos\Delta\alpha - sin\alpha * sin\Delta\alpha
and laws of cossines (l+\Delta l)^2=d1^2+d2^2-2d1d2*cos(\alpha+\Delta\alpha)
But I don't know how to continue. :(

So if Δα is small, then sin(Δα) = Δα and cos(Δα) =1.

So you can simplify cos(Δα+α) using that fact.

Now if you use the original triangle with sides d1,d2,l and angle α, what is l2 using the cosine rule?

(l+\Delta l)^2=d1^2+d2^2-2d1d2*cos(\alpha+\Delta\alpha

If you expand the left side and substitute for cos(α+Δα) and you know that Δl is small so (Δl)2 is negligible you should get it easily.
 
Thank you for answer I get it now.
 
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