Geometrical solution to static problem without friction

[maros522]

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 $\Delta$l equation
$\Delta$l=(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. :(

 

[rock.freak667]

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.

 

[maros522]

Thank you for answer I get it now.