Three rods A, B and C form an equilateral triangle at $0^{\circ} C$. Rods AB and BC have same coefficient of thermal expansion $\alpha_1$ And AC has $\alpha_2$. If the the temperature is increased to $T^{\circ}$C, what is the change in angle formed by rods AB and BC?
Intially, the angle is $60^{\circ}$
Final lenghts of rods AB and BC will be $l_1$ and $l_2$ while that of AC will be $l_3$
Using cosine rule
$$\cos \theta = \frac{l_1^2 + l_2^2 -l_3^3}{2l_1l_2}$$
$$\cos \theta =1-\frac{l_3^2}{2l_1^2}$$
$l_1$ and $l_2$ will be equal $$=1-\frac{l^2(1+\alpha_2 T)^2}{2l^2 (1+\alpha_1 T)^2}$$
$$=1-\frac{(1+\alpha_2 T)^2}{2(1+\alpha_1 T)^2}$$
Now that I have initial and final angles, how do I find the change in angle?
I know this question uses quite a bit of physics, but my problem ultimately boils down to the math of it. The question was not accepted on physics SE due it computational nature. I simply need to find the difference in angle, but I just need to know how to remove the cosine function.
In physics, it is understood that thermal expansion results in only small changes and is usually treated as such. So, consider $\alpha$’s as small quantities and simplify your expression
$$1-\frac{(1+\alpha_2 T)^2}{2(1+\alpha_1 T)^2}= \frac12- (\alpha_2-\alpha_1)T$$
Then, the change in angle is
$$\cos^{-1}\left( \frac12- (\alpha_2-\alpha_1)T\right) -\frac\pi3= \frac2{\sqrt3}(\alpha_2-\alpha_1)T $$