An equilateral triangle $ABC$ is formed by two copper rods $AB$ and $BC$ and one aluminium rod $AC$. The system is heated such that the temperature of each rod increases by $\Delta T$. Find the change in the angle $\angle ABC$. (Coefficient of linear expansion for copper is $\alpha_1$ and for aluminium is $\alpha_2$).

(N/A) Let the lengths of the sides be $AB = l_1$,$BC = l_3$,and $AC = l_2$. Initially,$l_1 = l_2 = l_3 = l$.
Using the law of cosines for $\angle ABC = \theta$:
$\cos \theta = \frac{l_1^2 + l_3^2 - l_2^2}{2 l_1 l_3}$
$2 l_1 l_3 \cos \theta = l_1^2 + l_3^2 - l_2^2$
Taking the differential of both sides:
$2(l_3 dl_1 + l_1 dl_3) \cos \theta - 2 l_1 l_3 \sin \theta d\theta = 2 l_1 dl_1 + 2 l_3 dl_3 - 2 l_2 dl_2$
Since $AB$ and $BC$ are copper $(\alpha_1)$ and $AC$ is aluminium $(\alpha_2)$:
$dl_1 = l_1 \alpha_1 \Delta T$,$dl_3 = l_3 \alpha_1 \Delta T$,$dl_2 = l_2 \alpha_2 \Delta T$
Substituting these into the equation and using $l_1 = l_2 = l_3 = l$ and $\theta = 60^\circ$:
$2(l^2 \alpha_1 \Delta T + l^2 \alpha_1 \Delta T) \cos 60^\circ - 2 l^2 \sin 60^\circ d\theta = 2 l^2 \alpha_1 \Delta T + 2 l^2 \alpha_1 \Delta T - 2 l^2 \alpha_2 \Delta T$
$4 l^2 \alpha_1 \Delta T (1/2) - 2 l^2 (\sqrt{3}/2) d\theta = 4 l^2 \alpha_1 \Delta T - 2 l^2 \alpha_2 \Delta T$
$2 l^2 \alpha_1 \Delta T - \sqrt{3} l^2 d\theta = 4 l^2 \alpha_1 \Delta T - 2 l^2 \alpha_2 \Delta T$
$-\sqrt{3} d\theta = 2 l^2 \alpha_1 \Delta T - 2 l^2 \alpha_2 \Delta T$
$d\theta = \frac{2(\alpha_2 - \alpha_1) \Delta T}{\sqrt{3}}$