# An equilateral triangle ABC is formed by two Cu rods AB and BC and one Al rod. It is heated in such a way that the temperature of each rod increases by T. Find the change in the angle ABC.

$\cos \theta = \frac{l_1^{2}+l_3^{2}-l_2^{2}}{2l_1l_3}$

$2l_1l_3\cos \theta =l_1^{2}+l_3^{2}-l_2^{2}$

Differentiating on both sides of the equation,

$2[d (l_1 l_3) \cos \theta + l_1 l_3 d (\cos \theta)] = 2 l_1 dl_1 + 2 l_3 dl_3 - 2 l_2 dl_2$

$(l_1 dl_3 + l_3 dl_1) \cos \theta - l_1 l_3 \cos \theta d\theta = l_1 dl_1 + l_3 dl_3 - l_2 dl_2$ --------------(1)

$dl_1 = l_1 \alpha_ 1 \Delta t$

$dl_2 = l_2 \alpha_ 2 \Delta t$

$dl_3 = l_3 \alpha _3 \Delta t$

L1 = L2 = L3

$\\dl_1=dl_3=l\alpha_1\Delta t$

$dl_2 = l\alpha _2 \Delta t$

substituting in 1, we get

$\cos \theta (l^{ 2} \alpha_ 1 \Delta t + l^{2} \alpha_ 1 \Delta t) - l^{2} \sin \theta d\theta = l^{2} \alpha_1 \Delta t + l^{2} \alpha_1 \Delta t - l^{2} \alpha_2 \Delta t$

$2 \alpha _1 \Delta t \times \frac{1}{2}- 2 \alpha_ 1 \Delta t + \alpha_ 2 \Delta t = \frac{\sqrt{3}}{2}d \theta$

$d\theta = 2 (\alpha 2-\alpha 1) \frac{\Delta t }{\sqrt{3}}$

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