Suppose, $\mathrm{AB}=l_{1}, \mathrm{AC}=l_{2}$ and $\mathrm{BC}=l_{3}$

$\therefore \cos \theta=\frac{l_{3}^{2}+l_{1}^{2}-l_{2}^{2}}{2 l_{3} l_{1}}$ where $\angle \mathrm{ABC}=\theta$ $\therefore 2 l_{3} l_{1} \cos \theta=l_{3}^{2}+l_{1}^{2}-l_{2}^{2}$

Integrating on both side,

$2\left(l_{3} d l_{1}+l_{1} d l_{3}\right) \cos \theta-2 l 3 l_{1} \sin \theta d \theta=2 l_{3} d l_{3}+2 l_{1} d l_{1}-2 l_{2} d l_{2}$

Dividing by $2$ ,

$\left(l_{3} d l_{1}+l_{1} \times d l_{3}\right) \cos \theta-l_{3} l_{1} \sin \theta d \theta=l_{3} d l_{3}+l_{1} d l_{1}-l_{2} d l_{2}$

Now $d l_{1}=l_{1} \alpha_{1} \Delta \mathrm{T}, d l_{2}=l_{2} \alpha_{2} \Delta \mathrm{T}, d l_{3}=l_{3} \alpha_{3} \Delta \mathrm{T}$ then,

$\left(l_{3} \times l_{1} \alpha_{1} \Delta \mathrm{T}+l_{1} \times l_{3} \alpha_{3} \Delta \mathrm{T}\right) \cos \theta-l_{3} l_{1} \sin \theta d \theta=l_{3} \times l_{3} \alpha_{3} \Delta \mathrm{T}+l_{1} \times l_{1} \alpha_{1} \Delta \mathrm{T}-l_{2} \times l_{2} \alpha_{2} \Delta \mathrm{T}$

Now let $l_{1}=l_{2}=l_{3}=l$ and $\alpha_{3}=\alpha_{1}$

$\therefore\left(l^{2} \alpha_{1} \Delta \mathrm{T}+l^{2} \alpha_{1} \Delta \mathrm{T}\right) \cos \theta-l^{2} \sin \theta d \theta=l^{2} \alpha_{1} \Delta \mathrm{T}+l^{2} \alpha_{1} \Delta \mathrm{T}-l^{2} \alpha_{2} \Delta \mathrm{T}$

$\quad \cos \theta=\cos 60^{\circ}=\frac{1}{2} \quad(\mathrm{Equilateral}$ triangle $)$

$\therefore 2 l^{2} \alpha_{1} \Delta \mathrm{T} \times \frac{1}{2}-l^{2} \sin \theta d \theta=2 l \alpha_{1} \Delta \mathrm{T}-l^{2} \alpha_{2} \Delta \mathrm{T}$

$\therefore l \alpha_{1} \Delta \mathrm{T}-l^{2} \sin \theta d \theta=2 l^{2} \alpha_{1} \Delta \mathrm{T}-l^{2} \alpha_{2} \Delta \mathrm{T}$