The figure below shows an isosceles triangular frame ABC made out...

According to condition of the problem, height of the isosceles triangle $ABC$ is unchanged. The dotted lines show configuration

After a temperature rise. Increase in length of rod $AB,$

$∆ {l}_{1} = {l}_{1}{\alpha}_{1} ∆ T$

Thus A{A}^{′} = 1 / 2{l}_{1}{\alpha}_{1} ∆ T

We draw a normal from $A$ to A′C (the final length of $AC$ ). Increase in length of $AC$ is A′N

{A}^{′}N = {l}_{2}{\alpha}_{2} ∆ T

Considering increase in angle $\theta$ to be very small.

{A}^{′}N ≃ A{A}^{′}cos \theta

Where $cos \theta = \dfrac{{l}_{1}}{2{l}_{2}}$

Thus, we have ${l}_{2}{\alpha}_{2} ∆ T =$\left(\dfrac{1}{2}{l}_{1}{\alpha}_{1} ∆ T\right)$\left(\dfrac{{l}_{1}}{2{l}_{2}}\right)$

Hence $\dfrac{{l}_{1}}{{l}_{2}} = 2\sqrt{\dfrac{{\alpha}_{2}}{{\alpha}_{2}}}$