Hint: The change in temperature produces strain in the rods.
Step 1: Find a relation between the lengths of the rods and the angle of the triangle.
Consider the diagram shown.

Let
l1=AB, l2=AC, l3=BCl1=AB, l2=AC, l3=BC∴∴ cosθ=l23+l21-l222l3l1 cosθ=l23+l21−l222l3l1 (assume
∠ABC=θ∠ABC=θ)
⇒ 2l3l1cosθ=l23+l21-l22⇒ 2l3l1cosθ=l23+l21−l22
Step 2: Differentiate on both sides.
Differentiating, 2(l3dl1+l1dl3)cosθ-2l1l3sinθdθ2(l3dl1+l1dl3)cosθ−2l1l3sinθdθ
=2l3dl1+2l1dl1-2l2dl2=2l3dl1+2l1dl1−2l2dl2
Now, dl1=l1α1Δtdl1=l1α1Δt (where ΔΔt = change in temperature)
and l1=l2=l3=ll1=l2=l3=l
(l2α1Δt+l2α1Δt)cosθ+l2sinθdθ=l2α1Δt+l2α1Δt-l2α2Δt(l2α1Δt+l2α1Δt)cosθ+l2sinθdθ=l2α1Δt+l2α1Δt−l2α2Δt
sinθdθ=2α1Δt(1-cosθ)-α2Δt
Putting, θ=60° (for equilateral triangle)
dθ×sin60°=2α1Δt(1-cos60°)-α2Δt
=2α1Δt×12-α2Δt=(α1-α2)Δt
Step 3: Find the change in angle.
⇒ dθ= change in the angle ∠ABC
=(α1-α2)ΔTsin60°=2(α1-α2)ΔT√3 (∵Δt=ΔT given)