Triangular Geometry and Sine Rule

In $\triangle ABC$: $AB = V_1 T$, $BC = V_2 T$. Interior angle $\angle B = 180^\circ - 105^\circ = 75^\circ$. $\angle C = 30^\circ$. Sum of angles = $180^\circ \implies \angle A = 180^\circ - 75^\circ - 30^\circ = 75^\circ$. Since $\angle A = \angle B = 75^\circ$, $\triangle ABC$ is isosceles with $AC = BC = V_2 T$ or using Sine Rule: $\frac{AC}{\sin 75^\circ} = \frac{BC}{\sin 75^\circ} \implies AC = BC = V_2 T$. Wait, using altitude $BD \perp AC$: $BD = V_2 T \sin 30^\circ = \frac{V_2 T}{2}$. $CD = V_2 T \cos 30^\circ = \frac{\sqrt{3}}{2} V_2 T$. In $\triangle ABD$, $\angle A = 45^\circ \implies AD = BD = \frac{V_2 T}{2}$. Total $AC = CD + AD = \frac{\sqrt{3} + 1}{2} V_2 T = 0.5(\sqrt{3} + 1) V_2 T$. Time taken at speed $V_2 = \frac{AC}{V_2} = 0.5(\sqrt{3} + 1)T$.