Let the positions of the two men are B and C. Let AP represent the tower in the figure. Join B, P, and C, P.We have, ∠ABP = 30° and ∠ACP = 45°. Given that AP = 50 m.We get two right-angled triangles ∆APB and ∆APC, both right angled at P. We use trigonometric ratio tan for both the triangles using BP as a base and AP as height(for ∆APB) and PC as base and AP as height(for ∆APC). We get BP and PC. The distance between the men is BC.

We have to find BC. Let BP = x. In ∆APB,

or,

Again, in ∆APC,

or,

PC = 50

So, distance between the two men is

BC = BP + PC = 50√3 + 50 = 50(√3 + 1) = 136.6m.