Answer :

Untitled9.jpg


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.


Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
caricature
view all courses