Question

Q-1) Two poles of equal height are standing opposite each others on either side of the road , which is 80m wide . from a point between them on the road , the angles of elevation of the top of the poles are 60° and 30° , respectively . find the height of the poles and the distance of the point from the poles .

With full explanation :​

Answer 1

$\text{Upon drawing the following scenario we get the image in the attachment}$

$\text{According to the question:-}$

$AB = DE$

$\text{BC + CD} = 80 \; m$

$\angle{ACB} = 30^{\circ}$

$\angle{ECD} = 60^{\circ}$

$\text{Let BC}= x$

$\implies CD = 80 - x$

$\implies \text{In }\triangle{ABC}$

$\implies tan(\angle{ACB}) = \dfrac{AB}{BC}$

$\implies tan(30) = \dfrac{AB}{BC}$

$\implies \dfrac{1}{\sqrt3} = \dfrac{AB}{x}$

$\implies x = AB\sqrt3 \; m \text{ ------ (i)}$

$\implies \text{In }\triangle{CDE}$

$\implies tan{(\angle{ECD})} = \dfrac{ED}{DC}$

$\implies tan{60} = \dfrac{ED}{DC}$

$\implies \sqrt3 = \dfrac{ED}{80 - x}$

$\implies \sqrt3(80-x) = ED$

$\implies 80\sqrt3 - x\sqrt3 = ED$

$\text{From (i)}$

$\implies 80\sqrt3 - (AB\sqrt3)\sqrt3 = ED$

$\implies 80\sqrt3 - 3AB = ED$

$\implies 80\sqrt3 = ED + 3AB$

$AB = DE \text{ (Given)}$

$\implies 80\sqrt3 = 4AB$

$\implies \boxed{AB = 20\sqrt3 \; m= DE}$

$\text{Substituting in (i)}$

$\implies x = (20\sqrt3)(\sqrt3)$

$\implies \boxed{x = 60 \; m \text{ from }AB}$

$\implies \boxed{80 - x = 80 - 60 = 20 \; m \text{ from }DE}$