Question

The angle of elevation of the top of a tower $$30\,m$$ high from the foot of another tower in the same plane is $$60^{\circ}$$ , and the angle of elevation of the top of second  tower from the foot of the first tower is $$30^{\circ}$$ . Find the distance between the two towers and also the height of the other tower.

Solution

Two vertical towers $$TW = 30 \,m$$ and $$ER = x \,m$$ (let) are standing on a horizontal plane $$RW = y$$ (let) . The angle of elevation from R to top $$90\,m$$ high tower is $$60^{\circ}$$ and the angle of elevation of second tower from $$W$$ is $$30^{\circ}$$ . In $$\Delta \, ERW$$ ,$$tan \, 30^{\circ} = \dfrac{x}{y}$$ $$\Rightarrow \, \dfrac{1}{\sqrt{3}} = \dfrac{x}{y}$$ $$\Rightarrow \, y = \sqrt{3} x \, (i)$$ Now , In $$\Delta \, TWR$$ , $$tan \, 60^{\circ} = \dfrac{30^{\circ}}{y} \, (from\ i)$$ $$\Rightarrow \, \sqrt{3} = \dfrac{30}{\sqrt{3}x}$$  [From $$(i)$$] $$\Rightarrow \, 3x = 30$$ $$\Rightarrow \, x = 10\,m$$  Now , $$y = \sqrt{3}x$$ $$\Rightarrow \, y = \sqrt{3}(10)$$ $$\Rightarrow \, y = 1.732 \times 10$$ $$\Rightarrow \, y = 17.32 \, m$$ Hence , the distance between the two towers is $$17.32 \,m$$ and the height of the second tower is $$10\,m$$. Mathematics

Suggest Corrections

0

People also searched for
View More