Two vertical poles of heights, 20 m and 80 m stand a part on a horizontal plane.

Question

Two vertical poles of heights, 20 m and 80 m stand
a part on a horizontal plane. The height (in meters)
of the point of intersection of the lines joining the
top of each pole to the foot of the other, from this
horizontal plane is :

Accepted Answer

From triangle BCD,

tan $$\alpha $$ = $${{80} \over x}$$

and from triangle BFE,

tan $$\alpha $$ = $${{h} \over y}$$

$$ \therefore $$ $${{80} \over x}$$ = $${{h} \over y}$$

$$ \Rightarrow $$ $$y = {{hx} \over {80}}$$ ........ (1)

From triangle ABC,

tan $$\beta $$ = $${{20} \over x}$$

and from triangle EFC,

tan $$\beta $$ = $${{h} \over {x-y}}$$

$$ \therefore $$ $${{20} \over x}$$ = $${{h} \over {x-y}}$$

$$ \Rightarrow $$ $$x-y = {{hx} \over {20}}$$ ........ (2)

By adding equation (1) and (2) we get,

x = $${{hx} \over {80}}$$ + $${{hx} \over {20}}$$

$$ \Rightarrow $$ 1 = $${{h} \over {80}}$$ + $${{h} \over {20}}$$

$$ \Rightarrow $$ h = 16 m

$$ \therefore $$ Height of intersection point is 16 m

Solutions · 1