AB is a vertical pole with B at the ground level and A at the top.

Question

$AB$ is a vertical pole with $B$ at the ground level and $A$ at the top. $A$ man finds that the angle of elevation of the point $A$ from a certain point $C$ on the ground is $${60^ \circ }$$. He moves away from the pole along the line $BC$ to a point $D$ such that $CD=7$ m. From $D$ the angle of elevation of the point $A$ is $${45^ \circ }$$. Then the height of the pole is :

Accepted Answer

In $$\Delta ABC$$

$${h \over x} = \tan {60^ \circ } = \sqrt 3 $$

$$ \Rightarrow x = {h \over {\sqrt 3 }}$$

In $$\Delta ABD{h \over {x + 7}}$$

$$ = \tan {45^ \circ } = 1$$

$$ \Rightarrow h = x + 7 \Rightarrow h - {h \over {\sqrt 3 }} = 7$$

$$ \Rightarrow h = {{7\sqrt 3 } \over {\sqrt 3  - 1}} \times {{\sqrt 3  + 1} \over {\sqrt 3  + 1}}$$

$$ \Rightarrow h = {{7\sqrt 3 } \over 2}\left( {\sqrt 3  + 1\,m} \right)$$

Solutions · 1