The angle of elevation of the top of a vertical tower standing on a horizontal plane is ob…

Question

The angle of elevation of the top of a vertical tower standing on a horizontal plane is observed to be 45o from
a point A on the plane. Let B be the point 30 m vertically above the point A. If the angle of elevation of the
top of the tower from B be 30o, then the distance (in m) of the foot of the tower from the point A is :

Accepted Answer

$${x \over d} = \tan {30^o} \Rightarrow {x \over d} = {1 \over {\sqrt 3 }} \Rightarrow d = \sqrt 3 x$$

and $${{x + 30} \over d} = \tan {45^o}$$ $$ \Rightarrow $$ d = x + 30

$$d = {d \over {\sqrt 3 }} + 30 \Rightarrow \left( {1 - {1 \over {\sqrt 3 }}} \right)d = 30$$

$$ \Rightarrow d = {{30\sqrt 3 } \over {\sqrt 3  - 1}} \Rightarrow d = {{30\sqrt 3 (\sqrt 3  + 1)} \over 2} = 15\sqrt 3 (\sqrt 3  + 1)$$

$$ \Rightarrow $$ d = 15 ( 3 + $$\sqrt 3 $$)

Solutions · 1