Angle of elevation of the top of the tower from 3 points (collinear) A, B and C on a road leading to the foot of the tower are 30° , 45° and 60° respectively. The ratio of AB and BC is

Concept:

If the angle of elevation from a point at distance D from tower of length L is θ 

tan θ = \(\rm L\over D\)

Calculation:

Let the length of the tower OP be x

From the figure,

OA = \(\rm {x\over\tan30}=\sqrt{3}x\)

OB = \(\rm {x\over\tan45}=x\)

OC = \(\rm {x\over\tan60}={x\over√{3}}\)

∴ AB = OA - OB = \(\rm \sqrt 3 x-x\)

BC = OB - OC = \(\rm x-\frac{x}{\sqrt 3}\)

⇒ \(\rm {AB\over BC}={\sqrt 3 x-x\over\rm x-\frac{x}{\sqrt 3}}\)

⇒ \(\rm {AB\over BC}={\sqrt 3(\sqrt 3 -1)\over \sqrt 3 -1} \)

⇒ \(\boldsymbol{\rm AB:BC=\sqrt{3} : 1}\)