The top of a 15 metre high tower makes an angle of elevation of 60° with the bottom of an electronic pole and angle of elevation of 30° with the top of the pole. What is the height of the electric pole?
Consider the diagram shown above. AC represents the tower and DE represents the pole.
Given that AC = 15 m, $\angle$ADB = 30°, $\angle$AEC = 60°
Let DE = h
Then, BC = DE = h,
AB = (15-h) (∵ AC=15 and BC = h),
BD = CE
tan60°=$\dfrac{AC}{CE}$
=>√3=$\dfrac{15}{CE}$
=>CE=$\dfrac{15}{\sqrt{3}}$ ⋯(1)
tan30°=$\dfrac{AB}{BD}$
=>$\dfrac{1}{\sqrt{3}}=\dfrac{15−h}{BD}$
=>$\dfrac{1}{\sqrt{3}}=\dfrac{15−h}{\left( \dfrac{15}{\sqrt{3}}\right)}$ (∵ BD = CE and substituted the value of CE from equation 1 )
=>(15−h)=$\dfrac{1}{\sqrt{3}}\times\dfrac{15}{\sqrt{3}}=\dfrac{15}{3}=5$
=>h=15−5=10 m
i.e., height of the electric pole = 10 m