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?
5 metres
8 metres
10 metres
12 metres
Explanation:
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
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