From the top of a hill 100 m high, the angles of depression of the top and bottom of a pole are 30° and 60° respectively. What is the height of the pole?
52 m
50 m
66.67 m
33.33 m
Explanation:
Consider the diagram shown above. AC represents the hill and DE represents the pole
Given that AC = 100 m
$\angle$XAD = $\angle$ADB = 30° (∵ AX || BD )
$\angle$XAE = $\angle$AEC = 60° (∵ AX || CE)
Let DE = h
Then, BC = DE = h,
AB = (100-h) (∵ AC=100 and BC = h),
BD = CE
tan60°=$\dfrac{AC}{CE}$
=>√3=$\dfrac{100}{CE}$
=>CE=$\dfrac{100}{\sqrt{3}} $ ⋯(1)
tan30°=$\dfrac{AB}{BD}$
=>$\dfrac{1}{\sqrt{3}} =\dfrac{100-h}{\left(\dfrac{100}{\sqrt{3}}\right)}$ (∵ BD = CE and substituted the value of CE from equation 1)
=>(100−h)=$\dfrac{1}{\sqrt{3}} \times \dfrac{100}{\sqrt{3}}$
=$\dfrac{100}{3}=33.33$
=>h=100−33.33=66.67 m
i.e., the height of the pole = 66.67 m
Consider the diagram shown above. AC represents the hill and DE represents the pole
Given that AC = 100 m
$\angle$XAD = $\angle$ADB = 30° (∵ AX || BD )
$\angle$XAE = $\angle$AEC = 60° (∵ AX || CE)
Let DE = h
Then, BC = DE = h,
AB = (100-h) (∵ AC=100 and BC = h),
BD = CE
tan60°=$\dfrac{AC}{CE}$
=>√3=$\dfrac{100}{CE}$
=>CE=$\dfrac{100}{\sqrt{3}} $ ⋯(1)
tan30°=$\dfrac{AB}{BD}$
=>$\dfrac{1}{\sqrt{3}} =\dfrac{100-h}{\left(\dfrac{100}{\sqrt{3}}\right)}$ (∵ BD = CE and substituted the value of CE from equation 1)
=>(100−h)=$\dfrac{1}{\sqrt{3}} \times \dfrac{100}{\sqrt{3}}$
=$\dfrac{100}{3}=33.33$
=>h=100−33.33=66.67 m
i.e., the height of the pole = 66.67 m