A vertical tower stands on ground and is surmounted by a vertical flagpole of height 18 m. At a point on the ground, the angle of elevation of the bottom and the top of the flagpole are 30° and 60° respectively. What is the height of the tower?
9 m
10.40 m
15.57 m
12 m
Explanation:
Let DC be the vertical tower and AD be the vertical flagpole. Let B be the point of observation.
Given that AD = 18 m, $\angle$ABC = 60°, $\angle$DBC = 30°
Let DC be h.
tan30°=$\dfrac{DC}{BC}$
=>$\dfrac{1}{\sqrt{3}}=\dfrac{h}{BC}$
=>h=$\dfrac{BC}{\sqrt{3}}$ ...(1)
tan60°=$\dfrac{AC}{BC}$
=>√3=$\dfrac{18+h}{BC}$
18+h=BC×√3 ⋯(2)
$ \dfrac{(1)}{(2)}=>\dfrac{h}{18+h}=\dfrac{\dfrac{BC}{\sqrt{3}}} {(BC×\sqrt{3})}=\dfrac{1}{3}$
=>3h=18+h
=>2h=18
=>h=9 m
i.e., the height of the tower = 9 m
Let DC be the vertical tower and AD be the vertical flagpole. Let B be the point of observation.
Given that AD = 18 m, $\angle$ABC = 60°, $\angle$DBC = 30°
Let DC be h.
tan30°=$\dfrac{DC}{BC}$
=>$\dfrac{1}{\sqrt{3}}=\dfrac{h}{BC}$
=>h=$\dfrac{BC}{\sqrt{3}}$ ...(1)
tan60°=$\dfrac{AC}{BC}$
=>√3=$\dfrac{18+h}{BC}$
18+h=BC×√3 ⋯(2)
$ \dfrac{(1)}{(2)}=>\dfrac{h}{18+h}=\dfrac{\dfrac{BC}{\sqrt{3}}} {(BC×\sqrt{3})}=\dfrac{1}{3}$
=>3h=18+h
=>2h=18
=>h=9 m
i.e., the height of the tower = 9 m