From a tower of 80 m high, the angle of depression of a bus is 30°. How far is the bus from the tower?
40 m
138.4 m
46.24 m
160 m
Explanation:
Let AC be the tower and B be the position of the bus.
Then BC = the distance of the bus from the foot of the tower.
Given that height of the tower, AC = 80 m and the angle of depression, $\angle$DAB = 30°
$\angle$ABC = $\angle$DAB = 30° (because DA || BC)
tan30°=$\dfrac{AC}{BC}$
=>tan30°=$\dfrac{80}{BC}$
=>BC=$\dfrac{80}{tan30°}=\dfrac{80}{\left(\dfrac{1}{\sqrt{3}}\right)}$
=80×1.73=138.4 m
i.e., Distance of the bus from the foot of the tower = 138.4 m
Let AC be the tower and B be the position of the bus.
Then BC = the distance of the bus from the foot of the tower.
Given that height of the tower, AC = 80 m and the angle of depression, $\angle$DAB = 30°
$\angle$ABC = $\angle$DAB = 30° (because DA || BC)
tan30°=$\dfrac{AC}{BC}$
=>tan30°=$\dfrac{80}{BC}$
=>BC=$\dfrac{80}{tan30°}=\dfrac{80}{\left(\dfrac{1}{\sqrt{3}}\right)}$
=80×1.73=138.4 m
i.e., Distance of the bus from the foot of the tower = 138.4 m