Let AB is tower and C is the point 30 m away from the foot of the tower. In Δ ABC, AB/BC= tan 30° ⇒ AB/30 = 1/√3 ⇒ AB = 30/√3 = 30√3/3 = 10√3 Hence, the height of the tower is 10√3m.
Let AB is tower and C is the point 30 m away from the foot of the tower.
In Δ ABC,
AB/BC= tan 30°
⇒ AB/30 = 1/√3
⇒ AB = 30/√3 = 30√3/3 = 10√3
Hence, the height of the tower is 10√3m.
The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30°. Find the height of the tower.
Let AB is tower and C is the point 30 m away from the foot of the tower. In Δ ABC, AB/BC= tan 30° ⇒ AB/30 = 1/√3 ⇒ AB = 30/√3 = 30√3/3 = 10√3 Hence, the height of the tower is 10√3m.
Let AB is tower and C is the point 30 m away from the foot of the tower.
See lessIn Δ ABC,
AB/BC= tan 30°
⇒ AB/30 = 1/√3
⇒ AB = 30/√3 = 30√3/3 = 10√3
Hence, the height of the tower is 10√3m.