As observed from the top of a 100 meters high lighthouse from the sea-level, the angles of depression of two ships are 30° and 60°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.
In ΔACE, |
tan 60° = AE/CE |
√3 = (AF – EF)/CE |
√3 = (121.5 – 1.5)/CE |
CE = 120/√3 m ---- (i) |
In ΔBCG, |
tan 45° = BG/CG |
1 = BG/CG |
BG = CG |
BG = CE + EG |
EG = BG – CE |
EG = (BH – GH) – CE |
EG = (121.5 – 1.5) – (120/√3) (∵ from eq. (i)) |
EG = 120 – (120/√3) |
EG = 120((√3 – 1)/√3) meters |
∴ The distance traveled by balloon is 120((√3 – 1)/√3) meters. |