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| 1. |
An observer 1.5 m tall is 20.5 m away f |
| Answer» let ht of observer=CD\xa0=1.5mht of tower = AB=22mdistance between observer and tower = AC=20.5mht of remaining tower from his eye D\'B=22-1.5=20.5mangle of elevation tan D=20.5/20.5=1 thus Angle=45o | |