the angle of elevation of the top of the tower from a point on same level at the foot of the tower is 30 degree. an advancing 130m towards the foot of tower the angle of elevation become 60 degree . show that the height of the tower is 129.9m?
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Call the top of the tower A. Let the foot of the tower be B. Let the initial point of observation be C. Let the next point of observation be C'. So CC' = 130 m
or, BC = 130 + BC' -- (1)
At point C: elevation:
Tan 30⁰ = AB / BC => AB = BC / √3 --- (2)
At point C': elevation angle:
Tan 60 = AB / BC' => AB = √3 BC' --- (3)
=> AB = √3 (BC - 130)
=> BC /√3 = √3 BC - √3 * 130
=> BC = 3 BC - 390
BC = 195 m
AB = BC/√3 = 112.58 m -- height of the tower.
Perhaps the answer you have is not right.
or, BC = 130 + BC' -- (1)
At point C: elevation:
Tan 30⁰ = AB / BC => AB = BC / √3 --- (2)
At point C': elevation angle:
Tan 60 = AB / BC' => AB = √3 BC' --- (3)
=> AB = √3 (BC - 130)
=> BC /√3 = √3 BC - √3 * 130
=> BC = 3 BC - 390
BC = 195 m
AB = BC/√3 = 112.58 m -- height of the tower.
Perhaps the answer you have is not right.
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