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Question
An observer, 1.5 m tall, is 28.5 m away from a 30 m high tower. Determine the angle of elevation of the top of the tower from the eye of the observer.
Solution
`"Let AB = 1.5 m be the observer and CD = 30 m be the tower."`
`"Let the angle of elevation of the top of the tower be α"`
`CD= CE+ED`
`⇒ CD=CE+AB`
`⇒30=CE+1.5`
`⇒CE=30-1.5=28.5 m`
In ΔCEB,
`tan ∝ = (CE)/(BE)=28.5/28.5`
`⇒ tan ∝ =1`
`⇒ tan ∝= tan 45° `
`⇒∝=45°`
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