A man observes the angle of elevation of the top of a building to be 30o. He walks towards it in a horizontal line through its base. On covering 60 m the angle of elevation changes to 60o. Find the height of the building correct to the nearest metre.
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Solution
Let the height of the building be AB = h and BC = x
In ΔABC
`tan 60^@ = h/x`
`=> xsqrt3 = h` ...(1)
In ΔADB
`tan 30^@ = h/(x + 60)`
`=> 1/sqrt3 = h/(x + 60)`
`=> x + 60 = hsqrt3` ... (2)
From (1)and (2)
`=> x + 60 = xsqrt3.sqrt3`
`=> x= 30`
Thus `h = 3sqrt3 = 51.96m ~~ "52 metre"`
Concept: Heights and Distances - Solving 2-D Problems Involving Angles of Elevation and Depression Using Trigonometric Tables
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