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Question
A man observes the angle of elevation of the top of a building to be 30°. He walks towards it in a horizontal line through its base. On covering 60 m, the angle of elevation changes to 60°. Find the height of the building correct to the nearest metre.
Solution
Let AB be a building and M and N are the two positions of the man which makes angle of elevation of top of buildings as 30° and 60° respectively.
MN = 60 m
Let AB = h and NB = x m
Now in right ΔAMB,
`tan 30^circ = (AB)/(MB)`
`=> tan 30^circ = h/(60 + x)`
`=> 1/sqrt(3) = h/(60 + x)`
`=> 60 + x = sqrt(3)h`
`=> x = sqrt(3)h - 60` ...(1)
Similarly in right ΔANB,
`tan 60^circ = (AB)/(NB)`
`tan 60^circ = h/(60 + x)`
`=> sqrt(3) = h/x`
`=> x = h/sqrt(3)` ...(2)
From (1) and (2), we have,
`sqrt(3)h - 60 = h/sqrt(3)`
`=> 3h - 60sqrt(3) = h`
`=> 3h - h = 60sqrt(3)`
`=> 2h = 60sqrt(3)`
`=> h= (60sqrt(3))/(2)`
`=> h = 30sqrt(3) = 30 xx 1.732`
`=>` h = 51.96 m
∴ Height of the building = 51.96 = 52 m (approx)
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