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Question
Two points A and B are on the same side of a tower and in the same straight line with its base. The angles of depression of these points from the top of the tower are 60° and 45° respectively. If the height of the tower is 15 m, then find the distance between the points.
Solution
Let CD be the tower. A and B are the two points on the same side of the tower.
In ΔDBC
`tan 60^@ = (DC)/(BC)`
`=> sqrt3 = 15/(BC)`
`=> BC = 15/sqrt3`
`=> BC = 5sqrt3 m`
In ΔDAC
`tan 45^@ = (DC)/(AC)`
`=> 1 = 15/"AC"`
`=> AC = 15 m`
Now
AC = AB + BC
`:. AB = AC - BC = 15 - 5sqrt3 = 5(3 - sqrt3) m`
Hence, the distance between the two points A and B is `5(3 - sqrt3) m`
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