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Question
A vertical tow er stands on a horizontal plane and is surmounted by a flagstaff of height 7m. From a point on the plane the angle of elevation of the bottom of the flagstaff is 30° and that of the top of the ft agstaff is 45°. Find the height of the tower.
Solution
Let AB be the flagstaff, BC be the tower and D be the point on ground from where elevation angles are measured.
In ΔBCD
`"BC"/"CD" = tan 30^circ`
`"BC"/"CD" = 1/sqrt(3)`
`sqrt(3)"BC" = "CD"`
In ΔACD
`("AB + BC")/"CD" = tan 45^circ`
`("AB + BC")/(sqrt(3)"BC") = 1`
`7 + "BC" = sqrt(3)"BC"`
`"BC"(sqrt(3)-1) = 7`
BC = `((7)(sqrt(3) + 1))/((sqrt(3) - 1)(sqrt(3) + 1)`
= `(7(sqrt(3)+ 1))/((sqrt(3))^2 - (1)^2)`
= `(7(sqrt(3) + 1))/2 = 3.5(sqrt(3) + 1) = 3.5 xx 2.732 = 9.562`
Thus , the height of the tower is 9.562 m = 9.56 m.
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