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Question
On a horizonal plane there is a vertical tower with a flagpole on the top of the tower. At a point, 9 meters away from the foot of the tower, the angle of elevation of the top and bottom of the flagpole are 60 and 30 respectively. Find the height of the tower and the flagpole mounted on it.
Solution
Let OX be the horizontal plane, AD be the tower and CD be the vertical flagpole
We have:
AB = 9m, ∠DBA= 30° and ∠CBA = 60°
Let:
AD = hm and CD = xm
In the right ΔABD,we have:
`(AD)/(AB) = tan 30° = 1/ sqrt(3)`
`⇒ h/8 = 1/sqrt(3)`
`⇒ h = 9/sqrt(3) = 5.19 `m
Now, in the right ΔABC,we have
`(AC)/(BA) = tan 60° = sqrt(3)`
`⇒ (h+x ) /9 = sqrt(3)`
`⇒ h + x = 9 sqrt(3)`
By putting h= `9/sqrt(3)` in the above equation, we get:
`9/sqrt(3) + x = 9 sqrt(3) `
`⇒ x = 9 sqrt(3) - 9/ sqrt(3) `
`⇒ x = (27-9)/ sqrt(3) = 18/ sqrt(3) = 18/ 1.73 = 10.4`
Thus, we have:
Height of the flagpole = 10. 4 m
Height of the tower = 5. 19 m
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