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Questions
The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.
The angle of elevation of the top of the building from the foot of the tower is 30° and the angle of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.
Solution 1
Let AB be the building and CD be the tower.
In ΔCDB,
`"CB"/"BD"` = tan 60°
`50/("BD") = sqrt3`
`"BD" = 50/sqrt3`
In ΔABD,
`("AB")/("BD") = tan 30°`
AB = `50/sqrt3 xx 1/sqrt3`
= `50/3`
= `16 2/3`
Therefore, the height of the building is `16 2/3` m.
Solution 2
Let AD be the building of height h m. and an angle of elevation of the top of building from the foot of the tower is 30° and an angle of the top of the tower from the foot of building is 60°.
Let AD = h, AB = x and BC = 50 and ∠DBA = 30°, ∠CAB = 60°
So we use trigonometric ratios.
In a triangle ABC
⇒ `tan 60° = 50/x`
⇒ `sqrt3 = 50/x`
⇒ `x = 50/sqrt3`
Again in a triangle ABD
⇒ `tan 30° = ("AD")/("AB")`
⇒ `1/sqrt3 xx h/x`
⇒ `h = x/sqrt3`
⇒ `h = 50/(sqrt3 xx sqrt3)`
⇒ `h = 50/3`
⇒ `h =16 2/3`
Therefore, the height of the building is `16 2/3` m.
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