BC is a tower, B is its base. A is a point on a horizontal line passing through B, the
angle of elevetion of C from A is 60° From another point Don AB, the angle of
elevation is found to be 30°, then BD=
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Given :-
- BC is a tower with B is its base. = Let height = h units.
- A is a point on a horizontal line passing through B.
- The angle of elevetion of C from A is 60° = ∠CAB.
- From another point D on AB, the angle of elevation is found to be 30° = ∠CDB .
To Find :-
- BD = ?
Solution :-
in Right ∆CBA ,
→ Tan60° = Perpendicular / Base
→ Tan60° = h / BA
→ √3 = h/BA
→ h = √3BA . ----------- (1)
Similarly,
→ in Right ∆CBD ,
→ Tan30° = Perpendicular / Base
→ Tan30° = h / BD
→ (1/√3) = h/BD
→ √3h = BD .
→ h = (BD/√3). -------------- (2)
From (1) and (2) , we get,
→ (BD/√3) = √3BA
→ BD = √3 * (√3BA)
→ BD = 3BA. (Ans.)
Hence, Length of BD is 3 times to the Length BA.
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