The angle of depression of a ship from the top of a tower 30 m height has measure 60. Then, the distance of the ship from the base of the tower is ......,select a proper option (a), (b), (c) or (d) from given options so that the statement becomes correct.
(a) 10
(b) 30
(c) 10√3
(d) 30√3
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We know height of tower h =30 meter.
angle of depression = 60°
angle OAC = angle ACB
After applying angle height and distance relation we got x which shows option c is correct.
Where x is distance of ship from the base of tower.
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From the figure attached above ,
A is the ship.
Height of the tower ( BC ) = h = 30 m
Angle of depression=<DCA = <BAC =60°
Distance from ship to the foot of
the tower = AB
In ∆ABC , <B = 90°
tan 60° = BC/AB
=> √3 = 30/AB
=> AB = 30/√3
=> AB = ( 30 ×√3 )/(√3×√3)
=> AB = ( 30 × √3 )/3
=> AB = 10√3 m
Option ( c ) is correct.
••••
A is the ship.
Height of the tower ( BC ) = h = 30 m
Angle of depression=<DCA = <BAC =60°
Distance from ship to the foot of
the tower = AB
In ∆ABC , <B = 90°
tan 60° = BC/AB
=> √3 = 30/AB
=> AB = 30/√3
=> AB = ( 30 ×√3 )/(√3×√3)
=> AB = ( 30 × √3 )/3
=> AB = 10√3 m
Option ( c ) is correct.
••••
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