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secondary 3 | E Maths
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Date Posted: 1 year ago
Views: 230
(a) Line AB = Hypotenuse of triangle ABC
sin (47) = bc / 258
Solve bc
(b) Line BC = Hypotenuse of triangle CPB
sin (29) = bc / cp
(c) Tower CT = Opposite of Angle CTP
CP = Adjacent of Angle CTP
tan (Angle ctp ) = 46 / cp
Hope it helps :)
sin (47) = bc / 258
Solve bc
(b) Line BC = Hypotenuse of triangle CPB
sin (29) = bc / cp
(c) Tower CT = Opposite of Angle CTP
CP = Adjacent of Angle CTP
tan (Angle ctp ) = 46 / cp
Hope it helps :)
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