triangle ABC right angled at b and bd is perpendicular to AC find cos angle CBD
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BD/BC =cos angleCBD
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The right angle ∆BCD so formed.
In ∆BCD
Angle BDC =90°
So,
Hypotenuse-->BC
Cos(angle CBD) = b/h = BD/BC
In ∆BCD
Angle BDC =90°
So,
Hypotenuse-->BC
Cos(angle CBD) = b/h = BD/BC
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