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secondary 3 | E Maths
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Date Posted: 1 year ago
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Using Pythagoras theorem to find,
BD = 12 cm
Angle ADB = Arcsin(5/13×sin 90) =22.62 deg
a) use sine rule,
Angle BCD = Arcsin (12/9 × sin 22.62) = 30.85 deg
b) use sine rule,
BC = Arcsin(9×sin 126.53÷sin 22.62)
= 18.8 cm
c) Using sine rule,
Shortest distance= 9xsin30.85÷sin90
= 4.62cm
BD = 12 cm
Angle ADB = Arcsin(5/13×sin 90) =22.62 deg
a) use sine rule,
Angle BCD = Arcsin (12/9 × sin 22.62) = 30.85 deg
b) use sine rule,
BC = Arcsin(9×sin 126.53÷sin 22.62)
= 18.8 cm
c) Using sine rule,
Shortest distance= 9xsin30.85÷sin90
= 4.62cm
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