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secondary 3 | E Maths
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Piranha
Piranha

secondary 3 chevron_right E Maths chevron_right Singapore

Need help asap please

Date Posted: 4 years ago
Views: 278
snell
Snell
4 years ago
ABC = ADC = CDE = 90°
so
DCE = DEC = 45°

in triangle ABE,
BAD = 180-90-45 = 45°

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Angle ABC= Angle CDA= 90° (triangle in a semicircle property)
Since Angle CDA=90°, Angle CDE=90° (sum of angles on a straight line is 180°)
Angle DCE= Angle DEC= (180-90)/2= 45° (sum of angles in a triangle is 180°+isosceles triangle)
As such, Angle DCB=180-45= 135° (sum of angles on a straight line is 180°)
So, Angle BAD= 360-90-135-90= 45° (sun of angles in a quadrilateral is 360°)
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Hafsa Sheikh
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