Hafsa Sheikh's answer to Piranha's Secondary 3 E Maths Singapore question.
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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°)
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°)
Date Posted:
4 years ago
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