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secondary 3 | E Maths

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Angle ABC = 90° (angle in a semicircle)

Angle OBC = Angle ABC - Angle OBA

= 90° - 64° = 26° (complementary angles)

OB = OC

(both are radius of the same circle)

So △ OBC is isosceles.

Angle OCB = Angle OBC = 26°

(base angles of isosceles triangle are equal ,OB = OC)

Angle ABC = angle AQB + angle BAQ

(Exterior angle = sum of 2 interior opposite angles)

90° = 35° + Angle BAQ

So Angle BAQ = 90° - 35° = 55°

Since PAB and DAQ are straight lines,

then angle DAP = angle BAQ

(Vertically opposite angles)

= 55°

(Sum of angles on a straight line add up to 180°)

Angle CAB

= 180° - angle ABC - angle OCB

= 180° - 90° - 26°

= 64°

(Angle sum of triangle ABC)

Or

OA = OB (both are radius of same circle)

So triangle OAB is isosceles.

Angle OAB = angle OBA = 64°

(Base angles of isosceles triangle OAB, OA = OB)

Angle CAB = Angle OAB = 64°

(Common angles)

Then,

Angle DAC = 180° - Angle BAQ - angle CAB

= 180° - 55° - 64°

= 61°

(Sum of angles on a line DAQ)

Angle COD = 2 x angle DAC

(angle at centre = 2 x angle at circumference, relevant segment is CD)

2 x angle DAO (since DAO and DAC are common angles)

= 2 x 61°

= 122°

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