 ## Question

secondary 3 | E Maths

Anyone can contribute an answer, even non-tutors.

##### Megan Yifei

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Date Posted: 2 months ago
Views: 12
J
2 months ago
a)

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)
J
2 months ago
b)

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°)
J
2 months ago
c)

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° 