Circle (1)

Problem 1a)

Look at triangle OCB, OC = OB make it an isosceles triangle. Therefore $ c = \frac{180 - 36}{2} = 72 $.

Similarly, look at triangle OAB, OA = OB make it an isosceles triangle. Therefore $ a = \frac{180 - 144}{2} = 18 $

Last but not least. b = angle OBC + angle OBA = 72 + 18 = 90. One can also argue that is true because it is an angle in a semi-circle.

Problem 1b)

For (i)

An angle inscribed in a semicircle is always a right angle:

<DCA is 90 degree because it is an angle subtended by the diameter (i.e. angle in a semi-circle)

For (ii)

The sum of the measures of the interior angles of a triangle is 180. 

<DAC = 180 - <ADC - < DCA = 180 - 60 - 90 = 30.

For (iii)

Opposite angles of a cyclic quadrilateral add to 180°

<ABC + < ADC = 180

Therefore < ABC = 180 - 60 = 120 [The answer on the sheet is wrong]

For (iv)

Opposite angles of a cyclic quadrilateral add to 180°

<DAB + < DCB = 180

Therefore <DAB = 180 - <DCB = 180 - <DCA - <ACB = 180 - 90 - 35 = 55  [The answer on the sheet is wrong]