Question

This diagram shows a cyclic quadrilateral ABCD, centre O.
Angle DAB = 110 and angle ABC = 100.
TA and TD are tangents at A and D.
TD is extended to U and angle UDC = 64.

Calculate:
1) Angle ODC?
2) Angle DAC?

Answer 1

Draw A Line From O to D
A line drawn from center to the tangent at the point where it is touching the circumference of the circle is perpendicular to it

Therefore,

∠ODT = 90'

∠UDC + ∠ODC + ∠ODT = 180' [since UDT is a straight line]

64 + ∠ODC + 90 = 180
∠ODC = 180 - 154
∠ODC = 26'

∠ADC + ∠ABC = 180' [opposite angles in the cyclic quadrilateral are supplementary]

∠ADC + 100 = 180
∠ADC = 80'

Draw a line from A to C

∠UDC = ∠DCA [alternate segment theorem]

Therefore 
∠DCA = 64'

In ΔDAC

∠CDA + ∠DAC + ∠ACD = 180 [Angle Sum Property]

80 + ∠DAC + 64 = 180

∠DAC = 180 - 144

∠DAC = 36'

Therefore,
∠ODC = 26'
∠DAC = 36'

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Answer 2

1.∠ODC = 26
2.∠DAC = 36