In the figure, PQ is a tangent to the circle at A. PQ//BD, BC=DC, ∠BCD=132° and ACD is a straight line.
(a) Find ∠DAQ.
(b) Find ∠BAC.

I have problem on part b.

Question

In the figure, PQ is a tangent to the circle at A. PQ//BD, BC=DC, ∠BCD=132° and ACD is a straight line.
(a) Find ∠DAQ.
(b) Find ∠BAC.

I have problem on part b.

anonymous · Answer

|dw:1359172897801:dw|

hartnn · Answer

how do you conclude that CAB = 90 ? 
how is that angle in semicircle ? (where is centre ?)

hartnn · Answer

|dw:1359173286074:dw|
angles in a triangle =180
so , 132 +x+x = 180 
also, x=DAQ (alternate angles)

anonymous · Answer

wait....
sorry I type the answer wrongly, I overlook my answer book...

132°+2∠CDB=180° (∠ sum of ∆)
∠CDB=24
∠DAQ=∠CDB=24 (alt.∠s, PQ//BD)

anonymous · Answer

is it?

hartnn · Answer

thats exactly what i wrote....yes, its correct.

anonymous · Answer

thx. how about part b?

hartnn · Answer

if centre lies on BD then i have an easy solution....else i am still thinking.

anonymous · Answer

I think BD is not passing through the center

anonymous · Answer

hey...