trigonometry

1. Let BP = PQ = QC = r ,
Consider △BAC ,
tan 44° = AC / BC = AC / 3r ... (1)
Consider △AQC ,
tan∠CAQ = QC / AC = r / AC ... (2)
(1)x(2) : (tan 44°) tan∠CAQ = 1/3 , ∠CAQ = 19.0°

2. (2sinA - 3cosA) / (3sinA + 2cosA)
= ( 2tanA - 3 ) / (3tanA + 2)
= [ 2(-5/4) - 3 ] / [ 3(-5/4) + 2 ]
= 22 / 7

2012-01-28 21:16:50 補充：
I think your answer is wrong , where did you find the answer ?

my ans is19 too...