F.5Maths (circle) 20POINTS!!!!

87Q53

From the the Angle at the Center Theorem,
a+b=(360- c )/ 2 = 180- c/2

90Q39

From Angle at the Center Theorem,
tan 30 = BC/AC =1/√3
and BC*AC=2√3
AC=√6

cos 30=AC/AB=√3/2=√6/AB
AB=√6/(√3/2)=2√2
radius=√2

2013-11-10 10:39:44 補充：
92Q50
By Alternate Segment Theorem
∠ADE= ∠ECD
∠CDE=180-24-3∠ADE=52
∠DCB=90-52=38 (Angle in semicircle)