推 yulaw:謝謝您的詳解！:) 11/16 10:25
(1)
∠DCA = ∠DBA = x (圓周角對同弧)
sin∠DCA = sin∠DBA = sin x = AD/2R ---1
cos x = BD/2R ---2
∠CDB = ∠CAB = y (對同弧)
sin∠CDB = sin∠CAB = sin y = BC/2R ---3
cos y = AC/2R ---4
(1) + (3) => 9/2R = sin x + sin y
(2) + (4) => 18/2R = cos x + cos y
把R消掉
2(sin x + sin y) = cos x + cos y
和差化積後
tan( (x+y)/2 ) = 1/2
tan (x+y) = 4/3
sin (x+y) = 4/5 (看圖可知 sin(x+y)為所求)
(2)
∠BCA = x , ∠DCA = y
2R = 13/sinx = 46/siny ---1
x + y = 60度 ---2
用和角 解(1) (2)可求 AC = 2R = 62
圖要自己畫一下
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