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F.5Maths (circle) 20POINTS!!!!
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please view the file below, thx. Give detailed steps please. https://www.dropbox.com/s/u1dypiz5h8v7cj6/circle%20mc.doc
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)
其他解答:
F.5Maths (circle) 20POINTS!!!!
發問:
please view the file below, thx. Give detailed steps please. https://www.dropbox.com/s/u1dypiz5h8v7cj6/circle%20mc.doc
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最佳解答: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)
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