ABCD is a parallelogram with angle ABC=50. E is a pointinside the parallelogram such that EC=BA and angle EAC=37 and angle ECA=13.Find angle ECB.
Thanks indeed.
Math F5
2012-03-05 3:10 am
回答 (1)
2012-03-05 3:40 am
✔ 最佳答案
let angle ECB=xtriangle AEC:
angle AEC=180-(37+13)=130
EC/sin37=AC/sin130
AC=ABsin130/sin37 ------(1)
triangle ABC
AB/sin(x+13)=AC/sin50
AC=ABsin50/sin(x+13) ---------(2)
By (1) and (2),
sin130/sin37=sin50/sin(x+13)
x=22
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