For triangle ABC ( order in anti-clockwise direction), D is a point on AC such that angle DBC = 50 degree and angle BDC = 105 degree. AB = DC. Find angle BAC ( to the nearest degree.)
Angle of triangle
2009-03-10 11:34 pm
回答 (2)
2009-03-11 12:37 am
✔ 最佳答案
Note : (d) = degreeLet x be the angle BAC.
In triangle BDC, by sine law,
DC / sin 50(d) = BC / sin 105(d) ---------- (1)
In triangle ABC, by sine law,
BC / sin x = AB / sin 25(d) ---------------- (2)
From (1) : DC = [BC sin 50(d)] / sin 105(d) ----- (3)
Since AB = DC, sub. (3) into (2)
BC / sin x = {[BC sin 50(d)] / sin 105(d)} / sin 25(d)
____sin x = sin 25(d) / [sin 50(d) / sin 105(d)]
____sin x = 0.532890614
_______x = 32.20097083(d)
Therefore, the angle BAC = 32(d) (to the nearest degree)
2009-03-11 12:17 am
Angle BAC=25 degree
Because angle BCA=25 degree(180-105-50), AB=DC(iss. triangle)
So angle BAC=25 degree
Because angle BCA=25 degree(180-105-50), AB=DC(iss. triangle)
So angle BAC=25 degree
收錄日期: 2021-04-13 16:29:37
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