數學三問

[happy520]

1.三角形,三中線長為3,5,6,則此三角形之面積為?

2.三角形ABC中,aa-cc=bc且角B=51度,求角C=?

3.[(√3+i)/2]^7+1=Z,求｜Z｜及ArgZ

[thepiano]

第 1 題
參考
http://forum.nta.org.tw/oldphpbb2/viewtopic.php?t=11910
http://forum.nta.org.tw/oldphpbb2/viewtopic.php?t=17368


第 2 題
a^2 - c^2 = bc
由正弦定理
(sinA)^2 - (sinC)^2 = sinBsinC
sin(A + C)sin(A - C) = sinBsinC
sin(A + C) = sinB
sin(A - C) = sinC
A - C = C 或 A - C = 180 - C(不合)
......


第 3 題
Z = [cos(π/6) + isin(π/6)]^7 + 1
= 1 + cos(7π/6) + isin(7π/6)
= 2cos(7π/12)[cos(7π/12) + isin(7π/12)]
= 2cos(7π/12)[-cos(5π/12) + isin(5π/12)]
= [-2cos(7π/12)][cos(5π/12) - isin(5π/12)]
= [-2cos(7π/12)][cos(19π/12) + isin(19π/12)]

[happy520]

第二題不懂吔

[armopen]

我來幫 thepiano 老師回一下，

第 2 題
a^2 - c^2 = bc
由正弦定理
(sinA)^2 - (sinC)^2 = sinBsinC
sin(A + C)sin(A - C) = sinBsinC
sin(A + C) = sinB
sin(A - C) = sinC
A - C = C 或 A - C = 180 - C(不合)
......

首先這邊用到二個觀念，第一是正弦定理： a/sinA = b/sinB = c/sinC = 2R, R 是三角形 ABC 的外接圓半徑

第二是用到和角公式展開 sin(A+C) sin(A-C) 並利用平方關係代換可以推得 (sinA)^2 - (sinC)^2.

第三是因為 π - (A+B) = C 所以 sinC = sin[π - (A+B)] = sin(A+B). 得到 A = 2C 時就可以用一元一次方程式得答案.