✔ 最佳答案
1.G=ΔABC重心，向量AB=a，AC=b，c=a+t*b，向量AG=x*a+y*b，(x，y)，向量BG=?a+?b
BG延長交AC於E點:BE = BA + AE = -a + b/2BG = BE*2/3 = (-a + b/2)*2/3= -2a/3 + b/3= Answer(x,y) = (-2/3,1/3) 2.正8邊形由最底邊最左邊逆時針分別是OABCDEFG，若每邊長為2，則向量AB，BC，CD=?Set O = (0,0)=> A = (2,0)=> B = A + 2*(√2/2, √2/2)= (2,0) + (√2,√2)= (2+√2,√2)=> C = B + (0,2) = (2+√2,2+√2) => D = C + (-√2,√2) = (2,2+2√2)=> AB = B - A = (√2,√2) = √2*(1,1)=> BC = C - B = (0,2) = 2*(0,1)=> CD = D - C = (-√2,√2) = √2*(-1,1) 3.ΔABC: A(3，3),B(2，2),C(1+√3，1-√3)，則向量BA•向量BC=?BA = B - A= (2,2) - (3,3)= (-1,-1)BC = C - B= (1+√3,1-√3) - (2,2)= (-1+√3,-1-√3)BA.BC = -1*(-1+√3) + 1*(1+√3) = 2 4.向量a=(√3，1)，向量b=(1，0)，向量c=(0，2)，向量d=(3，3)，求向量a&b，c&d之夾角cosA = a.b/|a|*|b|= √3 / 2*1= √3/2A = 30 deg
cosB = c.d / |c|*|d|= 6 / 2*3√2= 1/√2B = 45 deg
2015-07-14 19:45:33 補充：
5.向量AB=(6，1)，BC=(x，y)，CD=(-2，-3)，BC//DA,AC垂直BD，求(x，y)
0 = AB + BC + CD + DA
= (6,1) + (x,y) + (-2,-3) + (kx,ky)
= (4+(k+1)x,-2+(k+1)y)
=> x = -4/(k+1), y = 2/(k+1)
AC = AB + BC
= (x+6, y+1)
= (-4/(k+1)+6, 2/(k+1)+1)
= (6k+2,k+3)/(k+1)
2015-07-14 19:46:18 補充：
BD = BC + CD
= (x,y) + (-2,-3)
= (x-2, y-3)
= (-4/(k+1)-2, 2/(k+1)-3)
= (-2k-6,-3k-1)/(k+1)
0 = AC.BD
= (6k+2)(2k+6) + (k+3)(3k+1)
= 15k^2 + 50k +15
= 5*(3k^2 + 10k + 3)
= 5(3k+1)(k+3)
=> k = 3 or -1/3
=> x = -1, y = 1/2
2015-07-14 20:15:53 補充：
=> or x = -6, y = 3
6.sinA/(1+cosA) + sinA/(1-cosA) - 2cscA = ?
= [sinA(1-cosA) + sinA(1+cosA)]/(1-cos^2A) - 2/sinA
= sinA/sin^2A - 2/sinA
= (1-2)/sinA
= - 1/sinA
= - cscA
2015-07-14 20:17:26 補充：
6修改:
2/ainA - 2/sinA = 0
2015-07-14 20:21:32 補充：
7.x^-x+k=0之二根為tanA cotA，則sinA*cosA=?
k = tanA*cotA = 1
1 = 兩根和
= tanA + cotA
= sinA/cosA + cosA/sinA
= (sin^2A + cos^2A)/sinA*cosA
= 1/sinA*cosA
=> sinA*cosA = 1
2015-07-14 20:49:49 補充：
8.三角形ABC，C的度數為90，若2cosA+3cosB=3，則sinA/sinB=
射影定律:
b*cosA + a*cosB = c
b = 2k, a= 3k
正弦定律:
sinA/sinB = a/b
= 2k/3k
= 2/3
= Answer
2015-07-14 21:08:03 補充：
如果 C = 90 deg
則可以求取諸邊長,商高定理:
(4 + 9)k^2 = 13k^2
= c^2
= 3^2
=> k = 3/√13
=> b = 2k = 6/√13
=> a = 3k = 9/√13
2015-07-15 06:26:56 補充：
打顛倒修改: sinA/sinB = a/b = 3/2
第5題漏打負號修改:
(x,y) = (2,-1) for k = -3
(x,y) = (6,-3) for k = -1/3
2015-07-15 06:32:13 補充：
第3題位置顛倒修改:
BA = A - B = (1,1)
=> BA.BC = -2
