jika panjang vektor oa =1,2 dan vektor ob=4,2 dan sudut antara vektor oa dan ob maka tan

VEKTOR


cos@ = (1,2)•(4,2)/(akar(1^2+2^2)×akar(4^2+2^2)
cos@ = (1×4 + 2×2)/akar(1+4)×akar(16+4)
cos@ = (4+4)/(akar5 × akar20)
cos@ = 8/akar5×10
cos@ = 8/akar100
cos@ = 8/10
cos@ = 4/5
=> SA= 4
=> MI = 5
DE^2 = 5^2 - 4^2 = 25 -16 = 9
DE^2 = 3^2
DE = 3

tan@ = DE/SA = 3/4