I got -5 for i but what do i do from there to solve ii (1 Viewer)

[user18181818]

 

[scaryshark09]

BA = -1i + 2j
BC = -1i - 3j

BA.BC = |BA||BC|cosABC

(-1)(-1) + (2)(-3) = (root5)(root10) cosABC

-5/root50 = cosABC

angle ABC = 135 degrees

 

[user18181818]

the answer is 135

 

[user18181818]

ohh i get it it's the same as what u did but just considering cos being negative
thanks

 

[p711]

ur supposed to do BA.BC thats how u get 135

 

[p711]

same thing as using AB.CB (the vectors should be both pointing to B or both away from B) im pre sure

 

[scaryshark09]

the answer is 135

sorry, i just edited my answer