user18181818
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I got -5 for i but what do i do from there to solve ii (1 Viewer)
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scaryshark09
∞∆ who let 'em cook dis long ∆∞
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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
BC = -1i - 3j
BA.BC = |BA||BC|cosABC
(-1)(-1) + (2)(-3) = (root5)(root10) cosABC
-5/root50 = cosABC
angle ABC = 135 degrees
Last edited:
user18181818
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the answer is 135
user18181818
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ohh i get it it's the same as what u did but just considering cos being negative
thanks
thanks
scaryshark09
∞∆ who let 'em cook dis long ∆∞
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sorry, i just edited my answer