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January 18, 2017
Posted by **christina** on Friday, September 13, 2013 at 11:05pm.

A(2, 0),B(4, 5), C(-3,2)

ANGLE- CAB

ANGLE- ABC

ANGLE- BCA

In degrees.

I got the vectors AB <,5>

BC <1,3>

AC <-1,2>

and I know the equation b.c= |b| |c| cos t

- math -
**Anonymous**, Friday, September 13, 2013 at 11:53pmI have no idea how you were taught to calculate vectors.

For any two points P(a,b) and Q(c,d)

vector PQ = < c-a, d-b>

I got

vector AB = <2,5>

vector BC = <-7,-3) and

vector AC = <-5,2>

let angle CAB = Ø

then vector AC . vector AB = |AC| |AB| cosØ

<-5,2> . <2,5> = | <-5,2>| |<2,5>| cosØ

-10+10 = √29 √29 cosØ

cosØ = 0

Ø = 90°

well, that was obvious from looking at the two vectors, their dot product is zero, so they are perpendicular

for angle ABC

<-3,-7> . <-5,-2> = √58 √29 cos B

15+14 = √58√29cosB

cos B =.707106...

B = 45°

Well, how about that ?

Suppose we had taken the length of each vector

|AB = √29

|AC| = √29

|BC| = √58

so it is isosceles, and since

(√58) ^2 = (√29)^2 + (√29)^2

so it is also right-angled, as I showed in my first method. - math -
**Reiny**, Friday, September 13, 2013 at 11:55pmI have no idea how you were taught to calculate vectors.

For any two points P(a,b) and Q(c,d)

vector PQ = < c-a, d-b>

I got

vector AB = <2,5>

vector BC = <-7,-3) and

vector AC = <-5,2>

let angle CAB = Ø

then vector AC . vector AB = |AC| |AB| cosØ

<-5,2> . <2,5> = | <-5,2>| |<2,5>| cosØ

-10+10 = √29 √29 cosØ

cosØ = 0

Ø = 90°

well, that was obvious from looking at the two vectors, their dot product is zero, so they are perpendicular

for angle ABC

<-3,-7> . <-5,-2> = √58 √29 cos B

15+14 = √58√29cosB

cos B =.707106...

B = 45°

Well, how about that ?

Suppose we had taken the length of each vector

|AB = √29

|AC| = √29

|BC| = √58

so it is isosceles, and since

(√58) ^2 = (√29)^2 + (√29)^2

so it is also right-angled, as I showed in my first method. - math -
**christina**, Friday, September 13, 2013 at 11:59pmTHANK YOU SO MUCH!