math
posted by christina .
Find, correct to the nearest degree, the three angles of the triangle with the given vertices.?
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

I have no idea how you were taught to calculate vectors.
For any two points P(a,b) and Q(c,d)
vector PQ = < ca, db>
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 rightangled, as I showed in my first method. 
I have no idea how you were taught to calculate vectors.
For any two points P(a,b) and Q(c,d)
vector PQ = < ca, db>
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 rightangled, as I showed in my first method. 
THANK YOU SO MUCH!
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