Finding correct angle of triangle when using vectors

[Zulgok]

[SOLVED]Finding correct angle of triangle when using vectors

Hello, I have to find an angle of a triangle and determine its shape. For this task, I was given 3 points :
A(-2;5)
B(4;-3)
C(6;3)

Now, when I'm drawing these points on a paper, it is obviously an obtuse triangle, but when I am trying to calculate the angle using vectors, I am getting an acute triangle. And it seems when I'm not using vectors, AC is the longest line, when with vectors - AB is the longest. What I did :

Vectors:
AB(6;-8), magnitude 10
AC(8;-2), magnitude 2 * sqrt(17)
BC(2;6), magnitude 2 * sqrt(10)

Then I used a formula to find cosine:
(BA * BC) / (|BA| * |BC|) ( I used BA and BC because AC has a higher magnitude, and the largest angle is looking at it)

The cos should be negative, but it isn't. What am I doing wrong ?

 

[Zulgok]

Ok, I've found the problem. It seems I graphed it wrong, my bad

 

[Deleted member 4993]

Hello, I have to find an angle of a triangle and determine its shape. For this task, I was given 3 points :
A(-2;5)
B(4;-3)
C(6;3)

Now, when I'm drawing these points on a paper, it is obviously an obtuse triangle, but when I am trying to calculate the angle using vectors, I am getting an acute triangle. And it seems when I'm not using vectors, AC is the longest line, when with vectors - AB is the longest. What I did :

Vectors:
AB(6;-8), magnitude 10
AC(8;-2), magnitude 2 * sqrt(17)
BC(2;6), magnitude 2 * sqrt(10)

Then I used a formula to find cosine:
(BA * BC) / (|BA| * |BC|) ( I used BA and BC because AC has a higher magnitude, and the largest angle is looking at it)

The cos should be negative, but it isn't. What am I doing wrong ?

You are trying to calculate the angle whose vertex is C. As I plotted it, ACB is actually acute (~88°)

You need to consider vectors CA(-8,2) and CB(-2,-6).

Then CA\(\displaystyle \cdot\)CB = (-8*-2) + (2*-6) = 4

cos(Θ) = 4/(4*√170) → Θ = 85.6°