# Vector triangle help

• Jan 25th 2008, 01:12 PM
Centara
Vector triangle help
Okay, so first time on the site. Let's hope this goes well, lol.

The question: Find, correct to the nearest degree,the three angles of the triangle with the given vertices.
The vertices: A(1,0), B(3,6), C(-1,4)

So I tried to start it out. Drew it because visuals rock. Then I found the distances between the points.

AB= $\sqrt{40}$
BC= $\sqrt{20}$
CA= $\sqrt{20}$

That's about as far as I was able to go. Any help is appreciated. Thanks!

~Cen
• Jan 25th 2008, 02:45 PM
Plato
$m\left( {\angle ABC} \right) = \arccos \left( {\frac{{\overrightarrow {BA} \cdot \overrightarrow {BC} }}{{\left\| {\overrightarrow {BA} } \right\|\left\| {\overrightarrow {BC} } \right\|}}} \right)
$
• Jan 25th 2008, 03:05 PM
Centara
Ok, I slightly get what you're saying but how would I find the magnitude of BC and BA?
• Jan 25th 2008, 03:12 PM
Plato
Quote:

Originally Posted by Centara
Ok, I slightly get what you're saying but how would I find the magnitude of BC and BA?

You did it in the original post.
The length of the vector(BA) is the length of the line segment BA.

Do you know how to find the vector(BA)?
Do you know how to find dot products?
• Jan 25th 2008, 05:51 PM
Centara
I have no idea how to find vector BA. We kind of just started this stuff last week and I'm trying to understand it all but I'm getting so caught up it's insane.

I can do dot products. That's simple. Is that how you find BA? It'd be my luck it is, lol.
• Jan 25th 2008, 06:08 PM
Plato
Quote:

Originally Posted by Centara
I have no idea how to find vector BA. We kind of just started this stuff last week and I'm trying to understand it all but I'm getting so caught up it's insane. I can do dot products. That's simple. Is that how you find BA? It'd be my luck it is, lol.

It is a sad statment on education that you are asked to do this.
• Jan 26th 2008, 12:25 AM
Opalg
Quote:

Originally Posted by Centara
Okay, so first time on the site. Let's hope this goes well, lol.

The question: Find, correct to the nearest degree,the three angles of the triangle with the given vertices.
The vertices: A(1,0), B(3,6), C(-1,4)

So I tried to start it out. Drew it because visuals rock. Then I found the distances between the points.

AB= $\sqrt{40}$
BC= $\sqrt{20}$
CA= $\sqrt{20}$

That's about as far as I was able to go. Any help is appreciated. Thanks!

~Cen

Do you know the cosine rule? Since you know the lengths of the sides of the triangle, that will enable you to calculate the angles.
• Jan 26th 2008, 09:58 AM
Centara
Okay, I finally realized what I was missing. There was a whole section of notes I was skipping over that had the formulas I was looking for. Same one Plato gave me but with the notation that I needed to remember how to find the vectors for BA and BC. Thanks guys!!!