Week 3 Distance between vectors question

It was of my understand that cos(θ) from soh-CAH-toa only applied for square triangles?

Could you please help me to understand why does it applies in an example of the triangle generated by coordinates (1,5) and (6,2)?

It doesn’t seems like a rectangular triangle to me.

Thanks beforehand.

Sorry, I don’t quite understand what you mean by a “rectangular triangle”.

Can you give more details or specific references to the course materials?

Sorry, I meant right triangle.

Can you give a specific reference into the course materials?

Nevermind, I confused the direction of a vector with the cosine distance.
I can see they are related.

The direction it’s related to the usage of SoHCah---->[Toa].
Where Tan(opposite/adjacent) = tan(θ), and the usage of it’s inverse arctan(opposite/adjacent)=θ.

But then I notice that the cosine distance was being use in a not right triangle.
As SoHCaHToa it’s related to the cosine distance, then I thought that cosine distance could only be applied (just as SoHCaHToa) to right triangles.

But I can see now, that it can be applied to any kind of triangle.

Thanks.