Triangle, 3 angles of despair

Oh, sorry for the weird title. Anyway, the other day when I was walking back home with David Mosiun after the can rings hunt, he suggested to me about making the triangle a little bit more complicated. Yeah, so I wondered, why not?

Here's what I've done.

WARNING: If you an uncontrollable hatred towards Maths, I suggest you should look away.

First, let's place a triangle in the cartesian plane:

Simple enough? It's important that the labels for the vertices are in clockwise order. I'm gonna use vectors here, since vectors are so fun! No trigonometry will be involved (well, except for Pythagoras' Theorem)

First, run a line from point C down to line AB so that it is perpendicular to line AB.

That line (h) will divide line AB into 2 sections, a and b. What I'm gonna do now is find a, b and h. From there, I will be able to get the area of the triangle. Remember, I'm gonna do this without the help of trigo, since I'm on a trigo-hiatus. Hahaha.

We'll need the vector for line AB, so, it'll be

A-B.

Then, we'll need to normalize it.

W=(A-B)/(|A-B|)

Right, now, that'll be called W. To get the length of section a, we're gonna have to project line AC onto W. That's where dot product comes in handy. Because dot products can give negative numbers, which we don't really need, let's get the absolute value of it.

a=|(A-C) ּ W|

Same goes for section b, instead, it'll involve line BC.

b=|(B-C) ּ W|

Now for h. Since we know b. And c is just equal to |B-C|, we can get h:

h=sqrt(c²-b²)
h=sqrt[ (|B-C|)² - ((B-C) ּ W)² ]
Actually, since we are squaring those numbers, the absolute value sign is actually quite redundant, since squares are always positive. That's why I took out the absolute value sign. But note that I didn't touch the modulus sign for B-C in the (|B-C|) because that sign there represents the magnitude of vector (B-C).

Now we have a,b and h, let's put them together:

Area of triangle=1/2 x base x height

so,

Area of triangle
=1/2 (a+b) h
=1/2 (|(A-C) ּ W| + |(B-C) ּ W|) sqrt[ (|B-C|)² - ((B-C) ּ W)² ]

or, if you like a better view of this equation:

where,


Ok, so there's your "area of triangle" being turned into a garbled mess! Now, I don't guarantee that the equation above will be 100% accurate. There may be errors in there which I may have overlooked. You won't be seeing that equation in your textbooks anytime soon, so don't worry too much about it and blah. It's just for fun.