SOHCAHTOA works for any type of triangle

And there I was thinking that only right angle triangles had Hippopotamuses

The hours just fly by for you guys, huh?

So how do you work it out?

So how do you work it out?

I think the gist of it is that any triangle can be thought of as 2 triangles containing right angles (imagine drawing a dotted line through it). For both these triangles you know the length of the side opposite the right angle so you can use trig (SOCAHTOA) to find the lengths of the other 2 sides. Then for each triangle you do b*h*0.5 and add the results.

EDIT: Er, or maybe not!
EDIT AGAIN: No, that's right 🙂

I think the gist of it is that any triangle can be thought of as 2 triangles containing right angles (imagine drawing a dotted line through it).

That is how I would've done it…. But after reading this thread, I'm well confussed!!

MrSalmon: I think if you're going for that method then you have to use the Law of Cosines first to get the angles.

Do you need the angles? If you have split your triangle into two right angled triangles by dropping a perpendicular line from C in GrahamS diag to the side AB. Then you don't know the length of this line, say h. Likewise you dont know how far along it intersects along side AB, assume the distance is x from A. Using Pythagoras, you know b^2 = x^2 + h^2 and you also know a^2= (c-x)^2 + h^2, solve for h and x and work out the areas and add together.

On you go then…

mefty – for help, read the link in inigo's post.

The algebraic manipulation is stepped through very nicely in there.

If my algebra is right, and it has been a while since I've had to do stuff like this, then that should work as it should simplify to

2x^2+2cx+a^2-b^2-c^2 = 0.

You will need to solve the resultant quadratic equation which granted isn't as easy as that other formula but still doesn't require the use of any trig.