1. ## General question about a triangle's height

Hi Forum

I studied parallelograms and triangles and I got to the formula of the area of both of them.
Namely, 1absinA/2 or absinA.
Naturally, an area of a triangle is also 1bh/2.

So, if we mash them up, considering that sinA=c/b
1/2 x bh = 1/2 x ab x c/b

Could this mean that bh=ac?

But this formulas work for both right angled triangles and non-right angled triangles, right?
So we can't consider sinA=c/b at all times, since the angle is not always right.
Does this has -something- to do with the Law of Sines or Cosines?

Sorry if I haven't made myself clear.
Thanks.

2. Originally Posted by Zellator
Hi Forum

I studied parallelograms and triangles and I got to the formula of the area of both of them.
Namely, 1absinA/2 or absinA.
Naturally, an area of a triangle is also 1bh/2.

So, if we mash them up, considering that sinA=c/b
1/2 x bh = 1/2 x ab x c/b

Could this mean that bh=ac?

But this formulas work for both right angled triangles and non-right angled triangles, right?
So we can't consider sinA=c/b at all times, since the angle is not always right.
Does this has -something- to do with the Law of Sines or Cosines?

Sorry if I haven't made myself clear.
Thanks.
You have got everything confused. Even in a right triangle sin(A) !=c/b, assuming you have labelled the sides and angles in the conventional manner.

CB

3. Yes, that is what I meant.
I know that in a right triangle sin(A) is possible.
I'm considering b as the base of the triangle.

Yes I got confused on this one, now I get it.
bh=absin(a)
so sin(a)=h/a

Second opinions get me back to reality.
That's why they are important.

Thanks CaptainBlack