## Cosine Rule Proof

We can split the triangle into two right-angled triangles like so:

Using Pythagoras we know tha:

$①:h^2+x^2=b^2$

and $②:h^2+(c-x)^2=a^2$

We can eliminate $$h$$ by subtracting the equations:

$①-②: (h^2+x^2)-(h^2+(c-x)^2)=b^2-a^2$

$x^2-(c-x)^2=b^2-a^2$

$x^2-(c^2-2cx-c^2)=b^2-a^2$

Rearranging: $a^2=b^2+c^2-2cx$

We know that $$\cos A=\frac{x}{b}$$ and so $$x=b \cos A$$. Substituting this in gives:

$a^2=b^2+c^2-2bc \cos A$

## Sine Rule Proof

We can split the triangle again into two right-angled triangles as we did with the cosine rule:

(* diagram)

We know that $$\sin B=\frac{h}{a}$$ and $$\sin A=\frac{h}{b}$$, so we see that $$h=a\sin B$$ and $$h=b\sin A$$.

$\therefore a \sin B=b \sin A$

$\therefore \frac{a}{\sin A}=\frac{b}{\sin B}$

## Area of a Triangle Proof

Continuing to use the same triangle as before, we can easily see that the area of our triangle is $$\frac{1}{2}hc$$:

(* diagram)

We can see from basic trigonometry that $$h=b\sinA$$, and therefore we conclude that the area of the triangle is given by $$\frac{1}{2}bc\sinA$$