## Cosine Rule Proof

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

(*add diagram)

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 \(\frac{1}{2}bc\sinA\)