Alternative area of a triangle formula

Question

The problem is as follows: There is a triangle $ABC$ and I need to show that it's area is: $$\frac{1}{2} c^2 \frac{\sin A \sin B}{\sin (A+B)}$$ Since there is a half in front I decided that base*height is equivalent to $c^2 \frac{\sin A \sin B}{\sin (A+B)}$. So I made an assumption that base is $c$ and went on to prove that height is $c \frac{\sin A \sin B}{\sin (A+B)}$. But I end up expressing height in terms of itself.. i.e. $h \equiv \frac{ch}{a\cos B + b \cos A}$. How do I prove this alternative area of triangle formula?

Answer 1

Let's denote by $[ABC]$ the area of $\triangle ABC$, its known that $$[ABC]=\frac{1}{2}ab\sin C$$ From Sine Law we have $a\sin C=c\sin A$ and $b\sin C=c\sin B$, also $\sin (A+B)=\sin(\pi-C)=\sin C$, then \begin{align*} [ABC]&=\frac{1}{2}\frac{(a\sin C)(b\sin C)}{\sin C}\\ &=\frac{1}{2}\frac{(c\sin A)(c\sin B)}{\sin (A+B)}\\ &=\frac{1}{2}c^2\frac{\sin A\sin B}{\sin (A+B)} \end{align*}

Answer 2

Here is what I did:

$$\frac{1}{2}bh \equiv \frac{1}{2}c^2\frac{\sin A \sin B}{\sin (A+B)}$$

Assume that $b$ = $c$.

Then, $c\frac{\sin A \sin B}{\sin (A+B)} \equiv \frac{c}{\cot A + \cot B}$. But $\cot A \equiv \frac{b \cos A}{h}$ and $\cot B \equiv \frac{a\cos B}{h}$. Therefore $\frac{c}{\cot A + \cot B} \equiv \frac{ch}{a\cos B + b\cos A}$. Since $a\cos B + b\cos A \equiv c$, we have $h \equiv \frac{ch}{c}$. $h\equiv h$.

I realised this after I posted the question...

Answer 3

$$\triangle =\dfrac12bc\sin A=\dfrac12(2R\sin B)c\sin A\cdot\dfrac c{2R\sin C}$$

Now $A+B=\pi-C\implies\sin(A+B)=\sin(\pi-C)=?$