Hello arze Quote:

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**arze** 1. $\displaystyle \cot C=\frac{a}{c}\csc B-\cot B$

I have not idea where to start here.

2. $\displaystyle abc=4\bigtriangleup R$ where R is the radius of the circumcircle, and $\displaystyle \bigtriangleup$ is the area of the triangle.

I used $\displaystyle \bigtriangleup=\sqrt{s(s-a)(s-b)(s-c)}$ where $\displaystyle s=\frac{1}{2}(a+b+c)$ and got the value for $\displaystyle \bigtriangleup=\frac{1}{4}\sqrt{-a^4+2a^2b^2+2a^2c^2-b^4+2b^2c^2}$ which doesn't have a square root. Now I'm stuck

Thanks

1. Use the Sine Rule on the RHS, by writing$\displaystyle \frac{a}{c}=\frac{\sin A}{\sin C}$

and write $\displaystyle \csc B$ and $\displaystyle \cot B$ in terms of $\displaystyle \sin B$ and $\displaystyle \cos B$.

So:$\displaystyle \frac{a}{c}\csc B-\cot B =\frac{\sin A}{\sin C\sin B}-\frac{\cos B}{\sin B}$$\displaystyle =\frac{\sin A-\sin C\cos B}{\sin C\sin B}$

Now use the fact that $\displaystyle \sin A = \sin (180^o-A) = \sin(B+C)$. Expand this, and then simplify the numerator. Can you finish this now?

2. Use $\displaystyle \Delta = \tfrac12bc\sin A$, and $\displaystyle \frac{a}{\sin A}=2R$ and you're done.

Grandad