# Prove for trianlge ABC

• Feb 24th 2010, 06:20 PM
arze
Prove for trianlge ABC
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
• Feb 25th 2010, 12:44 AM
Hello arze
Quote:

Originally Posted by 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.