Question

Use the result of Problem 50 and the results of Problems 56 and 57 in Section 8.3 to show that

$\cot \frac{C}{2}=\frac{s-c}{r}$where$s=\frac{1}{2}\left(a+b+c\right)$.

Short answer

It is proved that$\cot \frac{C}{2}=\frac{s-c}{r}$.

Step-by-step solution

Step 1. Given Informtion

Given that $s=\frac{1}{2}\left(a+b+c\right)$

use localid="1650175628630" $r=\frac{c\sin {\displaystyle \frac{A}{2}}\sin {\displaystyle \frac{B}{2}}}{\cos {\displaystyle \frac{C}{2}}}$

localid="1650175390705" $$\begin{gathered} \sin \frac{C}{2}=\sqrt{\frac{\left(s-a\right)\left(s-b\right)}{ab}} \\ \cot =\frac{\sin }{\cos } \end{gathered}$$

Step 2. Proving the statement

$$\begin{gathered} r=\frac{c\sin {\displaystyle \frac{A}{2}}\sin {\displaystyle \frac{B}{2}}}{\cos {\displaystyle \frac{C}{2}}} \\ \cos \frac{C}{2}=\frac{c\sin \frac{A}{2}\sin \frac{B}{2}}{r} \end{gathered}$$

$$\begin{gathered} \frac{\cos {\displaystyle \frac{C}{2}}}{\sin {\displaystyle \frac{C}{2}}}=\frac{c}{r}\frac{\sin {\displaystyle \frac{A}{2}}\sin {\displaystyle \frac{B}{2}}}{\sin \frac{C}{2}} \\ \cot \frac{C}{2}=\frac{c}{r}\frac{\sin {\displaystyle \frac{A}{2}}\sin {\displaystyle \frac{B}{2}}}{\sin \frac{C}{2}} \end{gathered}$$

Now.

$$\begin{gathered} \cot \frac{C}{2}=\frac{c}{r}\frac{\sqrt{{\displaystyle \frac{\left(s-b\right)\left(s-c\right)}{bc}}}\sqrt{\frac{\left(s-a\right)\left(s-c\right)}{ac}}}{\sqrt{\frac{\left(s-a\right)\left(s-b\right)}{ab}}} \\ \cot \frac{C}{2}=\frac{c}{r}\sqrt{\frac{ab\left(s-a\right)\left(s-b\right){\left(s-c\right)}^2}{ab{c}^2\left(s-a\right)\left(s-b\right)}} \\ \cot \frac{C}{2}=\frac{c}{r}\sqrt{\frac{{\left(s-c\right)}^2}{c^2}} \\ \cot \frac{C}{2}=\frac{c}{r}\frac{s-c}{c}=\frac{s-c}{r} \end{gathered}$$