Chapter 8.4, Problem 50AYU

To determine

To Prove The relation $r=\frac{c\sin \frac{A}{2}\sin \frac{B}{2}}{\cos \frac{C}{2}}$ .

Answer to Problem 50AYU

  $\begin{array}{c}\frac{a\sin B\sin C}{\sin A}=\sin B\cdot c\\ =h\end{array}$

Explanation of Solution

Given:

The area of the triangle is $K$ .

The value of $s=\frac{1}{2}\left(a+b+c\right)$ .

The value of $h_1=\frac{2K}{a}$ .

The given diagram is shown in Figure 1

  

Figure 1

Calculation:

Consider the relation $\sin B$ ,

  $\begin{array}{l}\sin B=\frac{AH}{AB}\\ \sin B=\frac{h}{c}\\ c\sin =B\end{array}$

Then, by the law of sine.

  $\begin{array}{c}\frac{a}{\sin A}=\frac{c}{\sin C}\\ =\frac{a\sin C}{\sin A}\\ =c\end{array}$

From above consider the relation,

  $\frac{a\sin C}{\sin A}=c$

Then, solve further as,

  $\begin{array}{c}\frac{a\sin B\sin C}{\sin A}=\sin B\cdot c\\ =h\end{array}$

Hence, Proved.