Chapter 11, Problem 15PS

To determine

To proof the law of sine of the given triangle.

Answer to Problem 15PS

Law of sine is proved.

Explanation of Solution

Given:

Consider the following figure of triangle.

  

The law of sine is

  $\frac{\sin A}{\parallel a\parallel }=\frac{\sin B}{\parallel b\parallel }=\frac{\sin C}{|c\parallel }$

To prove this expression, use cross product method. According to the definition, half of the magnitude of cross product of any two vectors from the set of three vectors together making a triangle is equal to the area of that triangle. Therefore for above triangle

  $\begin{array}{l}\left|\left|\text{a×b}\right|\right|\text{=}\left|\left|\text{b×c}\right|\right|\hfill \\ \text{ =}\left|\left|\text{c×a}\right|\right|\hfill \\ \text{ =2}\left(\text{area}\right)\hfill \end{array}$

Now,

  $\begin{array}{l}\left|\left|\text{a×b}\right|\right|\text{=}\left|\left|\text{a}\right|\right|\left|\left|\text{b}\right|\right|\sin C\mathrm{......}\left(1\right)\hfill \\ \left|\left|\text{b×c}\right|\right|\text{=}\left|\left|\text{b}\right|\right|\left|\left|\text{c}\right|\right|\sin A\mathrm{....}\left(2\right)\hfill \\ \left|\left|\text{c×a}\right|\right|\text{=}\left|\left|\text{c}\right|\right|\left|\left|\text{a}\right|\right|\sin B\mathrm{......}\left(3\right)\hfill \end{array}$

On equating equation (1) and (2);

  $\begin{array}{c}\left|\text{a}\right|\parallel \text{b}\parallel \sin C=\parallel \text{b}\parallel \text{c}\parallel \sin A\\ \frac{\sin A}{\parallel \text{a}\parallel }=\frac{\sin C}{\parallel \text{c}\parallel }\end{array}$

Similarly from equation (2) and (3);

  $\begin{array}{cc}\parallel c\parallel a\parallel \sin B& =\parallel b\parallel \parallel c\parallel \sin A\\ \frac{\sin A}{\parallel a\parallel }& =\frac{\sin B}{\parallel b\parallel }\end{array}$

Therefore from the above two conclusion it is proved that

  $\frac{\sin A}{\left|a\right||}=\frac{\sin B}{\parallel b\parallel }=\frac{\sin C}{\left|c\right|}$

Hence Law of Sine is proved.