Chapter 9.6, Problem 42E

Solution

To prove: The Law of cosine for a triangle.

  

Given data: In the $\Delta ABC$ shown in Fig. 1,, the Law of Cosine states:

  $\cos \angle B=\frac{c^2+a^2-b^2}{2ca}$

Method/Formula used:

Use co-ordinate geometry to derive the value of $\cos \angle B$

Calculations:

The position of triangle ABC in x - y plane is shown in Fig. 1.

  

The vertex B of the triangle is taken as origin. AM is perpendicular to the side BC ,

In triangle $ABM$ ,

  $\begin{array}{c}\sin \angle B=\frac{h}{c}\\ h=c\sin \angle B\\ \cos \angle B=\frac{BM}{c}\\ BM=c\cos \angle B\end{array}$

Therefore, coordinates of point A are $\left(c\cos \angle B,c\sin \angle B\right)$ .

The side MC of triangle AMC is

  $MC=BC-BM$

Substitute $c\sin \angle B$ for BM and a for BC ,

  $MC=a-c\cos \angle B$

Now, from Pythagoras theorem in triangle AMC

  $A{C}^2=M{C}^2+M{A}^2$

Substitute b for AC , $a-c\cos \angle B$ for MC , and $c\sin \angle B$ for MA ,

  $\begin{array}{c}{b}^2={\left(a-c\cos \angle B\right)}^2+{\left(c\sin \angle B\right)}^2\\ =\left(a^2-2ca\cos \angle B+c^2{\cos }^2\angle B\right)+\left(c^{\text{2}}{\sin }^2\angle B\right)\\ 2ca\cos \angle B=a^2+c^2\left({\cos }^2\angle B+{\sin }^2\angle B\right)-b^2\\ \cos \angle B=\frac{a^2+c^2-b^2}{2ca}\end{array}$

Similarly, by symmetry $\cos \angle A=\frac{c^2+b^2-a^2}{2cb}$ and $\cos \angle C=\frac{a^2+b^2-c^2}{2ab}$ .

Therefore, Law of cosine is, $\cos \angle A=\frac{c^2+b^2-a^2}{2cb}$ , $\cos \angle B=\frac{a^2+c^2-b^2}{2ca}$ , and $\cos \angle C=\frac{a^2+b^2-c^2}{2ab}$ .

Consider the law $\cos \angle C=\frac{a^2+b^2-c^2}{2ab}$ . If $\angle C=90°$ , then

  $\begin{array}{c}\cos 90°=\frac{a^2+b^2-c^2}{2ab}\\ 0=\frac{a^2+b^2-c^2}{2ab}\\ a^2+b^2-c^2=0\cdot 2ab\\ a^2+b^2=c^2\end{array}$

Thus, in a right angle triangle ABC, in which $\angle C=90°$ , the relation between sides a, b, and c is

  $a^2+b^2=c^2$ , that is Pythagoras theorem.

Hence, proved.