Chapter 5.8B, Problem 1GCE

To determine

To Find: The coordinate of vertex B.

Answer to Problem 1GCE

  $B\left(9.94,7\right)$

Explanation of Solution

Given:A $\Delta ABC$ given that $AC=6,BC=8,A={35}^{\circ }$

Law of cosine:

  $\begin{array}{l}\cos A=\frac{b^2+c^2-a^2}{2bc}\\ \cos B=\frac{a^2+c^2-b^2}{2ac}\\ \cos C=\frac{a^2+b^2-c^2}{2ab}\end{array}$

First draw a triangle ABC place AC as base such that end points have coordinates $A\left(0,0\right)\text{ and }C\left(6,0\right)$ . Make a line $y=\left(\tan {35}^{\circ }\right)x$

The length of $BC=8$ , Make a circle with center at C and radius 8.

The intersection of circle and line is coordinate B as shown in below figure.

  

Using law of cosine:

  $\begin{array}{c}2\times 6\times c\times \cos {35}^{\circ }=c^2+6^2-8^2\\ c^2+9.83c+28=0\\ c=12.14\end{array}$

  $\begin{array}{l}\cos B=\frac{8^2+{12.14}^2-6^2}{2\cdot 8\cdot 12.14}=0.8195\\ B={35}^{\circ }\\ \cos C=\frac{8^2+6^2-{12.14}^2}{2\cdot 8\cdot 6}=-0.4935\\ C={110}^{\circ }\end{array}$

Hence, the coordinate $B\left(9.94,7\right)$ .