Chapter 5.6, Problem 15E

Solution

To solve: The triangle if $A={63}^{\circ },a=8.6$

  $b=11.1$ .

The required values are $B={106.37}^{\circ }$, $C={10.63}^{\circ }$ and $c=5$ .

Given information: The given angle and two sides of triangle are $A={63}^{\circ }$

  $a=8.6$ and $b=11.1$ .

Formula used: The law of cosines to find the angles and sides.

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

Calculation:

Substitute ${63}^{\circ }$ for $\cos A$ , $8.6$ for $a$ and $11.1$ for $b$ in the formula. $a^2=b^2+c^2-2bc\cos A$ .

  $\begin{array}{l}{\left(8.6\right)}^2={\left(11.1\right)}^2+{\left(c\right)}^2-2\left(11.1\right)\left(c\right)\cos {63}^{\circ }\\ 73.96=123.21+c^2-2\left(11.1\right)\left(c\right)\left(0.4539\right)\\ 73.96=123.21+c^2-10.07658c\\ c^2-10.07658c+49.25=0\end{array}$

Further simplify.

  $\begin{array}{l}c=\frac{-\left(10.7658\right)\pm \sqrt{{\left(-10.07658\right)}^2-4\cdot 1\cdot 49.25}}{2.1}\\ c=5\end{array}$

Substitute $8.6$ for $a$ and $11.1$ for $b$ and $5$ for $c$ in the formula $b^2=c^2+a^2-2ca\cos B$

  $\begin{array}{c}{\left(11.1\right)}^2={\left(5\right)}^2+{\left(8.6\right)}^2-2\left(5\right)\left(8.6\right)\cos B\\ 123.21=25+73.96-86\cos B\\ \cos B=\frac{24.25}{-86}\\ B={106.37}^{\circ }\end{array}$

To find the angle $C$ subtract the sum of angles $A$ and $B$ from sum of all three angles of a triangle.

  $\begin{array}{l}C={180}^{\circ }-\left({63}^{\circ }+{106.37}^{\circ }\right)\\ C={10.63}^{\circ }\end{array}$

Thus the required values are $B={106.37}^{\circ }$, $C={10.63}^{\circ }$and $c=5$ .