Chapter 5.6, Problem 1E

Solution

To find: The length of side $AC$, and included angles $A$ and $C$ in the given triangle.

The length of side $AC$ is $19.22$, and $\angle A=30.7°$ and $\angle C=18.3°$ .

Given information:

The length of side $AB=8$, $BC=13$ and $\angle B=131°$ .

Formula used:

Law of Cosines:

  $A{C}^2=A{B}^2+B{C}^2-2\times AB\times BC\cos B$

Law of Sine:

  $\frac{\sin A}{BC}=\frac{\sin B}{AC}=\frac{\sin C}{AB}$

Calculation:

Substitute $8$ for $AB$, $13$ for $BC$ and $131°$ for $\angle B$ in $A{C}^2=A{B}^2+B{C}^2-2\times AB\times BC\cos B$ to calculate the length of side $AC$ .

  $\begin{array}{c}A{C}^2=A{B}^2+B{C}^2-2\times AB\times BC\cos B\\ ={13}^2+8^2-2\times 13\times 8\cos 113\\ AC=19.22\end{array}$

Rewrite the Law of Sine for $\angle B$ and $\angle C$ .

  $\frac{\sin B}{AC}=\frac{\sin C}{AB}$

Substitute $19.22$ for $AC$, $13$ for $BC$ and $131°$ for $\angle B$ in $\frac{\sin B}{AC}=\frac{\sin C}{AB}$ to calculate the $\angle C$ .

  $\begin{array}{c}\frac{\sin 131}{19.22}=\frac{\sin C}{8}\\ C={\sin }^{-1}\frac{8\sin 131}{19.22}\\ =18.3°\end{array}$

Sum of all angles in a triangle is $180°$ .

  $\angle A+\angle B+\angle C=180$

Substitute $18.3°$ for $\angle B$ and $131°$ for $\angle C$ in $\angle A+\angle B+\angle C=180$ to calculate $\angle A$ .

  $\begin{array}{c}\angle A=180-\left(\angle B+\angle C\right)\\ =180-\left(18.3+131\right)\\ =30.7°\end{array}$

Therefore, length of side $AC$ is $19.22$, and $\angle A=30.7°$ and $\angle C=18.3°$ .