Chapter 6.1, Problem 8E

To determine

Tofind:Side $a$ , side $c$ and angle $B$ in the given triangle.

Answer to Problem 8E

Side $a$ , side $c$ and angle $B$ in the given triangle are $4.59$ , $7.03$ and $41°$ , respectively.

Explanation of Solution

Given information:

The given triangle is shown below.

  

Calculation:

In the given triangle, angle $A$ is $16°$ , angle $C$ is $123°$ and side $b$ is $5.5$ .

Since, the sum of internal angles of a triangle is $180°$ .Find angle $B$ .

  $\begin{array}{c}\angle A+\angle B+\angle C=180°\\ 16°+\angle B+123°=180°\\ \angle C=180°-16°-123°\\ \angle C=41°\end{array}$

Since, two angles and one side (AAS) is given. Apply law of Sines to calculate the remaining two sides.

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

Calculate side $c$ using the above law.

  $\begin{array}{c}\frac{b}{\sin B}=\frac{c}{\sin C}\\ c=\frac{b}{\sin B}\times \sin C\end{array}$

Substitute $41°$ for $B$ , $123°$ for $C$ and 5.5 for $b$ in the above equation.

  $\begin{array}{c}c=\frac{5.5}{\sin 41°}\times \sin 123°\\ \approx 7.03\end{array}$

Similarly, calculate side $a$ using the above law.

  $\begin{array}{c}\frac{a}{\sin A}=\frac{c}{\sin C}\\ a=\frac{c}{\sin C}\times \sin A\left\{\text{Rearrange}\text{the}\text{equation}\right\}\end{array}$

Substitute $123°$ for $C$ , $16°$ for $A$ and $7.03$ for $c$ in the above equation.

  $\begin{array}{c}a=\frac{7.03}{\sin 123°}\times \sin 16°\\ \approx 4.59\end{array}$

Therefore, side $a$ , side $c$ and angle $B$ in the given triangle are $4.59$ , $7.03$ and $41°$ , respectively.