Chapter 6.1, Problem 9E

To determine

Tofind: Side $b$ , side $c$ and angle $B$ in the triangle if $\angle A$ is $102.4°$ , $\angle C$ is $16.7°$ and side $a$ is $21.6$ .

Answer to Problem 9E

Side $b$ , side $c$ and angle $B$ are 19.32, 6.36 and $60.9°$ ,respectively.

Explanation of Solution

Given information:

Angle $A$ is $102.4°$ , angle $C$ is $16.7°$ and side $a$ is $21.6$ .

Calculation:

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

  $\begin{array}{c}\angle A+\angle B+\angle C=180°\\ 102.4°+\angle B+16.7°=180°\\ \angle B=180°-102.4°-16.7°\\ \angle B=60.9°\end{array}$

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.

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

Substitute $102.4°$ for $A$ , $60.9°$ for $B$ and 21.6for $a$ in the above equation.

  $\begin{array}{c}b=\frac{21.6}{\sin 102.4°}\times \sin 60.9°\\ \approx 19.32\end{array}$

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

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

Substitute $102.4°$ for $A$ , $16.7°$ for $C$ and $21.6$ for $a$ in the above equation.

  $\begin{array}{c}c=\frac{21.6}{\sin 102.4°}\times \sin 16.7°\\ \approx 6.36\end{array}$

Therefore, side $b$ , side $c$ and angle $B$ are 19.32, 6.36 and $60.9°$ , respectively.