Chapter 6.1, Problem 10E

To determine

Tofind: Side $a$ , side $b$ and angle $B$ in the triangle if $\angle A$ is $24.3°$ , $\angle C$ is $54.6°$ and side $c$ is $2.68$ .

Answer to Problem 10E

Side $a$ , side $b$ and angle $B$ are 1.35, 3.23 and $101.1°$ ,respectively.

Explanation of Solution

Given information:

In the given triangle, angle $A$ is $24.3°$ , angle $C$ is $54.6°$ and side $c$ is $2.68$ .

Calculation:

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

  $\begin{array}{c}\angle A+\angle B+\angle C=180°\\ 24.3°+\angle B+54.6°=180°\\ \angle B=180°-24.3°-54.6°\\ \angle B=101.1°\end{array}$

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 $a$ using the above law.

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

Substitute $24.3°$ for $A$ , $54.6°$ for $C$ and 2.68for $c$ in the above equation.

  $\begin{array}{c}a=\frac{2.68}{\sin 54.6°}\times \sin 24.3°\\ \approx 1.35\end{array}$

Similarly, calculate side $b$ 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 $24.3°$ for $A$ , $101.1°$ for $B$ and 1

3 $1.35$ for

  $a$ in the above equation.

  $\begin{array}{c}b=\frac{1.35}{\sin 24.3°}\times \sin 101.1°\\ \approx 3.23\end{array}$

Therefore, side $a$ , side $b$ and angle $B$ are 1.35, 3.23 and $101.1°$ , respectively.