Chapter 6.1, Problem 5E

To determine

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

Answer to Problem 5E

Side $a$ , side $c$ and angle $A$ in the given triangle are $14.14$ , $27.32$ and $30°$ , respectively.

Explanation of Solution

Given information:

The given triangle is shown below.

  

Calculation:

In the given triangle, angle $B$ is $45°$ , angle $C$ is $105°$ and side $b$ is 20.

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

  $\begin{array}{c}\angle A+\angle B+\angle C=180°\\ \angle A+45°+105°=180°\\ \angle A=180°-45°-105°\\ \angle A=30°\end{array}$

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.

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

Substitute $30°$ for $A$ , $45°$ for $B$ and 20 for $b$ in the above equation.

  $\begin{array}{c}a=\frac{20}{\sin 45°}\times \sin 30°\\ =\frac{20}{\left(\frac{1}{\sqrt{2}}\right)}\times \left(\frac{1}{2}\right)\\ =10\sqrt{2}\\ \simeq 14.14\end{array}$

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

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

Substitute $45°$ for $B$ , $105°$ for $C$ and 20 for $b$ in the above equation.

  $\begin{array}{c}c=\frac{20}{\sin 45°}\times \sin 105°\\ =\frac{20}{\left(\frac{1}{\sqrt{2}}\right)}\times 0.9659\\ \approx 27.32\end{array}$

Therefore, side $a$ , side $c$ and angle $A$ in the given triangle are $14.14$ , $27.32$ and $30°$ , respectively.