Chapter 5.7, Problem 20E

To determine

To find: all the solutions of the triangle and round to the nearest tenth. In case no solutions exist, write none.

Answer to Problem 20E

  $B=90°,C=60°,b=8\text{ and }c=4\sqrt{3}$

Explanation of Solution

Given information:

  $A=30°,a=4,b=8$

Calculation:

Since the angle given is acute i.e. $A<90°$ and $a<b$ , so case I will be considered. Evaluate $b\sin A$ and compare with $a$ as shown below.

  $\begin{array}{c}b\sin A=8\sin 30°\\ =4\end{array}$

Since $b\sin A=a$ , so there is one triangle and it is right triangle.

Use laws of sine to evaluate $B$ as follows:

  $\begin{array}{c}\frac{a}{\sin A}=\frac{b}{\sin B}\\ \frac{4}{\sin 30°}=\frac{8}{\sin B}\left[\text{Substitute the values}\right]\\ \sin B=\frac{8\left(\sin 30°\right)}{4}\\ \sin B=1\\ B={\sin }^{-1}\left(1\right)\left[\text{Taking sin inverse each side}\right]\\ B=90°\end{array}$

The third angle $\left(C\right)$ of this triangle can be evaluated as follows:

  $\begin{array}{c}\angle C=180°-\left(\angle A+\angle B\right)\\ =180°-\left(30°+90°\right)\left[\text{Put the values}\right]\\ =60°\end{array}$

Use laws of sine to evaluate $c$ for first triangle as follows:

  $\begin{array}{c}\frac{a}{\sin A}=\frac{c}{\sin C}\\ \frac{4}{\sin 30°}=\frac{c}{\sin 60°}\left[\text{Substitute the values}\right]\\ c=\frac{4\left(\sin 60°\right)}{\sin 30°}\\ c=4\sqrt{3}\end{array}$

Use laws of sine to evaluate $B$ for the first triangle as follows:

  $\begin{array}{c}\frac{a}{\sin A}=\frac{b}{\sin B}\\ \frac{4}{\sin 30°}=\frac{b}{\sin 60°}\left[\text{Substitute the values}\right]\\ b=\frac{4\left(\sin 60°\right)}{\sin 30°}\\ =8\end{array}$

Thus angle $B$ and $C$ will be $90°$ and $60°$ respectively. Since $A$ is the smallest angle, so the opposite smallest side is $a=4$ . The hypotenuse is $b$ and middle side is $c$ . The value of side $c=4\sqrt{3}$ and $b=8$ .