Chapter 6.5, Problem 15E

To determine

To calculate: The complete triangle with the help of law of sines and sketch the triangle.

Answer to Problem 15E

The complete triangle `with sketch is,

  $\begin{array}{c}\angle \text{A}={30}^{\circ }\\ \angle B={85}^{\circ }\\ \angle C={65}^{\circ }\end{array}$ and $\begin{array}{c}a=5\\ b=10\\ c=9.1\end{array}$

Explanation of Solution

Given information:

  $\angle {\text{A=30}}^{\circ },\angle \text{C}={65}^{\circ }\text{ and b}=10$ .

Formula used:

Sum of interior angles of any triangle is ${180}^{\circ }$ . So, in any $\text{ΔABC}$ , $\angle \text{A}+\angle \text{B}+\angle \text{C}={180}^{\circ }$ .

According to law of sines, the lengths of the sides of the triangle are proportional to the sines of the corresponding opposite angles. In any triangle, $\Delta \text{XYZ}$

  $\frac{\text{sin X}}{\text{x}}=\frac{\text{sin Y}}{\text{y}}\text{=}\frac{\text{sin Z}}{\text{z}}$

Calculation:

Consider the given values, $\angle {\text{A=30}}^{\circ },\angle \text{C}={65}^{\circ }\text{ and b}=10$ .

Recall that the sum of interior angles of any triangle is ${180}^{\circ }$ . So, in any $\text{ΔABC}$ , $\angle \text{A}+\angle \text{B}+\angle \text{C}={180}^{\circ }$ .

So, $\angle \text{B}$ will be calculated as,

  $\begin{array}{c}\angle \text{A}+\angle \text{B}+\angle \text{C}={180}^{\circ }\\ {30}^{\circ }+\angle {\text{B + 65}}^{\circ }={180}^{\circ }\\ \angle \text{B}={180}^{\circ }-{95}^{\circ }\\ \angle \text{B}={85}^{\circ }\end{array}$

Recall that according to law of sines, the lengths of the sides of the triangle are proportional to the sines of the corresponding opposite angles. In any triangle, $\Delta \text{XYZ}$

  $\frac{\text{sin X}}{\text{x}}=\frac{\text{sin Y}}{\text{y}}\text{=}\frac{\text{sin Z}}{\text{z}}$

So, the value of $a$ will be calculated as,

  $\begin{array}{c}\frac{\text{sin A}}{a}=\frac{\text{sin B}}{10}\\ \frac{\sin 30}{a}=\frac{\sin 85}{10}\\ a=\frac{10\sin 30}{\sin 85}\\ a\approx 5\end{array}$

Now, the value of $c$ will be calculated as,

  $\begin{array}{c}\frac{\text{sin B}}{10}=\frac{\text{sin C}}{c}\\ \frac{\sin 85}{10}=\frac{\sin 65}{c}\\ c=\frac{10\sin 65}{\sin 85}\\ c\approx 9.1\end{array}$

Thus, the complete triangle `with sketch is,

  $\begin{array}{c}\angle \text{A}={30}^{\circ }\\ \angle B={85}^{\circ }\\ \angle C={65}^{\circ }\end{array}$ and $\begin{array}{c}a=5\\ b=10\\ c=9.1\end{array}$