Chapter 6.5, Problem 16E

To determine

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

Answer to Problem 16E

The complete triangle `with sketch is,

  $\begin{array}{c}\angle \text{A}={22}^{\circ }\\ \angle B={95}^{\circ }\\ \angle C={63}^{\circ }\end{array}$ and $\begin{array}{c}a=420\\ b=1116.9\\ c=999\end{array}$

Explanation of Solution

Given information:

  $\angle {\text{A=22}}^{\circ },\angle \text{B}={95}^{\circ }\text{ and a}=420$ .

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, $\angle {\text{A=22}}^{\circ },\angle \text{B}={95}^{\circ }\text{ and a}=420$ .

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, $\angle \text{C}$ will be calculated as,

  $\begin{array}{c}\angle \text{A}+\angle \text{B}+\angle \text{C}={180}^{\circ }\\ {22}^{\circ }+{95}^{\circ }\text{ + }\angle \text{C}={180}^{\circ }\\ \angle C={180}^{\circ }-{117}^{\circ }\\ \angle C={63}^{\circ }\end{array}$

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 $b$ will be calculated as,

  $\begin{array}{c}\frac{\text{sin A}}{420}=\frac{\text{sin B}}{b}\\ \frac{\sin 22}{420}=\frac{\sin 95}{b}\\ b=\frac{420\sin 95}{\sin 22}\\ b\approx 1116.9\end{array}$

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

  $\begin{array}{c}\frac{\text{sin A}}{420}=\frac{\text{sin C}}{c}\\ \frac{\sin 22}{420}=\frac{\sin 63}{c}\\ c=\frac{420\sin 63}{\sin 22}\\ c\approx 999\end{array}$

Thus, the complete triangle `with sketch is,

  $\begin{array}{c}\angle \text{A}={22}^{\circ }\\ \angle B={95}^{\circ }\\ \angle C={63}^{\circ }\end{array}$ and $\begin{array}{c}a=420\\ b=1116.9\\ c=999\end{array}$