Chapter 6.5, Problem 27E

To determine

To calculate: The complete triangle with the help of law of sines if possible with the given conditions.

Answer to Problem 27E

The two triangles that satisfy the given conditions are,

Triangle 1,

  $\begin{array}{c}\angle {\text{A}}_1={57.14}^{\circ }\\ \angle B_1={93.86}^{\circ }\\ \angle C={29}^{\circ }\end{array}$ and $\begin{array}{c}a=26\\ b_1=30.9\\ c=15\end{array}$

Triangle 2,

  $\begin{array}{c}\angle {\text{A}}_2={122.86}^{\circ }\\ \angle B_2={28.14}^{\circ }\\ \angle C={29}^{\circ }\end{array}$ and $\begin{array}{c}a=26\\ b_2=14.6\\ c=15\end{array}$

Explanation of Solution

Given information:

  $a=26,c=15\text{ and }\angle {\text{C=29}}^{\circ }$ .

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, $a=26,c=15\text{ and }\angle {\text{C=29}}^{\circ }$ .

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

  $\begin{array}{c}\frac{\text{sin A}}{a}=\frac{\text{sin C}}{c}\\ \frac{\sin \text{A}}{26}=\frac{\sin 29}{15}\\ \sin \text{A}=\frac{26\sin 29}{15}\\ \text{sin A}\approx 0.84\end{array}$

Since, $0^{\circ }<\text{A}<{180}^{\circ }$ so, there can be two angles A,

  ${\text{A}}_1={57.14}^{\circ }{\text{ and A}}_2={122.86}^{\circ }$

Since, $\text{C}+{\text{A}}_2={151.86}^{\circ }<{180}^{\circ }$ , therefore, there are two triangles possible that will satisfy the given conditions.

To solve first triangle with $\text{C}={29}^{\circ }{\text{ and A}}_1={57.14}^{\circ }$ ,

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, $\angle {\text{B}}_1$ will be calculated as,

  $\begin{array}{c}\angle \text{A}+\angle \text{B}+\angle \text{C}={180}^{\circ }\\ {57.14}^{\circ }+\angle {\text{B}}_1+{29}^{\circ }={180}^{\circ }\\ \angle {\text{B}}_1={180}^{\circ }-\left({29}^{\circ }+{57.14}^{\circ }\right)\\ \angle {\text{B}}_1\approx {93.86}^{\circ }\end{array}$

Now, the value of $b_1$ will be calculated as,

  $\begin{array}{c}\frac{\text{sin A}}{a}=\frac{\text{sin B}}{b}\\ \frac{\sin \left(57.14\right)}{26}=\frac{\sin \left(93.86\right)}{b_1}\\ b_1=\frac{26\sin \left(93.86\right)}{\sin \left(57.14\right)}\\ b_1\approx 30.9\end{array}$

Now, to solve second triangle with $\text{C}={29}^{\circ }{\text{ and A}}_2={122.86}^{\circ }$ ,

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, $\angle {\text{B}}_2$ will be calculated as,

  $\begin{array}{c}\angle \text{A}+\angle \text{B}+\angle \text{C}={180}^{\circ }\\ {122.86}^{\circ }+\angle {\text{B}}_2+{29}^{\circ }={180}^{\circ }\\ \angle {\text{B}}_2={180}^{\circ }-\left({29}^{\circ }+{122.86}^{\circ }\right)\\ \angle {\text{B}}_2\approx {28.14}^{\circ }\end{array}$

Now, the value of $b_2$ will be calculated as,

  $\begin{array}{c}\frac{\text{sin A}}{a}=\frac{\text{sin B}}{b}\\ \frac{\sin 122.86}{26}=\frac{\sin \left(28.14\right)}{b_2}\\ b_2=\frac{26\sin \left(28.14\right)}{\sin 122.86}\\ b_2\approx 14.6\end{array}$

Thus, the two triangles that satisfy the given conditions are,

Triangle 1,

  $\begin{array}{c}\angle {\text{A}}_1={57.14}^{\circ }\\ \angle B_1={93.86}^{\circ }\\ \angle C={29}^{\circ }\end{array}$ and $\begin{array}{c}a=26\\ b_1=30.9\\ c=15\end{array}$

Triangle 2,

  $\begin{array}{c}\angle {\text{A}}_2={122.86}^{\circ }\\ \angle B_2={28.14}^{\circ }\\ \angle C={29}^{\circ }\end{array}$ and $\begin{array}{c}a=26\\ b_2=14.6\\ c=15\end{array}$