Chapter 6.5, Problem 26E

To determine

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

Answer to Problem 26E

The complete triangle is,

  $\begin{array}{c}\angle \text{A}={135}^{\circ }\\ \angle B={34.45}^{\circ }\\ \angle C={10.55}^{\circ }\end{array}$ and $\begin{array}{c}a=100\\ b=80\\ c=25.89\end{array}$ .

Explanation of Solution

Given information:

  $a=100,b=80\text{ and }\angle {\text{A=135}}^{\circ }$ .

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, $a=100,b=80\text{ and }\angle {\text{A=135}}^{\circ }$ .

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 $\angle B$ will be calculated as,

  $\begin{array}{c}\frac{\text{sin A}}{a}=\frac{\text{sin B}}{b}\\ \frac{\sin 135}{100}=\frac{\sin \text{B}}{80}\\ \sin \text{B}=\frac{80\sin 135}{100}\\ \text{sinB}\approx 0.566\end{array}$

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

  $B_1={34.45}^{\circ }{\text{ and B}}_2={145.55}^{\circ }$

Since, $A+B_2={2280.55}^{\circ }>{180}^{\circ }$ , therefore, $B_2$ will be rejected and $B_1$ will be used to further solve the triangle.

Thus, $B={34.45}^{\circ }$ .

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

  $\begin{array}{c}\angle \text{A}+\angle \text{B}+\angle \text{C}={180}^{\circ }\\ {135}^{\circ }+{34.45}^{\circ }+\angle \text{C}={180}^{\circ }\\ \angle \text{C}={180}^{\circ }-\left({135}^{\circ }+{34.45}^{\circ }\right)\\ \angle \text{C}\approx {10.55}^{\circ }\end{array}$

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

  $\begin{array}{c}\frac{\text{sin A}}{a}=\frac{\text{sin C}}{c}\\ \frac{\sin 135}{100}=\frac{\sin \left(10.55\right)}{c}\\ c=\frac{100\sin \left(10.55\right)}{\sin 135}\\ c\approx 25.89\end{array}$

Thus, the complete triangle is,

  $\begin{array}{c}\angle \text{A}={135}^{\circ }\\ \angle B={34.45}^{\circ }\\ \angle C={10.55}^{\circ }\end{array}$ and $\begin{array}{c}a=100\\ b=80\\ c=25.89\end{array}$