Chapter 8.2, Problem 26AYU

In Problems 25-36, two sides and an angle are given. Determine whether the given information results in one triangle, two triangles, or no triangle at all. Solve any resulting triangle(s).

$b=4$, $c=3$, $B={40}^{\circ }$

To determine

To find: The given information results in one triangle, two triangles, or no triangle at all.

Answer to Problem 26AYU

Solution:

One triangle will be formed.

Explanation of Solution

Given:

$\text{b} = 4, \text{c} = 3, \text{B} = {40}^{\circ }$

Formula used:

$\frac{\text{sin}\text{A}}{\text{a}} = \frac{\text{sin}\text{B}}{\text{b}} = \frac{\text{sin}\text{C}}{\text{c}}$

$\angle \text{A} + \angle \text{B} + \angle \text{C} = {180}^{\circ }$

Calculation:

From sine law we know that

$\frac{\text{sin}\text{A}}{\text{a}} = \frac{\text{sin}\text{B}}{\text{b}} = \frac{\text{sin}\text{C}}{\text{c}}$

$\frac{\text{sin}\text{A}}{\text{a}} = \frac{\text{sin}40°}{4} = \frac{\text{sin}\text{C}}{3}$

$\frac{\text{sin}40°}{4} = \frac{\text{sin}\text{C}}{3}$

$\frac{0.64}{4} = \frac{\text{sin}\text{C}}{3}$

$\text{SinC} = \frac{3 \times 0.64}{4} \approx 0.4821$

$\begin{array}{l}\text{SinC} = 0.4821\\ \text{C} = {\text{sin}}^{ -1}(0.48)\\ \text{C} \approx {28.8}^{\circ }\end{array}$

Explanation: There are two angles $\text{C}, 0^{\circ } < \text{C} < {180}^{\circ }$, for which $\text{sinC} \approx 0.48$

${\text{C}}_1 \approx {28.8}^{\circ }$ and ${\text{C}}_2 \approx {180}^{\circ } - {28.8}^{\circ } = {151.2}^{\circ }$

The second possibility, ${\text{C}}_2 \approx {151.2}^{\circ }$ is ruled out because $\text{B} = {40}^{\circ }$ makes $\text{B} + {\text{C}}_2 \approx {191.2}^{\circ } > {180}^{\circ }$, now use ${\text{C}}_1 \approx {28.8}^{\circ }$ to find that

$\text{A} = {180}^{\circ } - \text{B} - {\text{C}}_1 \approx {180}^{\circ } - {40}^{\circ } - {28.8}^{\circ } = {111.2}^{\circ }$

The third side $\text{c}$ may now be determined using the Law of Sines.

$\frac{\text{sin}\text{B}}{\text{b}} = \frac{\text{sin}\text{A}}{\text{a}}$

$\frac{\text{sin}40°}{4} = \frac{\text{sin}111.2°}{\text{a}}$

$\text{a} = \frac{\text{sin}111.2° \times 4}{\text{sin}40°} \approx 5.80$

One triangle $\text{A} = {111.2}^{\circ }, \text{C} = {28.8}^{\circ }, \text{a} = 5.80$