Chapter 8.2, Problem 30AYU

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).

$b=2$, $c=3$, $B={40}^{\circ }$

To determine

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

Answer to Problem 30AYU

Solution:

Two triangles will form

Explanation of Solution

Given:

$\text{b} = 2, \text{c} = 3$ and $\text{B} = {40}^{\circ }$

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:

Two sides and one angle are known, use the law of sines to find the angle C

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

$\frac{\text{sin}40°}{2} = \frac{\text{sin}\text{C}}{3}$

$\text{sin}\text{C} = \frac{3\text{sin}40°}{2} \approx 0.9642$

$\text{C} = {\text{sin}}^{ -1}(0.9642) = {74.6}^{\circ }$

There are two angles $\text{C}, 0^{\circ } < \text{C} < {180}^{\circ }$, for which $\text{B} = {40}^{\circ }$

${\text{C}}_1 = {74.62}^{\circ }$ and ${\text{C}}_2 = {180}^{\circ } - {74.62}^{\circ } = {105.38}^{\circ }$

Both choices of C result in $\text{C} + \text{B} < {180}^{\circ }$. There are two triangles, one containing the angle ${\text{C}}_1 \approx {74.62}^{\circ }$ and the other containing the angle ${\text{C}}_2 \approx {105.38}^{\circ }$ so angle ${\text{A}}_1 = {180}^{\circ } - \text{B} - {\text{C}}_1 = {180}^{\circ } - {40}^{\circ } - {74.6}^{\circ } = {65.4}^{\circ }$ and ${\text{A}}_2 = {180}^{\circ } - {40}^{\circ } - {105.4}^{\circ } = {34.62}^{\circ }$

Case 1

When A₁

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

$\frac{\text{sin}65.4°}{\text{a}} = \frac{\text{sin}40°}{2}$

$\text{a} = 2 * \frac{\text{sin}74.62°}{\text{sin}65.4°}$

$\text{a} \approx 2.83$

case 2

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

$\frac{\text{sin}34.62°}{\text{a}} = \frac{\text{sin}40°}{2}$

$\text{a} = \frac{\text{sin}34.62° \times 2}{\text{sin}40°}$

$\text{a} \approx 1.77$

two triangles ${\text{A}}_1 \approx {65.4}^{\circ }, {\text{C}}_1 \approx {74.62}^{\circ }, a_1 \approx 2.83$ and ${\text{A}}_2 \approx {34.6}^{\circ }, {\text{C}}_2 \approx {105.4}^{\circ }, a_1 \approx 1.77$ will be formed.