Chapter 8.2, Problem 36AYU

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).

$b=4$, $c=5$, $B={40}^{\circ }$

To determine

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

Answer to Problem 36AYU

Solution:

Two triangles will be formed from the given information.

Explanation of Solution

Given:

$\text{b} = 4, \text{c} = 5, \text{B} = {40}^{\circ }$

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 $\text{C}$

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

$\frac{\text{sin}40°}{4} = \frac{\text{sin}\text{C}}{5}$

$\text{sin}\text{C} = \frac{5\text{sin}40°}{4} \approx 0.8035$

$\text{C} = {\text{sin}}^{ -1}(0.80) \approx {53.5}^{\circ }$

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

${\text{C}}_1 = {53.46}^{\circ }$ and ${\text{C}}_2 = {180}^{\circ } - {53.46}^{\circ } = {126.5}^{\circ }$

Both choices of C result in $\text{C} + \text{B} < {180}^{\circ }$. There are two triangles, one containing the angle ${\text{C}}_1 \approx {53.46}^{\circ }$ and the other containing the angle ${\text{C}}_2 \approx {126.5}^0$ so angle ${\text{A}}_1 = {180}^{\circ } - \text{B} - {\text{C}}_1 = {86.5}^0$ and

${\text{A}}_2 = {180}^{\circ } - {40}^{\circ } - {126.54}^{\circ } = {13.5}^{\circ }$

Case 1

When A₁

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

$\frac{\text{sin}86.54°}{\text{a}} = \frac{\text{sin}40°}{4}$

$\text{a} = \frac{4\text{sin}86.54°}{\text{sin}40°} \approx 6.21$

case 2

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

$\frac{\text{sin}13.46°}{\text{a}} = \frac{\text{sin}40°}{4}$

$\text{a} = \frac{\text{sin}13.46° \times 4}{\text{sin}40°}$

$\text{a} \approx 1.45$

two triangles will be formed from the above values

${\text{A}}_1 \approx {86.5}^{\circ }, {\text{C}}_1 \approx 53.5$ and ${\text{a}}_1 \approx 6.21$

${\text{A}}_2 \approx {13.5}^{\circ }, {\text{C}}_2 \approx {126.5}^{\circ }$ and ${\text{a}}_2 \approx 1.45$