Chapter 8.2, Problem 31AYU

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).

$b=4$, $c=6$, $B={20}^{\circ }$

To determine

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

Answer to Problem 31AYU

Solution:

Two triangles will form.

Explanation of Solution

Given:

$\text{b} = 4, \text{c} = 6$ and $\text{B} = {20}^{\circ }$

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 $\text{C}$.

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

$\frac{\text{sin}20°}{4} = \frac{\text{sin}\text{C}}{6}$

$\text{sin}\text{C} = \frac{6\text{sin}20°}{4} \approx 0.513$

$\text{C} = {\text{sin}}^{ -1}(0.51) = {30.9}^{\circ }$

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

${\text{C}}_1 = {30.9}^{\circ }$ and ${\text{C}}_2 = {180}^{\circ } - {30.86}^{\circ } = {149.1}^{\circ }$

Both choices of C result in $\text{C} + \text{B} < {180}^{\circ }$. There are two triangles, one containing the angle ${\text{C}}_1 \approx {30.9}^{\circ }$ and the other containing the angle $\text{C}{}_2 \approx {149.1}^{\circ }$ so angle ${\text{A}}_1 = {180}^{\circ } - \text{B} - {\text{C}}_1 = {129.1}^0$ and ${\text{A}}_2 = {180}^{\circ } - {20}^{\circ } - {149.14}^0 = {10.9}^{\circ }$.

Case 1

When A₁

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

$\frac{\text{sin}129.1°}{\text{a}} = \frac{\text{sin}20°}{4}$

$\text{a} = \frac{2.05}{0.34}$

$\text{a} \approx 9.08$

case 2

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

$\frac{\text{sin}10.86°}{\text{a}} = \frac{\text{sin}20°}{4}$

$\text{a} = \frac{\text{sin}10.9° \times 4}{\text{sin}20°}$

$\text{a} \approx 2.20$

two triangles ${\text{A}}_1 \approx {129.1}^{\circ }, {\text{C}}_1 \approx {30.9}^{\circ }, a_{} \approx 9.08$

and ${\text{A}}_2 \approx {10.9}^{\circ }, {\text{C}}_2 \approx {149.1}^{\circ }, a_1 \approx 2.2$ will be formed