Chapter 8.2, Problem 25AYU

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).

$a=3$, $b=2$, $A={50}^{\circ }$

To determine

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

Answer to Problem 25AYU

Solution:

One triangle will be formed.

Explanation of Solution

Given:

$\text{a} = 3, \text{b} = 2, \text{A} = {50}^{\circ }$

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:

$\frac{\text{sin}50°}{3} = \frac{\text{sin}\text{B}}{2} = \frac{\text{sin}\text{C}}{\text{c}}$

$\frac{\text{sin}50°}{3} = \frac{\text{sin}\text{B}}{2}$

$\frac{0.766}{3} = \frac{\text{sin}\text{B}}{2}$

$\text{sin}\text{B} = \frac{2 \times 0.766}{3} \approx 0.51$

$\text{Sin}\text{B} = 0.51$

$\text{B} = {\text{sin}}^{ -1}(0.51)$

$\text{B} \approx 30.71$ There are two angles $\text{B}, 0^{\circ } < \text{B} < {180}^{\circ }$, for which $\text{sinB} \approx 0.51$

${\text{B}}_1 \approx {30.7}^{\circ }$ and ${\text{B}}_2 \approx {180}^{\circ } - {30.71}^{\circ } = {149.29}^{\circ }$

The second possibility, ${\text{B}}_2 \approx {149.29}^{\circ }$ is ruled out because $\text{A} = {50}^{\circ }$ makes $\text{A} + {\text{B}}_2 \approx {199.29}^{\circ } > {180}^{\circ }$, now use ${\text{B}}_1 \approx {30.71}^{\circ }$ to find that

$\text{C} = {180}^{\circ } - \text{A} - {\text{B}}_1 \approx {180}^{\circ } - {50}^{\circ } - {30.7}^{\circ } = {99.3}^{\circ }$

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

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

$\frac{\text{sin}50°}{3} = \frac{\text{sin}99.29°}{\text{c}}$

$\text{c} = \frac{\text{sin}99.29° \times 3}{\text{sin}50°}$

$\text{c} = \frac{0.986 \times 3}{0.766} \approx 3.86$