Chapter 8.2, Problem 27AYU

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).

$b=5$, $c=3$, $B={100}^{\circ }$

To determine

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

Answer to Problem 27AYU

Solution:

One triangle will be formed

Explanation of Solution

Given:

$\text{b} = 5, \text{c} = 3, \text{B} = {100}^{\circ }$

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

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

$\frac{\text{sin}\text{A}}{\text{a}} = \frac{\text{sin}100°}{5} = \frac{\text{sin}\text{C}}{3}$.

$\frac{\text{sin}100°}{5} = \frac{\text{sin}\text{C}}{3}$

$\frac{\text{sin}100}{5} = \frac{\text{sin}\text{C}}{3}$

$\begin{array}{l}\text{SinC} \approx 0.5909\\ \text{SinC} = 0.5909\\ \text{C} = {\text{sin}}^{ -1}(0.5909)\\ \text{C} \approx {36.2}^{\circ }\end{array}$

There are two angles $\text{C}, 0^{\circ } < \text{C} < {180}^{\circ }$, for which $\text{sinC} \approx 0.5909$

${\text{C}}_1 \approx {36.2}^{\circ }$ and ${\text{C}}_2 \approx {180}^{\circ } - {36.01}^{\circ } = {143.8}^{\circ }$

The second possibility, ${\text{C}}_2 \approx {143.8}^{\circ }$ is ruled out because $\text{B} = {100}^{\circ }$ makes $\text{B} + {\text{C}}_2 > {180}^{\circ }$, now use ${\text{C}}_1 \approx {36.2}^{\circ }$ to find that

$\text{A} = {180}^{\circ } - \text{B} - {\text{C}}_1 \approx {180}^{\circ } - {100}^{\circ } - {36.2}^{\circ } = {43.8}^{\circ }$

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

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

$\frac{\text{sin}100°}{5} = \frac{\text{sin}43.99°}{\text{a}}$

$\text{a} = \frac{\text{sin}43.99° \times 5}{\text{sin}100°}$

$\text{a} \approx .351$

One triangle $\text{A} \approx {43.8}^{\circ }, \text{C} \approx {36.2}^{\circ }, \text{a} \approx 3.51$