Chapter 8.2, Problem 35AYU

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).

$a=2$, $c=1$, $C={25}^{\circ }$

To determine

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

Answer to Problem 35AYU

Solution:

Two triangles will be formed from the given information.

Explanation of Solution

Given:

$\text{a} = 2, \text{c} = 1$ and $\text{C} = {25}^{\circ }$

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 $\text{C}$

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

$\frac{\text{sin}\text{A}}{2} = \frac{\text{sin}25°}{1}$

$\text{sin}\text{A} = 2 \times \text{sin}25° \approx 0.84$

$\text{A} = {\text{sin}}^{ -1}(0.84) \approx {57.70}^0$

There are two angles $\text{A}, 0^{\circ } < \text{A} < {180}^{\circ }$, for which $\text{C} = {25}^{\circ }$

${\text{A}}_1 = {57.70}^{\circ }$ and ${\text{A}}_2 = {180}^{\circ } - {57.70}^{\circ } = {122.30}^{\circ }$

Both choices of A result in $\text{C} + \text{A} < {180}^{\circ }$. There are two triangles, one containing the angle ${\text{A}}_1 \approx {57.70}^{\circ }$ and the other containing the angle ${\text{A}}_2 \approx {122.30}^{\circ }$ so angle ${\text{B}}_1 = {180}^{\circ } - \text{C} - {\text{A}}_1 = {97.30}^0$ and ${\text{B}}_2 = {180}^{\circ } - {25}^{\circ } - {122.30}^0 = {32.70}^{\circ }$

Case 1

When B₁

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

$\frac{\text{sin}25°}{1} = \frac{\text{sin}97.30°}{\text{b}}$

$\text{b} = \frac{\text{sin}97.30°}{\text{sin}25°}$

$\text{b} \approx \frac{0.99}{0.42} \approx 2.35$

case 2

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

$\frac{\text{sin}32.70°}{\text{b}} = \frac{\text{sin}25°}{1}$

$\text{b} = \frac{\text{sin}32.70° \times 1}{\text{sin}25°}$

$\text{b} \approx 1.28$

two triangles will be formed from the given information.

${\text{A}}_1 \approx {57.7}^{\circ }, {\text{B}}_1 \approx 97.30$ and ${\text{b}}_1 \approx 2.35$

${\text{A}}_2 \approx {122.3}^{\circ }, {\text{B}}_2 \approx {32.7}^{\circ }$ and ${\text{b}}_2 \approx 1.28$