Two sides and an angle are given below. Determine whether the given information results in one triangle, two triangles, or no triangle at all. So \( \mathrm{b}=8, \mathrm{c}=7, \mathrm{~B}=100^{\circ} \) Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice. (Type an integer or decimal rounded to two decimal places as needed.) A. A single triangle is produced, where \( \mathrm{C} \approx \square^{\circ}, \mathrm{A} \approx \square^{\circ} \), and \( \mathrm{a} \approx \square \). B. Two triangles are produced, where the triangle with the smaller angle C has \( \mathrm{C}_{1} \approx \square^{\circ}, \mathrm{A}_{1} \approx \square^{\circ} \), and \( \mathrm{a}_{1} \approx \square \), and the triangle with C. No triangles are produced.

To determine whether the given information results in one triangle, two triangles, or no triangle at all, we can use the Law of Sines and the properties of triangles. Given: - \( b = 8 \) - \( c = 7 \) - \( B = 100^\circ \) ### Step 1: Check for possible triangles Using the Law of Sines, we can find angle \( C \): \[ \frac{b}{\sin B} = \frac{c}{\sin C} \] Substituting the known values: \[ \frac{8}{\sin(100^\circ)} = \frac{7}{\sin C} \] Calculating \( \sin(100^\circ) \): \[ \sin(100^\circ) \approx 0.9848 \] Now substituting this value into the equation: \[ \frac{8}{0.9848} = \frac{7}{\sin C} \] Calculating \( \frac{8}{0.9848} \): \[ \frac{8}{0.9848} \approx 8.12 \] Now we can set up the equation: \[ 8.12 = \frac{7}{\sin C} \] Rearranging gives: \[ \sin C = \frac{7}{8.12} \approx 0.861 \] ### Step 2: Find angle \( C \) Now we can find angle \( C \): \[ C \approx \sin^{-1}(0.861) \approx 60.01^\circ \] ### Step 3: Find angle \( A \) Now we can find angle \( A \): \[ A = 180^\circ - B - C \approx 180^\circ - 100^\circ - 60.01^\circ \approx 19.99^\circ \] ### Step 4: Check for the possibility of a second triangle Since angle \( C \) is acute, we can check if there is a possibility of a second triangle by considering the supplementary angle: \[ C' = 180^\circ - C \approx 180^\circ - 60.01^\circ \approx 119.99^\circ \] Now we check if this angle \( C' \) is valid: \[ A' = 180^\circ - B - C' \approx 180^\circ - 100^\circ - 119.99^\circ \approx -39.99^\circ \] Since angle \( A' \) is negative, this means that there cannot be a second triangle. ### Conclusion Thus, we have determined that there is only one triangle possible with the given information. The answer is: **A. A single triangle is produced, where \( C \approx 60.01^{\circ} \), \( A \approx 19.99^{\circ} \), and \( a \) can be calculated using the Law of Sines:** \[ \frac{a}{\sin A} = \frac{b}{\sin B} \] Calculating \( a \): \[ \frac{a}{\sin(19.99^\circ)} = \frac{8}{\sin(100^\circ)} \] Calculating \( \sin(19.99^\circ) \): \[ \sin(19.99^\circ) \approx 0.342 \] Now substituting: \[ \frac{a}{0.342} = \frac{8}{0.9848} \] Calculating \( \frac{8}{0.9848} \approx 8.12 \): \[ a \approx 0.342 \times 8.12 \approx 2.78 \] So the final values are: - \( C \approx 60.01^{\circ} \) - \( A \approx 19.99^{\circ} \) - \( a \approx 2.78 \) Thus, the complete answer is: **A. A single triangle is produced, where \( C \approx 60.01^{\circ} \), \( A \approx 19.99^{\circ} \), and \( a \approx 2.78 \).**