Question

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°. select the correct choice and, if necessary, fill in the answer boxes to complete your choice. a. a single triangle is produced, where c ≈ □°, a ≈ □°, and a ≈ □ (type integers or decimals rounded to two decimal places as needed.) b. two triangles are produced, where the triangle with the smaller angle c has c1 ≈ □°, a1 ≈ □°, and a1 ≈ □, and the triangle with the larger angle c has c2 ≈ □°, a2 ≈ □°, and a2 ≈ □ (type integers or decimals rounded to two decimal places as needed.) c. no triangles are produced

Explanation:

Step1: Use the Law of Sines

By the Law of Sines, $\frac{\sin B}{b}=\frac{\sin C}{c}$. Substitute $b = 4$, $c = 6$, and $B=20^{\circ}$: $\sin C=\frac{c\sin B}{b}=\frac{6\sin20^{\circ}}{4}=\frac{6\times0.3420}{4}= 0.5130$.

Step2: Find angle C

Since $\sin C = 0.5130$, then $C=\sin^{- 1}(0.5130)\approx30.87^{\circ}$ or $C = 180^{\circ}-30.87^{\circ}=149.13^{\circ}$. But when $C = 149.13^{\circ}$, $B + C=20^{\circ}+149.13^{\circ}=169.13^{\circ}<180^{\circ}$, so there are two possible values for $C$.
For the smaller $C_1\approx30.87^{\circ}$:

Step3: Find angle A1

$A_1=180^{\circ}-B - C_1=180^{\circ}-20^{\circ}-30.87^{\circ}=129.13^{\circ}$.

Step4: Find side a1

Using the Law of Sines $\frac{a_1}{\sin A_1}=\frac{b}{\sin B}$, so $a_1=\frac{b\sin A_1}{\sin B}=\frac{4\times\sin129.13^{\circ}}{\sin20^{\circ}}=\frac{4\times0.7771}{0.3420}\approx9.08$.
For the larger $C_2 = 149.13^{\circ}$:

Step5: Find angle A2

$A_2=180^{\circ}-B - C_2=180^{\circ}-20^{\circ}-149.13^{\circ}=10.87^{\circ}$.

Step6: Find side a2

Using the Law of Sines $\frac{a_2}{\sin A_2}=\frac{b}{\sin B}$, so $a_2=\frac{b\sin A_2}{\sin B}=\frac{4\times\sin10.87^{\circ}}{\sin20^{\circ}}=\frac{4\times0.1887}{0.3420}\approx2.21$.

Answer:

B. Two triangles are produced, where the triangle with the smaller angle C has $C_1\approx30.87^{\circ}$, $A_1\approx129.13^{\circ}$, and $a_1\approx9.08$, and the triangle with the larger angle C has $C_2\approx149.13^{\circ}$, $A_2\approx10.87^{\circ}$, and $a_2\approx2.21$