Question

two sides and an angle (ssa) of a triangle are given. determine whether the given measurements produce one triangle, two triangles, or no triangle at all. solve each triangle that results.

a = 14, c = 14.6, a = 57°

select the correct choice below and, if necessary, fill in the answer boxes to complete your choice. (round side lengths to the nearest tenth and angle measurements to the nearest degree as needed.)

a. there is only one possible solution for the triangle.
the measurements for the remaining side b and angles b and c are as follows.
b ≈ □° c ≈ □° b ≈ □

b. there are two possible solutions for the triangle.
the measurements for the solution with the the smaller angle c are as follows.
c₁ ≈ 61° b₁ ≈ 62° b₁ ≈ 14.7
the measurements for the solution with the the larger angle c are as follows.
c₂ ≈ 119° b₂ ≈ 4° b₂ ≈ 1.2

c. there are no possible solutions for this triangle.

Explanation:

Step1: Check SSA Case with Law of Sines

First, use the Law of Sines: $\frac{\sin C}{c} = \frac{\sin A}{a}$. Plug in $a = 14$, $c = 14.6$, $A = 57^\circ$. So $\sin C = \frac{c \sin A}{a} = \frac{14.6 \sin 57^\circ}{14}$. Calculate $\sin 57^\circ \approx 0.8387$, so $\sin C \approx \frac{14.6 \times 0.8387}{14} \approx \frac{12.245}{14} \approx 0.8746$.

Step2: Determine Possible Angles for C

Since $\sin C \approx 0.8746$, $C$ could be $\arcsin(0.8746) \approx 61^\circ$ or $180^\circ - 61^\circ = 119^\circ$ (because $\sin \theta = \sin(180^\circ - \theta)$). Now check if these angles are valid with the triangle angle sum.

Step3: Case 1: $C_1 = 61^\circ$

Angle sum: $B_1 = 180^\circ - 57^\circ - 61^\circ = 62^\circ$. Use Law of Sines for $b_1$: $\frac{b_1}{\sin B_1} = \frac{a}{\sin A}$, so $b_1 = \frac{14 \sin 62^\circ}{\sin 57^\circ} \approx \frac{14 \times 0.8829}{0.8387} \approx \frac{12.361}{0.8387} \approx 14.7$.

Step4: Case 2: $C_2 = 119^\circ$

Angle sum: $B_2 = 180^\circ - 57^\circ - 119^\circ = 4^\circ$. Use Law of Sines for $b_2$: $b_2 = \frac{14 \sin 4^\circ}{\sin 57^\circ} \approx \frac{14 \times 0.0698}{0.8387} \approx \frac{0.977}{0.8387} \approx 1.2$. Both cases are valid (angles sum to $180^\circ$), so two triangles.

Answer:

B. There are two possible solutions for the triangle.
The measurements for the solution with the smaller angle C are as follows.
$C_1 \approx 61^\circ$ $B_1 \approx 62^\circ$ $b_1 \approx 14.7$
The measurements for the solution with the larger angle C are as follows.
$C_2 \approx 119^\circ$ $B_2 \approx 4^\circ$ $b_2 \approx 1.2$