To solve the triangle using the Law of Sines, we will start by calculating the possible values for angle $B$ and then determine the other sides and angles of the triangle.

Given:

• $a = 36$
• $c = 48$
• $\angle A = 31^\circ$

Step 1: Use the Law of Sines

The Law of Sines states: $\frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)}$ Substitute the known values for $a$, $c$, and $\angle A$: $\frac{36}{\sin(31^\circ)} = \frac{48}{\sin(C)}$ First, calculate the value of $\sin(31^\circ)$: $\sin(31^\circ) \approx 0.515$ Thus, $\frac{36}{0.515} \approx 69.9$ Now, apply the Law of Sines to find $\sin(C)$: $\frac{69.9}{48} = \frac{1}{\sin(C)}$ This gives: $\sin(C) = \frac{48}{69.9} \approx 0.686$

Step 2: Solve for $\angle C$

Using the inverse sine function, calculate $\angle C$: $\angle C_1 = \sin^{-1}(0.686) \approx 43.4^\circ$ Now, calculate the other possible solution for $\angle C$ (since sine has two possible values in the range of 0 to 180 degrees): $\angle C_2 = 180^\circ - 43.4^\circ \approx 136.6^\circ$

Step 3: Solve for $\angle B$

Using the fact that the sum of angles in a triangle is $180^\circ$, calculate $\angle B_1$ and $\angle B_2$:

For $\angle B_1$: $\angle B_1 = 180^\circ - 31^\circ - 43.4^\circ = 105.6^\circ$

For $\angle B_2$: $\angle B_2 = 180^\circ - 31^\circ - 136.6^\circ = 12.4^\circ$

Step 4: Solve for $b_1$ and $b_2$

Using the Law of Sines again, solve for $b_1$ and $b_2$:

For $b_1$: $\frac{b_1}{\sin(105.6^\circ)} = \frac{36}{\sin(31^\circ)}$ $b_1 = \frac{36 \cdot \sin(105.6^\circ)}{0.515} \approx \frac{36 \cdot 0.965}{0.515} \approx 67.4$

For $b_2$: $\frac{b_2}{\sin(12.4^\circ)} = \frac{36}{\sin(31^\circ)}$ $b_2 = \frac{36 \cdot \sin(12.4^\circ)}{0.515} \approx \frac{36 \cdot 0.215}{0.515} \approx 15.0$

Final Answers:

• $\angle B_1 = 105.6^\circ$
• $\angle B_2 = 12.4^\circ$
• $\angle C_1 = 43.4^\circ$
• $\angle C_2 = 136.6^\circ$
• $b_1 = 67.4$
• $b_2 = 15.0$

Let me know if you'd like further details or have any questions.

Here are five related questions for further exploration:

1. How does the Law of Cosines differ from the Law of Sines?
2. What happens when there are no possible triangles?
3. How do we determine if a triangle is ambiguous using the Law of Sines?
4. Can the Law of Sines be used in right triangles?
5. How would the calculations change if $\angle A$ was larger?

Tip: The Law of Sines can yield two possible triangles, but one must always check the validity of both using the angle sum property of triangles.