We are given the following information about the triangle:

• Angle $A = 22^\circ 34'$
• Side $a = 9.5$
• Side $b = 22$

We are tasked with solving for angle $B$, angle $C$, and side $c$ using the Law of Sines. Let's break this into steps.

Step 1: Convert Angle A to Decimal Form

To begin, we need to convert angle $A$ from degrees and minutes to decimal form.

• $34' = \frac{34}{60} = 0.567$

Thus: $A = 22^\circ + 0.567^\circ = 22.567^\circ$

Step 2: Apply the Law of Sines

The Law of Sines states: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$

Using $\frac{a}{\sin A} = \frac{b}{\sin B}$, we can solve for $\sin B$: $\frac{9.5}{\sin 22.567^\circ} = \frac{22}{\sin B}$ First, calculate $\sin 22.567^\circ$: $\sin 22.567^\circ \approx 0.3838$ Now solve for $\sin B$: $\frac{9.5}{0.3838} = \frac{22}{\sin B} \implies \sin B = \frac{22 \times 0.3838}{9.5} \approx 0.889$

Step 3: Find Possible Values of $B$

Using the inverse sine function, we find: $B_1 = \sin^{-1}(0.889) \approx 62.77^\circ$

Since $\sin(\theta) = \sin(180^\circ - \theta)$, there is also a second possible solution for angle $B$: $B_2 = 180^\circ - 62.77^\circ = 117.23^\circ$

Step 4: Check for Triangle Validity

For each value of $B$, check if a valid triangle is possible:

• For $B_1 = 62.77^\circ$, the sum of angles $A + B_1$ is: $A + B_1 = 22.567^\circ + 62.77^\circ = 85.34^\circ$ The third angle $C_1$ is: $C_1 = 180^\circ - 85.34^\circ = 94.66^\circ$ This is a valid triangle.

• For $B_2 = 117.23^\circ$, the sum of angles $A + B_2$ is: $A + B_2 = 22.567^\circ + 117.23^\circ = 139.8^\circ$ The third angle $C_2$ is: $C_2 = 180^\circ - 139.8^\circ = 40.2^\circ$ This is also a valid triangle.

Step 5: Solve for Side $c$

Using the Law of Sines again to find side $c$: $\frac{c}{\sin C} = \frac{a}{\sin A}$

• For the first solution ($C_1 = 94.66^\circ$): $\frac{c_1}{\sin 94.66^\circ} = \frac{9.5}{\sin 22.567^\circ}$ $c_1 \approx \frac{9.5 \times \sin 94.66^\circ}{\sin 22.567^\circ} \approx \frac{9.5 \times 0.9986}{0.3838} \approx 24.73$

• For the second solution ($C_2 = 40.2^\circ$): $\frac{c_2}{\sin 40.2^\circ} = \frac{9.5}{\sin 22.567^\circ}$ $c_2 \approx \frac{9.5 \times \sin 40.2^\circ}{\sin 22.567^\circ} \approx \frac{9.5 \times 0.6456}{0.3838} \approx 15.96$

Final Answers:

• First Solution:

  • $B_1 \approx 62.77^\circ$
  • $C_1 \approx 94.66^\circ$
  • $c_1 \approx 24.73$

• Second Solution:

  • $B_2 \approx 117.23^\circ$
  • $C_2 \approx 40.2^\circ$
  • $c_2 \approx 15.96$

Let me know if you need further clarifications or details!

Here are 5 follow-up questions for further practice:

1. How do you determine if a triangle is possible given its sides and angles?
2. Can the Law of Sines be used for right triangles?
3. What is the significance of the ambiguous case in the Law of Sines?
4. How do you handle obtuse angles when using the Law of Sines?
5. How does the Law of Cosines differ from the Law of Sines?

Tip: When using the Law of Sines, always check for the ambiguous case, as more than one solution might exist if two angles aren't fully defined.