Let's use the Law of Sines to solve the triangle. Here’s the given information:

• Angle $A = 37^\circ$
• Side opposite $A$ (i.e., $a$) = 9
• Side opposite $B$ (i.e., $b$) = 8

Step 1: Apply the Law of Sines

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

We'll first find angle $B$ using: $\frac{\sin B}{b} = \frac{\sin A}{a}$

Step 2: Solve for $\sin B$

$\sin B = \frac{b \cdot \sin A}{a} = \frac{8 \cdot \sin 37^\circ}{9}$

Using $\sin 37^\circ \approx 0.6018$: $\sin B = \frac{8 \cdot 0.6018}{9} \approx \frac{4.8144}{9} \approx 0.535$

Now, find $B$: $B = \sin^{-1}(0.535) \approx 32.41^\circ$

Step 3: Find Angle $C$

Since the sum of the angles in a triangle is $180^\circ$: $C = 180^\circ - A - B = 180^\circ - 37^\circ - 32.41^\circ \approx 110.59^\circ$

Step 4: Find Side $c$

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

So: $c = \frac{a \cdot \sin C}{\sin A} = \frac{9 \cdot \sin 110.59^\circ}{\sin 37^\circ}$

Using $\sin 110.59^\circ \approx 0.9397$: $c = \frac{9 \cdot 0.9397}{0.6018} \approx \frac{8.4573}{0.6018} \approx 14.05$

Final Answers

• $B \approx 32.41^\circ$
• $C \approx 110.59^\circ$
• $c \approx 14.05$

Would you like me to clarify or elaborate on any step?

Here are five follow-up questions:

1. How would the calculations change if angle $A$ was obtuse?
2. How does the Law of Sines apply to non-right triangles?
3. Can you solve a triangle if only two sides and no angles are known?
4. What happens if $\sin^{-1}(x)$ yields two possible angles?
5. What are some real-world applications of the Law of Sines?

Tip: Always check if the triangle is ambiguous (SSA case) when using the Law of Sines.