The image shows a triangle with angles labeled and two sides given. We can solve for side $b$ using the Law of Sines.

In triangle $ABC$:

• Angle $C = 40^\circ$
• Angle $B = 60^\circ$
• Angle $A = 80^\circ$ (since the sum of angles in a triangle is $180^\circ$)
• Side $c = 9$ units opposite angle $B = 60^\circ$

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

Using the known values: $\frac{b}{\sin(60^\circ)} = \frac{9}{\sin(40^\circ)}$

First, calculate the sines:

• $\sin(60^\circ) = \frac{\sqrt{3}}{2} \approx 0.866$
• $\sin(40^\circ) \approx 0.643$

Now substitute these values into the equation: $\frac{b}{0.866} = \frac{9}{0.643}$

Solve for $b$: $b = \frac{9 \times 0.866}{0.643} \approx 12.13$

Thus, the length of side $b$ is approximately $12.13$ units.

Would you like more details on the steps? Here are some related questions:

1. How would you calculate side $a$ in this triangle?
2. What are the applications of the Law of Sines in real-world problems?
3. Can you apply the Law of Cosines in this triangle, and when would that be necessary?
4. How does knowing all angles help in solving for triangle sides?
5. How do you use a calculator to find sine and cosine values for non-standard angles?

Tip: Always check the sum of angles in a triangle before applying any formula to ensure the angles are correct.