To solve for $b$ in the triangle $\triangle ABC$ using the Law of Sines, follow these steps:

Step 1: Recall the Law of Sines formula

The Law of Sines states:

$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$

Step 2: Plug in the known values

You are given:

• $A = 180^\circ - B - C = 180^\circ - 60^\circ - 90^\circ = 30^\circ$
• $B = 60^\circ$
• $C = 90^\circ$
• $a = 66$

Thus, the Law of Sines becomes:

$\frac{66}{\sin 30^\circ} = \frac{b}{\sin 60^\circ}$

Step 3: Solve for $b$

1. Evaluate the sines of the angles:
   • $\sin 30^\circ = 0.5$
   • $\sin 60^\circ = \sqrt{3}/2$

$\frac{66}{0.5} = \frac{b}{\sqrt{3}/2}$

2. Simplify $\frac{66}{0.5}$: $\frac{66}{0.5} = 132$

So:

$132 = \frac{b}{\sqrt{3}/2}$

3. Multiply both sides by $\sqrt{3}/2$ to isolate $b$: $b = 132 \cdot \frac{\sqrt{3}}{2}$

4. Simplify: $b = 66\sqrt{3}$

Final Answer:

$b = 66\sqrt{3}$

If you need further steps or a detailed explanation, feel free to ask!

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5 Related Questions:

1. How can the Law of Sines be used to find the third side or angle in other triangles?
2. What is the role of the Law of Cosines in solving triangles?
3. How does the value of $\sin 30^\circ$ and $\sin 60^\circ$ relate to their geometric significance in a unit circle?
4. What are the conditions under which the Law of Sines is not applicable?
5. How can the exact value of $c$ in this triangle be found using trigonometry?

Tip:

Always double-check whether the triangle is a right triangle before applying the Law of Sines, as simpler trigonometric methods may sometimes suffice.