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The problem involves solving for the angle \( \angle Q \) in triangle \( \triangle PQR \) where \( q = 420 \) inches, \( p = 400 \) inches, and \( \angle P = 64^\circ \).

To solve for \( \angle Q \), we'll use the Law of Sines:

\[
\frac{p}{\sin(\angle P)} = \frac{q}{\sin(\angle Q)}
\]

Substituting the known values:

\[
\frac{400}{\sin(64^\circ)} = \frac{420}{\sin(\angle Q)}
\]

Now, solve for \( \sin(\angle Q) \):

\[
\sin(\angle Q) = \frac{420 \times \sin(64^\circ)}{400}
\]

Calculating \( \sin(64^\circ) \) and then solving for \( \sin(\angle Q) \):

\[
\sin(64^\circ) \approx 0.8988
\]

\[
\sin(\angle Q) = \frac{420 \times 0.8988}{400} \approx 0.9447
\]

Now, find \( \angle Q \) by taking the inverse sine:

\[
\angle Q \approx \sin^{-1}(0.9447) \approx 71^\circ
\]

However, due to the ambiguous case in the Law of Sines (since \( \sin(\theta) \) can give two different angles in the range of \( 0^\circ \) to \( 180^\circ \)), we must consider the possibility of a second solution:

The other possible angle \( \angle Q \) is:

\[
\angle Q' = 180^\circ - 71^\circ \approx 109^\circ
\]

Thus, the possible values for \( \angle Q \) to the nearest degree are \( 71^\circ \) and \( 109^\circ \).

Would you like any additional details or have any questions?

Here are five related questions to deepen your understanding:
1. What is the Law of Sines, and how is it applied in different types of triangles?
2. How does the ambiguous case in the Law of Sines arise, and why does it lead to two possible angles?
3. Can the Law of Sines be used in non-acute triangles? Why or why not?
4. How do you determine which angle is correct when solving real-world problems using the Law of Sines?
5. What are some scenarios where only one solution exists when using the Law of Sines?

**Tip:** Always check for the possibility of two solutions when using the Law of Sines, especially in cases involving non-right triangles.

In ∆PQR, q = 420 inches, p = 400 inches, and ∠P = 64°. Find all possible values of ∠Q, to the nearest degree.

Learn how to solve for angle Q in triangle PQR using the Law of Sines. Given q = 420 inches, p = 400 inches, and angle P = 64°, the problem explores trigonometric concepts and the ambiguous case of the Law of Sines. Suitable for high school students in Grades 10-12.

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