In the figure below, the distance from A to D is y, the distance from D to C is x, and the distance from C to B is h. Find x, then y. Assume A = 45°, ∠BDC = 54°, AB = 16 and DB = 14. (Round each answer to one decimal place.)

To solve this problem, we need to apply trigonometric relationships. Let's break down the steps, using the given information:

### Step 1: Analyze the triangle
We are given the following:

- \( \angle A = 45^\circ \)
- \( \angle BDC = 54^\circ \)
- \( AB = 16 \)
- \( DB = 14 \)
- Distance from \( A \) to \( D = y \)
- Distance from \( D \) to \( C = x \)
- Distance from \( C \) to \( B = h \)

We are tasked with finding \( x \) (the distance from \( D \) to \( C \)) and \( y \) (the distance from \( A \) to \( D \)).

### Step 2: Use the Law of Sines
To find \( x \), we need to use the Law of Sines in triangle \( BDC \). In any triangle, the Law of Sines states:

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

In \( \triangle BDC \), we have \( \angle BDC = 54^\circ \), and we can calculate \( \angle DBC \) using the fact that the sum of angles in a triangle is \( 180^\circ \):

\[
\angle DBC = 180^\circ - 45^\circ - 54^\circ = 81^\circ
\]

### Step 3: Solve for \( x \)
Now apply the Law of Sines in triangle \( BDC \) to solve for \( x \):

\[
\frac{x}{\sin \angle DBC} = \frac{DB}{\sin \angle BDC}
\]

Substitute the known values:

\[
\frac{x}{\sin 81^\circ} = \frac{14}{\sin 54^\circ}
\]

Solve for \( x \):

\[
x = \frac{14 \times \sin 81^\circ}{\sin 54^\circ}
\]

### Step 4: Solve for \( y \)
Once \( x \) is found, we can calculate \( y \) using trigonometric relationships in triangle \( ABD \), since \( \triangle ABD \) involves \( \angle A = 45^\circ \), and we can use either the Law of Cosines or another trigonometric relation to find \( y \).

Let me compute the exact values for \( x \) and \( y \).The calculated distances are:

- \( x \) (distance from \( D \) to \( C \)): 17.1 units
- \( y \) (distance from \( A \) to \( D \)): 11.6 units

Would you like more details or have any questions? Here are some related questions you might explore:

1. How is the Law of Sines applied in other types of triangles?
2. What conditions are necessary for using the Law of Cosines?
3. How can the Law of Sines and Cosines be used in solving real-world problems?
4. How does changing angle values affect the side lengths in a triangle?
5. Can these methods be extended to solve non-right triangles?

**Tip:** The Law of Sines is especially useful when you know two angles and one side (AAS or ASA) in a triangle.

Solve a trigonometric problem involving the Law of Sines to find the distances x and y in a triangle. Given angles include A = 45° and ∠BDC = 54°, with side lengths AB = 16 and DB = 14. The step-by-step solution helps learners understand the use of the Law of Sines and triangle properties. Suitable for high school students (Grades 9-12).

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