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We are given a triangle with two marked angles (45° and 30°), one side labeled \( 13\sqrt{2} \), and two unknowns \( b \), \( c \), \( d \), and \( a \) where \( a \) is the altitude of the triangle.

We can approach solving this using trigonometry and the properties of right triangles. Here's how we'll proceed step by step:

### Step 1: Calculate side \( c \)
The angle adjacent to side \( c \) is 45°, and the hypotenuse of this right triangle is \( 13\sqrt{2} \). We use the cosine function:

\[
\cos(45^\circ) = \frac{c}{13\sqrt{2}}
\]

Since \( \cos(45^\circ) = \frac{1}{\sqrt{2}} \), we can solve for \( c \):

\[
\frac{1}{\sqrt{2}} = \frac{c}{13\sqrt{2}}
\]
\[
c = (13\sqrt{2}) \times \frac{1}{\sqrt{2}} = 13
\]

So, \( c = 13 \).

### Step 2: Calculate side \( d \)
Next, we use the sine function to find \( d \), since the opposite angle to \( d \) is 45°:

\[
\sin(45^\circ) = \frac{d}{13\sqrt{2}}
\]

Since \( \sin(45^\circ) = \frac{1}{\sqrt{2}} \), we solve for \( d \):

\[
\frac{1}{\sqrt{2}} = \frac{d}{13\sqrt{2}}
\]
\[
d = (13\sqrt{2}) \times \frac{1}{\sqrt{2}} = 13
\]

Thus, \( d = 13 \).

### Step 3: Use the 30° angle to calculate side \( b \)
For the other right triangle, with hypotenuse \( b \) and the 30° angle, we use the cosine function:

\[
\cos(30^\circ) = \frac{d}{b}
\]

We already know that \( d = 13 \) and \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \), so:

\[
\frac{\sqrt{3}}{2} = \frac{13}{b}
\]
\[
b = \frac{13 \times 2}{\sqrt{3}} = \frac{26}{\sqrt{3}}
\]

To rationalize the denominator:

\[
b = \frac{26 \times \sqrt{3}}{3} = \frac{26\sqrt{3}}{3}
\]

### Step 4: Calculate the altitude \( a \)
To find the altitude \( a \), we use the area formula for both triangles. The area of the whole triangle can be expressed as:

\[
\text{Area} = \frac{1}{2} \times c \times a = \frac{1}{2} \times d \times a
\]

Setting the two areas equal:

\[
13a = 13a
\]

Since both sides are equal, we can solve for \( a \).

---

In summary, we found:
- \( c = 13 \)
- \( d = 13 \)
- \( b = \frac{26\sqrt{3}}{3} \)
- \( a \) can be computed directly once further constraints are added.

Would you like to explore further details or ask about other concepts here?

Here are 5 related questions to expand on this:
1. How would we compute the exact area of the triangle?
2. Can we generalize the solution to any angle triangle?
3. What happens if one angle is changed, say from 45° to 60°?
4. How do trigonometric identities help in verifying solutions?
5. What role does the altitude play in scaling the triangle?

**Tip**: When solving right triangles, always check if simpler trigonometric ratios can reduce the complexity.

Explore step-by-step solutions for finding unknown variables in a right triangle with angles 45° and 30°, including calculations for side lengths using trigonometric functions. This problem demonstrates core trigonometry concepts and is suitable for Grades 10-12.

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