In Triangle ABC, AD is perpendicular to BC, ` tan B "" = (3)/(4) tan C = "" ( 5)/(12) ` and BC = 56 cm . Calculate the lengths of AD.

Answer

To solve the problem, we need to find the length of AD in triangle ABC, where AD is perpendicular to BC. We are given the following information:

- \( \tan B = \frac{3}{4} \)
- \( \tan C = \frac{5}{12} \)
- \( BC = 56 \, \text{cm} \)

Let's denote:
- \( BD = x \)
- \( DC = y \)
- \( AD = h \)

Since \( BC = BD + DC \), we have:
\[ x + y = 56 \quad \text{(1)} \]

### Step 1: Express \( h \) in terms of \( x \) using \( \tan B \)
From triangle ADB, we can express \( \tan B \):
\[ \tan B = \frac{AD}{BD} = \frac{h}{x} \]
Given \( \tan B = \frac{3}{4} \), we can write:
\[ \frac{h}{x} = \frac{3}{4} \]
Cross-multiplying gives:
\[ 4h = 3x \quad \text{(2)} \]

### Step 2: Express \( h \) in terms of \( y \) using \( \tan C \)
From triangle ADC, we can express \( \tan C \):
\[ \tan C = \frac{AD}{DC} = \frac{h}{y} \]
Given \( \tan C = \frac{5}{12} \), we can write:
\[ \frac{h}{y} = \frac{5}{12} \]
Cross-multiplying gives:
\[ 12h = 5y \quad \text{(3)} \]

### Step 3: Solve equations (2) and (3)
From equation (2):
\[ h = \frac{3}{4}x \]

From equation (3):
\[ h = \frac{5}{12}y \]

Now we can set these two expressions for \( h \) equal to each other:
\[ \frac{3}{4}x = \frac{5}{12}y \]

### Step 4: Express \( y \) in terms of \( x \)
Cross-multiplying gives:
\[ 3 \cdot 12x = 4 \cdot 5y \]
\[ 36x = 20y \]
\[ y = \frac{36}{20}x = \frac{9}{5}x \quad \text{(4)} \]

### Step 5: Substitute equation (4) into equation (1)
Substituting \( y \) in equation (1):
\[ x + \frac{9}{5}x = 56 \]
Combining terms:
\[ \frac{5}{5}x + \frac{9}{5}x = 56 \]
\[ \frac{14}{5}x = 56 \]

### Step 6: Solve for \( x \)
Multiplying both sides by \( \frac{5}{14} \):
\[ x = 56 \cdot \frac{5}{14} = 20 \]

### Step 7: Find \( y \)
Using equation (4) to find \( y \):
\[ y = \frac{9}{5}x = \frac{9}{5} \cdot 20 = 36 \]

### Step 8: Find \( h \) using \( x \)
Now we can find \( h \) using equation (2):
\[ h = \frac{3}{4}x = \frac{3}{4} \cdot 20 = 15 \]

### Conclusion
The length of \( AD \) is:
\[ \boxed{15 \, \text{cm}} \]

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