In the following figure , angle ` B = 90 ^(@) angle ADB = 30 ^(@) and angle ACB = 60 ^(@) `
If ` angle CD = m` find AB .
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Answer

To solve for the length of \( AB \) in the given triangle configuration, we will follow these steps:

### Step 1: Identify the given information
We have:
- \( \angle B = 90^\circ \)
- \( \angle ADB = 30^\circ \)
- \( \angle ACB = 60^\circ \)
- \( CD = 40 \, \text{meters} \)

### Step 2: Set up the triangle dimensions
Let:
- \( BC = X \)
- \( AB = Y \)

### Step 3: Use the tangent function in triangle \( ABD \)
In triangle \( ABD \), we can use the tangent of angle \( ADB \):
\[ \tan(30^\circ) = \frac{AB}{BD} \]
Since \( BD = BC + CD = X + 40 \):
\[ \tan(30^\circ) = \frac{Y}{X + 40} \]
We know that \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \), so we have:
\[ \frac{1}{\sqrt{3}} = \frac{Y}{X + 40} \]
Cross-multiplying gives us:
\[ Y = \frac{1}{\sqrt{3}}(X + 40) \]
This can be rewritten as:
\[ \sqrt{3}Y = X + 40 \quad \text{(Equation 1)} \]

### Step 4: Use the tangent function in triangle \( ABC \)
In triangle \( ABC \), we can use the tangent of angle \( ACB \):
\[ \tan(60^\circ) = \frac{AB}{BC} \]
Thus:
\[ \tan(60^\circ) = \frac{Y}{X} \]
Since \( \tan(60^\circ) = \sqrt{3} \), we have:
\[ \sqrt{3} = \frac{Y}{X} \]
Cross-multiplying gives us:
\[ Y = \sqrt{3}X \quad \text{(Equation 2)} \]

### Step 5: Substitute Equation 2 into Equation 1
Substituting \( Y \) from Equation 2 into Equation 1:
\[ \sqrt{3}(\sqrt{3}X) = X + 40 \]
This simplifies to:
\[ 3X = X + 40 \]
Subtracting \( X \) from both sides:
\[ 2X = 40 \]
Dividing by 2:
\[ X = 20 \]

### Step 6: Find \( Y \) using Equation 2
Now substituting \( X \) back into Equation 2 to find \( Y \):
\[ Y = \sqrt{3} \times 20 = 20\sqrt{3} \]

### Step 7: Calculate the numerical value of \( Y \)
Calculating \( Y \):
\[ Y \approx 20 \times 1.732 = 34.64 \, \text{meters} \]

### Final Answer
Thus, the length of \( AB \) is:
\[ AB \approx 34.64 \, \text{meters} \]

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