In `DeltaABC`,
In a triangle ABC, `angleC=60^(@) and angleA=75^(@)`. If D is a point on AC such that the area of the `DeltaBAD` is `sqrt(3)` times the area of the `DeltaBCD`, find the `angleABD`.

Answer

To solve the problem step by step, we will follow the reasoning presented in the video transcript.

### Step 1: Understand the given information
We have a triangle \( \Delta ABC \) with:
- \( \angle C = 60^\circ \)
- \( \angle A = 75^\circ \)

We need to find \( \angle ABD \) given that the area of triangle \( \Delta BAD \) is \( \sqrt{3} \) times the area of triangle \( \Delta BCD \).

### Step 2: Set up the area relationship
Let the area of triangle \( \Delta BCD \) be \( S \). Then, the area of triangle \( \Delta BAD \) is \( \sqrt{3} S \).

### Step 3: Use the area formula
The area of a triangle can be expressed as:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]
For triangle \( \Delta BAD \):
\[ \text{Area of } \Delta BAD = \frac{1}{2} \times AD \times BE \]
For triangle \( \Delta BCD \):
\[ \text{Area of } \Delta BCD = \frac{1}{2} \times CD \times BE \]

### Step 4: Set up the equation
From the area relationship:
\[ \frac{1}{2} \times AD \times BE = \sqrt{3} \times \left( \frac{1}{2} \times CD \times BE \right) \]
We can cancel \( \frac{1}{2} \) and \( BE \) (assuming \( BE \neq 0 \)):
\[ AD = \sqrt{3} \times CD \]

### Step 5: Express \( AD \) and \( CD \) in terms of a variable
Let \( CD = x \). Then:
\[ AD = \sqrt{3} x \]

### Step 6: Use the section formula
Since \( D \) divides \( AC \) in the ratio \( AD : CD = \sqrt{3} : 1 \), we can apply the section formula. The total length \( AC \) can be expressed as:
\[ AC = AD + CD = \sqrt{3}x + x = (\sqrt{3} + 1)x \]

### Step 7: Apply the MN theorem
According to the MN theorem, we have:
\[ \frac{AD}{CD} = \frac{\sqrt{3}}{1} \]
This leads to:
\[ \cot \theta = \frac{1 \cdot \cot A - \sqrt{3} \cdot \cot C}{\sqrt{3} + 1} \]
Where \( \cot A = \cot 75^\circ \) and \( \cot C = \cot 60^\circ \).

### Step 8: Calculate \( \cot 75^\circ \) and \( \cot 60^\circ \)
Using known values:
- \( \cot 60^\circ = \frac{1}{\sqrt{3}} \)
- \( \cot 75^\circ = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \)

### Step 9: Substitute and simplify
Substituting these values into the equation:
\[ \cot \theta = \frac{1 \cdot \frac{\sqrt{3} - 1}{\sqrt{3} + 1} - \sqrt{3} \cdot \frac{1}{\sqrt{3}}}{\sqrt{3} + 1} \]
This simplifies to:
\[ \cot \theta = \frac{\sqrt{3} - 1 - 1}{\sqrt{3} + 1} = \frac{\sqrt{3} - 2}{\sqrt{3} + 1} \]

### Step 10: Find \( \theta \)
After further simplification, we find that \( \theta = 105^\circ \).

### Step 11: Find \( \angle ABD \)
Using the exterior angle property:
\[ \angle ADB = 105^\circ \]
Thus:
\[ \angle ABD = \angle ADB - \angle A = 105^\circ - 75^\circ = 30^\circ \]

### Final Answer
\[ \angle ABD = 30^\circ \]