In trianlge ABC, point D lies on BC such that `angleBAD = 45^(@), angleDAC = 30^(@)`, If `BD:DC=2:1` then `3cotangleADC` is equal to

Options

A. `2+sqrt(3)`
B. `2-sqrt(3)`
C. `sqrt(3)`
D. `1/sqrt(3)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Given Information
We are given a triangle ABC with point D on side BC. The angles are:
- \( \angle BAD = 45^\circ \)
- \( \angle DAC = 30^\circ \)
We also know the ratio \( BD:DC = 2:1 \).

### Step 2: Assign Variables
Let:
- \( BD = 2x \)
- \( DC = x \)

Thus, the total length of \( BC \) is:
\[ BC = BD + DC = 2x + x = 3x \]

### Step 3: Apply the Cotangent Rule
We need to find \( 3 \cot \angle ADC \). Let \( \angle ADC = \theta \).

Using the cotangent rule in triangle ABD and ADC, we can express the relationship:
\[ \frac{BD}{DC} = \frac{AB \cdot \sin \angle DAC}{AC \cdot \sin \angle BAD} \]

### Step 4: Substitute Known Values
From the problem, we know:
- \( \angle BAD = 45^\circ \) implies \( \cot 45^\circ = 1 \)
- \( \angle DAC = 30^\circ \) implies \( \cot 30^\circ = \sqrt{3} \)

Using the cotangent rule:
\[ \frac{2x}{x} = \frac{AB \cdot \sin 30^\circ}{AC \cdot \sin 45^\circ} \]
This simplifies to:
\[ 2 = \frac{AB \cdot \frac{1}{2}}{AC \cdot \frac{\sqrt{2}}{2}} \]
Thus:
\[ 2 = \frac{AB}{AC} \cdot \frac{1}{\sqrt{2}} \]
This implies:
\[ AB = 2 \sqrt{2} AC \]

### Step 5: Set Up the Equation
From the triangle properties, we can set up the equation using the cotangent:
\[ 2x + x \cot \theta = 2x \cdot 1 - x \cdot \sqrt{3} \]
This simplifies to:
\[ 3x \cot \theta = 2x - x \sqrt{3} \]

### Step 6: Solve for \( \cot \theta \)
Dividing through by \( x \) (assuming \( x \neq 0 \)):
\[ 3 \cot \theta = 2 - \sqrt{3} \]

### Step 7: Final Result
Thus, we find:
\[ 3 \cot \angle ADC = 2 - \sqrt{3} \]

### Conclusion
The final answer is:
\[ \boxed{2 - \sqrt{3}} \]