In a triangle ABC, `/_ABC=45^@` and a point D is on BC such that `2BD=CD` and `/_DAB=15^@` then `/_ACB=`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the given information
We have triangle ABC with:
- \( \angle ABC = 45^\circ \)
- A point D on BC such that \( 2BD = CD \)
- \( \angle DAB = 15^\circ \)

### Step 2: Set up the triangle and angles
Let:
- \( \angle DAB = 15^\circ \)
- \( \angle ABC = 45^\circ \)

Since \( D \) divides \( BC \) in the ratio \( 2:1 \), we can denote:
- \( BD = x \)
- \( CD = 2x \)

### Step 3: Use the exterior angle theorem
According to the exterior angle theorem, we know that:
\[ \angle ADC = \angle DAB + \angle ABC \]
Substituting the known angles:
\[ \angle ADC = 15^\circ + 45^\circ = 60^\circ \]

### Step 4: Apply the MN theorem
Using the MN theorem, we have:
\[ \frac{M + N}{\cot \theta} = N \cot B + M \cot C \]
Where:
- \( M = 2 \) (for \( CD \))
- \( N = 1 \) (for \( BD \))
- \( \theta = \angle ADC = 60^\circ \)
- \( B = \angle ABC = 45^\circ \)
- \( C = \angle ACB \)

Substituting these values into the MN theorem:
\[ \frac{2 + 1}{\cot 60^\circ} = 1 \cdot \cot 45^\circ + 2 \cdot \cot C \]

### Step 5: Calculate cotangent values
We know:
- \( \cot 60^\circ = \frac{1}{\sqrt{3}} \)
- \( \cot 45^\circ = 1 \)

Substituting these values:
\[ \frac{3}{\frac{1}{\sqrt{3}}} = 1 + 2 \cot C \]
This simplifies to:
\[ 3\sqrt{3} = 1 + 2 \cot C \]

### Step 6: Solve for \( \cot C \)
Rearranging the equation gives:
\[ 2 \cot C = 3\sqrt{3} - 1 \]
Thus:
\[ \cot C = \frac{3\sqrt{3} - 1}{2} \]

### Step 7: Find angle \( C \)
To find \( C = \angle ACB \), we can find the angle corresponding to the cotangent:
\[ \cot C = 2 - \sqrt{3} \]
This corresponds to:
\[ C = 75^\circ \]

### Final Answer
Thus, the angle \( \angle ACB \) is:
\[ \angle ACB = 75^\circ \]

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