In a `DeltaABC` right angled at A, a line is drawn through A to meet BC at D dividing BC in `2:1`. If `tan(angleADC)=3` then `angleBAD` is : (a) `30^(@)` (b) `45^(@)` (c) `60^(@)` (d) `75^(@)`

Options

A. `30^(@)`
B. `45^(@)`
C. `60^(@)`
D. `75^(@)`

Answer

To solve the problem step by step, we will follow the given information and apply the necessary mathematical concepts.

### Step 1: Understand the Triangle and Given Information
We have a triangle ABC right-angled at A. A line is drawn through A to meet BC at D, dividing BC in the ratio of 2:1. We are given that \( \tan(\angle ADC) = 3 \).

### Step 2: Set Up the Ratios
Let \( BD = 2x \) and \( DC = x \) (since D divides BC in the ratio 2:1). Therefore, the total length \( BC = BD + DC = 2x + x = 3x \).

### Step 3: Use the M-N Theorem
According to the M-N theorem, we have:
\[ (m+n) \cdot \cot(\theta) = m \cdot \cot(\alpha) - n \cdot \cot(\beta) \]
Where:
- \( m = 2 \)
- \( n = 1 \)
- \( \theta = \angle ADC \)
- \( \alpha = \angle BAD \)
- \( \beta = \angle DAC \)

### Step 4: Substitute Known Values
We know \( \tan(\angle ADC) = 3 \), so \( \cot(\angle ADC) = \frac{1}{3} \). Thus, we can write:
\[ (2 + 1) \cdot \frac{1}{3} = 2 \cdot \cot(\alpha) - 1 \cdot \cot(\beta) \]
This simplifies to:
\[ 1 = 2 \cdot \cot(\alpha) - \cot(\beta) \]

### Step 5: Express \( \cot(\beta) \)
Since \( \beta = 90^\circ - \alpha \), we have:
\[ \cot(\beta) = \tan(\alpha) \]
Substituting this into our equation gives:
\[ 1 = 2 \cdot \cot(\alpha) - \tan(\alpha) \]

### Step 6: Use the Identity for Cotangent and Tangent
Recall that \( \cot(\alpha) = \frac{1}{\tan(\alpha)} \). Let \( t = \tan(\alpha) \). Then:
\[ 1 = 2 \cdot \frac{1}{t} - t \]
Multiplying through by \( t \) gives:
\[ t = 2 - t^2 \]
Rearranging this gives:
\[ t^2 + t - 2 = 0 \]

### Step 7: Solve the Quadratic Equation
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ t = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-2)}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 + 8}}{2} = \frac{-1 \pm 3}{2} \]
This gives us two solutions:
\[ t = 1 \quad \text{or} \quad t = -2 \]
Since \( t = \tan(\alpha) \) must be positive, we have \( \tan(\alpha) = 1 \).

### Step 8: Determine \( \alpha \)
If \( \tan(\alpha) = 1 \), then:
\[ \alpha = 45^\circ \]

### Conclusion
Thus, the angle \( \angle BAD \) is \( 45^\circ \).

### Answer
(b) \( 45^\circ \)