In a `Delta ABC,` If angles A, B and C are in A.P and angle A exceeds lowest angle by `30^(@)` and d divides BC internally in the ratio `1:3` thenfind the value of `(sin (lt BAD))/(sin (lt CAD))`

Answer

To solve the problem, we need to find the value of \(\frac{\sin(\angle BAD)}{\sin(\angle CAD)}\) given that angles \(A\), \(B\), and \(C\) of triangle \(ABC\) are in arithmetic progression (A.P.), and that \(\angle A\) exceeds the smallest angle by \(30^\circ\). Additionally, point \(D\) divides \(BC\) internally in the ratio \(1:3\).

### Step-by-Step Solution:

1. **Define the Angles**:
Let the angles be defined as follows:
- \(\angle A = a\)
- \(\angle B = b\)
- \(\angle C = c\)

Since \(A\), \(B\), and \(C\) are in A.P., we can express them as:
\[ b = \frac{a + c}{2} \]

2. **Use the Given Condition**:
We know that \(\angle A\) exceeds the smallest angle by \(30^\circ\). Assuming \(C\) is the smallest angle, we have:
\[ a = c + 30^\circ \]

3. **Sum of Angles in Triangle**:
The sum of angles in triangle \(ABC\) is \(180^\circ\):
\[ a + b + c = 180^\circ \]

4. **Substituting for \(b\)**:
Substitute \(b\) from the A.P. condition:
\[ a + \frac{a + c}{2} + c = 180^\circ \]
Multiply through by \(2\) to eliminate the fraction:
\[ 2a + a + c + 2c = 360^\circ \]
Simplifying gives:
\[ 3a + 3c = 360^\circ \quad \Rightarrow \quad a + c = 120^\circ \]

5. **Substituting for \(c\)**:
Now substitute \(c = a - 30^\circ\) into \(a + c = 120^\circ\):
\[ a + (a - 30^\circ) = 120^\circ \]
This simplifies to:
\[ 2a - 30^\circ = 120^\circ \quad \Rightarrow \quad 2a = 150^\circ \quad \Rightarrow \quad a = 75^\circ \]

6. **Finding \(c\)**:
Now substitute back to find \(c\):
\[ c = a - 30^\circ = 75^\circ - 30^\circ = 45^\circ \]

7. **Finding \(b\)**:
Now we can find \(b\):
\[ b = 180^\circ - a - c = 180^\circ - 75^\circ - 45^\circ = 60^\circ \]

8. **Using the Sine Rule**:
Since \(D\) divides \(BC\) in the ratio \(1:3\), we can use the sine rule:
\[ \frac{AD}{DC} = \frac{AB \cdot \sin C}{AC \cdot \sin B} \]

Let \(AD = x\) and \(DC = 3x\). Then:
\[ \frac{x}{3x} = \frac{AB \cdot \sin 45^\circ}{AC \cdot \sin 60^\circ} \]

9. **Setting Up the Ratios**:
From the sine rule, we have:
\[ \frac{1}{3} = \frac{AB \cdot \frac{1}{\sqrt{2}}}{AC \cdot \frac{\sqrt{3}}{2}} \]

10. **Finding the Final Ratio**:
Rearranging gives:
\[ \frac{AB}{AC} = \frac{1}{3} \cdot \frac{\sqrt{3}}{2} \cdot \sqrt{2} = \frac{\sqrt{6}}{6} \]

Therefore, we need to find:
\[ \frac{\sin(\angle BAD)}{\sin(\angle CAD)} = \frac{3 \cdot \sin 60^\circ}{\sin 45^\circ} = \frac{3 \cdot \frac{\sqrt{3}}{2}}{\frac{1}{\sqrt{2}}} = \frac{3\sqrt{3}}{2} \cdot \sqrt{2} = \frac{3\sqrt{6}}{6} = \frac{1}{\sqrt{6}} \]

### Final Answer:
\[ \frac{\sin(\angle BAD)}{\sin(\angle CAD)} = \frac{1}{\sqrt{6}} \]