In a `DeltaABC, angleB=(pi)/(3) and angleC=(pi)/(4)` let D divide BC internally in the ratio `1:3`, then ` (sin(angleBAD))/(sin(angleCAD))` is equal to :

Options

A. `(1)/(sqrt(6))`
B. `(1)/(3)`
C. `(1)/(sqrt(3))`
D. `(sqrt(2))/(3)`

Answer

To solve the problem, we need to find the ratio \(\frac{\sin(\angle BAD)}{\sin(\angle CAD)}\) in triangle \(ABC\) where \(\angle B = \frac{\pi}{3}\) and \(\angle C = \frac{\pi}{4}\), and point \(D\) divides \(BC\) internally in the ratio \(1:3\).

### Step-by-Step Solution:

1. **Identify Angles**:
We have:
\[ \angle B = \frac{\pi}{3}, \quad \angle C = \frac{\pi}{4} \]
Therefore, we can find \(\angle A\) using the fact that the sum of angles in a triangle is \(\pi\):
\[ \angle A = \pi - \angle B - \angle C = \pi - \frac{\pi}{3} - \frac{\pi}{4} \]
To calculate this, we need a common denominator:
\[ \angle A = \pi - \left(\frac{4\pi}{12} + \frac{3\pi}{12}\right) = \pi - \frac{7\pi}{12} = \frac{5\pi}{12} \]

2. **Apply the Sine Rule**:
In triangle \(ABD\), by the sine rule:
\[ \frac{AD}{\sin B} = \frac{BD}{\sin(\angle BAD)} \]
In triangle \(ACD\), by the sine rule:
\[ \frac{AD}{\sin C} = \frac{CD}{\sin(\angle CAD)} \]

3. **Express \(BD\) and \(CD\)**:
Since \(D\) divides \(BC\) in the ratio \(1:3\), we can express \(BD\) and \(CD\) as:
\[ BD = x, \quad CD = 3x \]
Hence, \(BC = BD + CD = x + 3x = 4x\).

4. **Set up the Ratios**:
From the sine rules, we can write:
\[ \frac{AD}{\sin B} = \frac{x}{\sin(\angle BAD)} \quad \text{and} \quad \frac{AD}{\sin C} = \frac{3x}{\sin(\angle CAD)} \]

5. **Divide the Two Equations**:
Dividing the two equations gives:
\[ \frac{\sin C}{\sin B} = \frac{3x \cdot \sin(\angle BAD)}{x \cdot \sin(\angle CAD)} \]
Simplifying this leads to:
\[ \frac{\sin C}{\sin B} = 3 \cdot \frac{\sin(\angle BAD)}{\sin(\angle CAD)} \]

6. **Substitute Values**:
We know:
\[ \sin B = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}, \quad \sin C = \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \]
Substituting these values into the equation gives:
\[ \frac{\frac{1}{\sqrt{2}}}{\frac{\sqrt{3}}{2}} = 3 \cdot \frac{\sin(\angle BAD)}{\sin(\angle CAD)} \]
Simplifying the left side:
\[ \frac{2}{\sqrt{6}} = 3 \cdot \frac{\sin(\angle BAD)}{\sin(\angle CAD)} \]

7. **Solve for the Ratio**:
Rearranging gives:
\[ \frac{\sin(\angle BAD)}{\sin(\angle CAD)} = \frac{2}{3} \cdot \frac{1}{\sqrt{6}} = \frac{2}{3\sqrt{6}} \]

### Final Result:
Thus, the ratio \(\frac{\sin(\angle BAD)}{\sin(\angle CAD)}\) is:
\[ \frac{\sin(\angle BAD)}{\sin(\angle CAD)} = \frac{1}{\sqrt{6}} \]