In a triagnle ABC, `angle B=pi/3 " and " angle C = pi/4` let D divide BC internally in the ratio 1:3 .Then `(sin (angle BAD))/((Sin (angle CAD))` 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 of \( \frac{\sin(\angle BAD)}{\sin(\angle CAD)} \) in triangle ABC, where \( \angle B = \frac{\pi}{3} \) and \( \angle C = \frac{\pi}{4} \). Point D divides side BC in the ratio 1:3.

### Step-by-step Solution:

1. **Identify Angles**:
- Given \( \angle B = \frac{\pi}{3} = 60^\circ \)
- Given \( \angle C = \frac{\pi}{4} = 45^\circ \)
- To find \( \angle A \):
\[ \angle A = \pi - \angle B - \angle C = \pi - \frac{\pi}{3} - \frac{\pi}{4} \]
To calculate this, we need a common denominator:
\[ \angle A = \pi - \frac{4\pi}{12} - \frac{3\pi}{12} = \pi - \frac{7\pi}{12} = \frac{5\pi}{12} \]

2. **Set Up the Ratio**:
- Let \( D \) divide \( BC \) internally in the ratio \( 1:3 \). This means if \( BD = x \) and \( DC = 3x \), then \( BC = BD + DC = x + 3x = 4x \).

3. **Apply the Sine Rule in Triangle ABD**:
- According to the sine rule:
\[ \frac{\sin(\angle BAD)}{BD} = \frac{\sin(\angle ABD)}{AD} \]
- Here, \( \angle ABD = \angle B = 60^\circ \):
\[ \frac{\sin(\angle BAD)}{x} = \frac{\sin(60^\circ)}{AD} \]
- Therefore,
\[ \sin(\angle BAD) = \frac{\sqrt{3}}{2} \cdot \frac{x}{AD} \quad \text{(Equation 1)} \]

4. **Apply the Sine Rule in Triangle ADC**:
- Again using the sine rule:
\[ \frac{\sin(\angle CAD)}{DC} = \frac{\sin(\angle ACD)}{AD} \]
- Here, \( \angle ACD = \angle C = 45^\circ \):
\[ \frac{\sin(\angle CAD)}{3x} = \frac{\sin(45^\circ)}{AD} \]
- Therefore,
\[ \sin(\angle CAD) = \frac{1}{\sqrt{2}} \cdot \frac{3x}{AD} \quad \text{(Equation 2)} \]

5. **Find the Ratio**:
- Now, we need to find \( \frac{\sin(\angle BAD)}{\sin(\angle CAD)} \):
\[ \frac{\sin(\angle BAD)}{\sin(\angle CAD)} = \frac{\frac{\sqrt{3}}{2} \cdot \frac{x}{AD}}{\frac{1}{\sqrt{2}} \cdot \frac{3x}{AD}} \]
- The \( AD \) and \( x \) cancel out:
\[ = \frac{\frac{\sqrt{3}}{2}}{\frac{3}{\sqrt{2}}} \]
- Simplifying this gives:
\[ = \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{3} = \frac{\sqrt{6}}{6} \]

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