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 will use the concept of the sine rule in triangle geometry. We need to find the ratio of the sines of angles BAD and CAD in triangle ABC, where angle B = π/3 and angle C = π/4, and point D divides side BC in the ratio 1:3.

### Step-by-Step Solution:

1. **Calculate Angle A**:
Since the angles in a triangle sum up to π, we can find angle A:
\[ \text{Angle A} = \pi - \text{Angle B} - \text{Angle C} = \pi - \frac{\pi}{3} - \frac{\pi}{4} \]
To combine these fractions, we find a common denominator (which is 12):
\[ \text{Angle A} = \pi - \left(\frac{4\pi}{12} + \frac{3\pi}{12}\right) = \pi - \frac{7\pi}{12} = \frac{5\pi}{12} \]

2. **Use the Sine Rule**:
According to the sine rule:
\[ \frac{AD}{DC} = \frac{\sin(\angle BAD)}{\sin(\angle CAD)} \]
Since D divides BC in the ratio 1:3, we have:
\[ \frac{AD}{DC} = \frac{1}{3} \]

3. **Set Up the Ratio**:
From the sine rule, we can set up the equation:
\[ \frac{1}{3} = \frac{\sin(\angle BAD)}{\sin(\angle CAD)} \]
Rearranging gives us:
\[ \frac{\sin(\angle BAD)}{\sin(\angle CAD)} = \frac{1}{3} \]

4. **Final Result**:
Thus, we find that:
\[ \frac{\sin(\angle BAD)}{\sin(\angle CAD)} = \frac{1}{3} \]

### Conclusion:
The final answer is:
\[ \frac{\sin(\angle BAD)}{\sin(\angle CAD)} = \frac{1}{3} \]