Consider a `DeltaABC` and let a, b and c denote the lengths of the sides opposite to vertices A, B and C, repectively. `if a=1,b=3 and C=60^(@),` then `sin ^(2)` B is equal to

Options

A. `27/28`
B. `3/28`
C. `81/28`
D. `1/3`

Answer

To solve the problem, we need to find \( \sin^2 B \) in triangle \( ABC \) where \( a = 1 \), \( b = 3 \), and \( C = 60^\circ \).

### Step-by-Step Solution:

1. **Use the Cosine Rule to find \( c^2 \)**:
The cosine rule states:
\[ c^2 = a^2 + b^2 - 2ab \cos C \]
Here, we know \( a = 1 \), \( b = 3 \), and \( C = 60^\circ \). We can substitute these values into the formula:
\[ c^2 = 1^2 + 3^2 - 2 \cdot 1 \cdot 3 \cdot \cos(60^\circ) \]
Since \( \cos(60^\circ) = \frac{1}{2} \):
\[ c^2 = 1 + 9 - 2 \cdot 1 \cdot 3 \cdot \frac{1}{2} \]
Simplifying this gives:
\[ c^2 = 1 + 9 - 3 = 7 \]

2. **Use the Sine Rule to find \( \sin B \)**:
The sine rule states:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \]
We can express \( \sin B \) in terms of \( \sin C \):
\[ \sin B = \frac{b \cdot \sin C}{c} \]
Substituting the known values \( b = 3 \), \( C = 60^\circ \), and \( c^2 = 7 \) (thus \( c = \sqrt{7} \)):
\[ \sin B = \frac{3 \cdot \sin(60^\circ)}{\sqrt{7}} \]
Since \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \):
\[ \sin B = \frac{3 \cdot \frac{\sqrt{3}}{2}}{\sqrt{7}} = \frac{3\sqrt{3}}{2\sqrt{7}} \]

3. **Calculate \( \sin^2 B \)**:
Now, we need to find \( \sin^2 B \):
\[ \sin^2 B = \left(\frac{3\sqrt{3}}{2\sqrt{7}}\right)^2 = \frac{(3\sqrt{3})^2}{(2\sqrt{7})^2} = \frac{27}{4 \cdot 7} = \frac{27}{28} \]

### Final Answer:
Thus, \( \sin^2 B = \frac{27}{28} \).