Two sides of a triangle are of lengths `sqrt6 and 4` and the angle opposite to smaller side is 30. How many such triangles are possible? Find the length of their third side and area.

Options

A. 0
B. 1
C. 2
D. infinite

Explanation

To solve the problem step by step, we will follow the given information about the triangle and apply the Law of Cosines and area formulas.

### Step 1: Identify the triangle and given values
We have a triangle with two sides:
- Side \( a = \sqrt{6} \) (opposite the angle of 30 degrees)
- Side \( b = 4 \)

The angle opposite the smaller side \( a \) is given as \( 30^\circ \).

### Step 2: Use the Law of Cosines
According to the Law of Cosines:
\[ c^2 = a^2 + b^2 - 2ab \cos(C) \]
where \( C \) is the angle opposite side \( c \).

Here, we need to find the length of side \( c \) (the third side). We can rearrange the formula to solve for \( c \):
\[ c^2 = (\sqrt{6})^2 + 4^2 - 2 \cdot \sqrt{6} \cdot 4 \cdot \cos(30^\circ) \]
Substituting the values:
\[ c^2 = 6 + 16 - 2 \cdot \sqrt{6} \cdot 4 \cdot \frac{\sqrt{3}}{2} \]
\[ c^2 = 22 - 4\sqrt{6}\sqrt{3} \]
\[ c^2 = 22 - 4\sqrt{18} \]
\[ c^2 = 22 - 12 = 10 \]
Thus, we have:
\[ c = \sqrt{10} \]

### Step 3: Determine the number of triangles
Since we have an angle of \( 30^\circ \) opposite the side \( \sqrt{6} \), we can check if there are two possible triangles. The angle opposite the longer side (4) can be calculated using the Law of Sines:
\[ \frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)} \]
Given that \( A = 30^\circ \), we can find \( B \) and check if it leads to a valid triangle.

### Step 4: Calculate the area of the triangle
The area \( A \) of a triangle can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \cdot a \cdot b \cdot \sin(C) \]
Substituting the known values:
\[ \text{Area} = \frac{1}{2} \cdot \sqrt{6} \cdot 4 \cdot \sin(30^\circ) \]
Since \( \sin(30^\circ) = \frac{1}{2} \):
\[ \text{Area} = \frac{1}{2} \cdot \sqrt{6} \cdot 4 \cdot \frac{1}{2} = \sqrt{6} \]

### Summary of Results
- The length of the third side \( c \) is \( \sqrt{10} \).
- Two triangles are possible due to the ambiguous case of the triangle.
- The area of each triangle is \( \sqrt{6} \).