A L cm long wire is bent to form a triangle with one of it's angle as `60^(@)`.Find the sides of the triangle for which area is largest.

Answer

To find the sides of a triangle formed by a wire of length \( L \) cm, with one angle measuring \( 60^\circ \), that maximizes the area, we can follow these steps:

### Step 1: Define the sides of the triangle
Let the sides of the triangle be \( a \), \( b \), and \( c \). According to the problem, we have:
\[ a + b + c = L \]

### Step 2: Use the angle information
Given that one of the angles is \( 60^\circ \), we can use the formula for the area of a triangle:
\[ \text{Area} = \frac{1}{2}ab \sin C \]
where \( C \) is the angle between sides \( a \) and \( b \). Here, \( C = 60^\circ \), and \( \sin 60^\circ = \frac{\sqrt{3}}{2} \). Thus, the area becomes:
\[ \text{Area} = \frac{1}{2}ab \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4} ab \]

### Step 3: Express one variable in terms of the others
From the perimeter equation \( a + b + c = L \), we can express \( c \) in terms of \( a \) and \( b \):
\[ c = L - a - b \]

### Step 4: Use the cosine rule
To express \( c \) in terms of \( a \) and \( b \), we can apply the cosine rule:
\[ c^2 = a^2 + b^2 - 2ab \cos(60^\circ) \]
Since \( \cos(60^\circ) = \frac{1}{2} \), we have:
\[ c^2 = a^2 + b^2 - ab \]

### Step 5: Substitute \( c \) into the area formula
We can substitute \( c = L - a - b \) into the area formula:
\[ \text{Area} = \frac{\sqrt{3}}{4} ab \]
We want to maximize this area under the constraint \( a + b + c = L \).

### Step 6: Use Lagrange multipliers or substitution
To maximize the area, we can use the method of Lagrange multipliers or substitute \( c \) back into the area equation. However, for simplicity, we can also analyze the case when \( a = b \) (since the area will be maximized when the triangle is equilateral).

Let \( a = b \). Then:
\[ 2a + c = L \implies c = L - 2a \]

### Step 7: Set up the area in terms of one variable
Substituting \( b = a \) into the area formula gives:
\[ \text{Area} = \frac{\sqrt{3}}{4} a^2 \]
Now we have:
\[ c = L - 2a \]

### Step 8: Find the maximum area
To find the maximum area, we can differentiate the area with respect to \( a \) and set it to zero:
\[ \frac{d(\text{Area})}{da} = \frac{\sqrt{3}}{2} a = 0 \]
This implies \( a = 0 \), which is not useful. Instead, we can analyze the constraints.

### Step 9: Solve for the sides
To maximize the area, we can set:
\[ a = b = c \]
Thus:
\[ 3a = L \implies a = \frac{L}{3}, \quad b = \frac{L}{3}, \quad c = \frac{L}{3} \]

### Conclusion
The sides of the triangle that maximize the area when one angle is \( 60^\circ \) are:
\[ a = b = c = \frac{L}{3} \text{ cm} \]