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 solve the problem of finding the sides of a triangle formed by bending a wire of length \( L \) cm, where one of the angles is \( 60^\circ \), and to maximize the area of the triangle, 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 know that:
\[ a + b + c = L \]

### Step 2: Use the Area Formula
The area \( A \) of a triangle can be calculated using the formula:
\[ A = \frac{1}{2}ab \sin C \]
where \( C \) is the angle between sides \( a \) and \( b \). In our case, \( C = 60^\circ \), and we know that \( \sin 60^\circ = \frac{\sqrt{3}}{2} \). Therefore, the area becomes:
\[ A = \frac{1}{2}ab \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4} ab \]

### Step 3: Express One Side in Terms of Others
From the perimeter equation \( a + b + c = L \), we can express \( c \) as:
\[ c = L - a - b \]

### Step 4: Substitute \( c \) into Area Formula
Using the Law of Cosines, we can express \( c^2 \) in terms of \( a \) and \( b \):
\[ c^2 = a^2 + b^2 - 2ab \cos(60^\circ) \]
Since \( \cos(60^\circ) = \frac{1}{2} \), we have:
\[ c^2 = a^2 + b^2 - ab \]

### Step 5: Maximize the Area
To maximize the area \( A \), we can use the method of Lagrange multipliers or simply observe that for a fixed perimeter, the area of a triangle is maximized when it is equilateral. Therefore, we set \( a = b = c \).

### Step 6: Set the Sides Equal
If \( a = b = c \), then:
\[ 3a = L \implies a = \frac{L}{3} \]
Thus, we have:
\[ a = b = c = \frac{L}{3} \]

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