In a `Delta ABC` if `b+c=2a` and `/_A=60^@` then `Delta ABC` is

Options

A. equilateral
B. right angled
C. isosceles
D. scalene

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Given Information
We are given that in triangle \( ABC \):
- \( b + c = 2a \)
- \( \angle A = 60^\circ \)

### Step 2: Use the Law of Sines
From the Law of Sines, we know:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \]
Given \( \angle A = 60^\circ \), we can express \( b \) and \( c \) in terms of \( a \):
\[ b = k \sin B \quad \text{and} \quad c = k \sin C \]
where \( k = \frac{a}{\sin A} = \frac{a}{\sin 60^\circ} = \frac{a}{\frac{\sqrt{3}}{2}} = \frac{2a}{\sqrt{3}} \).

### Step 3: Substitute Values
Substituting for \( b \) and \( c \):
\[ b = \frac{2a}{\sqrt{3}} \sin B \quad \text{and} \quad c = \frac{2a}{\sqrt{3}} \sin C \]
Now, substituting these into the equation \( b + c = 2a \):
\[ \frac{2a}{\sqrt{3}} \sin B + \frac{2a}{\sqrt{3}} \sin C = 2a \]
Dividing through by \( 2a \):
\[ \frac{1}{\sqrt{3}} (\sin B + \sin C) = 1 \]
Multiplying both sides by \( \sqrt{3} \):
\[ \sin B + \sin C = \sqrt{3} \]

### Step 4: Use the Angle Sum Property
We know that:
\[ B + C = 180^\circ - A = 180^\circ - 60^\circ = 120^\circ \]
Using the identity \( \sin(120^\circ - B) = \sin C \):
\[ \sin B + \sin(120^\circ - B) = \sqrt{3} \]

### Step 5: Solve for Angles
Using the sine addition formula:
\[ \sin(120^\circ - B) = \sin 120^\circ \cos B - \cos 120^\circ \sin B \]
Substituting \( \sin 120^\circ = \frac{\sqrt{3}}{2} \) and \( \cos 120^\circ = -\frac{1}{2} \):
\[ \sin B + \left(\frac{\sqrt{3}}{2} \cos B + \frac{1}{2} \sin B\right) = \sqrt{3} \]
Combining terms:
\[ \frac{3}{2} \sin B + \frac{\sqrt{3}}{2} \cos B = \sqrt{3} \]

### Step 6: Solve the Equation
Multiplying through by 2:
\[ 3 \sin B + \sqrt{3} \cos B = 2\sqrt{3} \]
This is a linear combination of sine and cosine, which can be solved using the method of coefficients or by recognizing that \( B \) and \( C \) must be equal due to the symmetry of the triangle.

### Step 7: Conclude the Type of Triangle
Since \( B + C = 120^\circ \) and \( B = C \):
\[ 2B = 120^\circ \implies B = 60^\circ \quad \text{and} \quad C = 60^\circ \]
Thus, \( \angle A = 60^\circ \), \( \angle B = 60^\circ \), and \( \angle C = 60^\circ \).

Therefore, triangle \( ABC \) is an **equilateral triangle**.

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