If `cos B=(sinA)/(2sinC)`, find the nature of triangle :

Options

A. isosceles
B. scalene
C. can't be determined
D. none of these

Answer

To determine the nature of the triangle given the relationship \( \cos B = \frac{\sin A}{2 \sin C} \), we can follow these steps:

### Step 1: Use the Sine Rule
We know from the Sine Rule that:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k \]
where \( k \) is a constant.

### Step 2: Express \( \sin A \) and \( \sin C \)
From the Sine Rule, we can express \( \sin A \) and \( \sin C \) as:
\[ \sin A = \frac{a}{k} \quad \text{and} \quad \sin C = \frac{c}{k} \]

### Step 3: Substitute into the Cosine Formula
We know that:
\[ \cos B = \frac{a^2 + c^2 - b^2}{2ac} \]
Now, substituting the values of \( \sin A \) and \( \sin C \) into the given equation \( \cos B = \frac{\sin A}{2 \sin C} \):
\[ \cos B = \frac{\frac{a}{k}}{2 \cdot \frac{c}{k}} = \frac{a}{2c} \]

### Step 4: Set the Two Expressions for \( \cos B \) Equal
Now we have two expressions for \( \cos B \):
\[ \frac{a^2 + c^2 - b^2}{2ac} = \frac{a}{2c} \]

### Step 5: Cross Multiply to Eliminate the Denominator
Cross multiplying gives:
\[ (a^2 + c^2 - b^2) \cdot 2c = a \cdot 2ac \]
This simplifies to:
\[ 2c(a^2 + c^2 - b^2) = 2a^2c \]

### Step 6: Simplify the Equation
Dividing both sides by \( 2c \) (assuming \( c \neq 0 \)):
\[ a^2 + c^2 - b^2 = a^2 \]
This simplifies to:
\[ c^2 - b^2 = 0 \]

### Step 7: Factor the Result
Factoring gives:
\[ (c - b)(c + b) = 0 \]
This implies either:
1. \( c - b = 0 \) which means \( c = b \)
2. \( c + b = 0 \) which is not possible since lengths cannot be negative.

### Step 8: Conclusion
Since \( c = b \), we conclude that the triangle has at least two equal sides. Therefore, the triangle is an **isosceles triangle**.

### Final Answer
The nature of the triangle is **isosceles**.

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