There are two regular polygons with number of sides equal to `(n-1) and (n+ 2)`. Their exterior angles differ by `6^@`. The value of n is

Options

A. 14
B. 12
C. 13
D. 11

Answer

To solve the problem, we need to find the value of \( n \) given that the exterior angles of two regular polygons with \( n-1 \) and \( n+2 \) sides differ by \( 6^\circ \).

### Step-by-Step Solution:

1. **Understand the formula for exterior angles**:
The exterior angle of a regular polygon with \( s \) sides is given by the formula:
\[ \text{Exterior Angle} = \frac{360^\circ}{s} \]
Therefore, for our two polygons:
- For the polygon with \( n-1 \) sides:
\[ \text{Exterior Angle}_1 = \frac{360^\circ}{n-1} \]
- For the polygon with \( n+2 \) sides:
\[ \text{Exterior Angle}_2 = \frac{360^\circ}{n+2} \]

2. **Set up the equation based on the difference of exterior angles**:
According to the problem, the difference between the two exterior angles is \( 6^\circ \):
\[ \left| \frac{360^\circ}{n-1} - \frac{360^\circ}{n+2} \right| = 6^\circ \]
We can simplify this to:
\[ \frac{360^\circ}{n-1} - \frac{360^\circ}{n+2} = 6^\circ \]

3. **Eliminate the degrees**:
Divide the entire equation by \( 360^\circ \):
\[ \frac{1}{n-1} - \frac{1}{n+2} = \frac{1}{60} \]

4. **Combine the fractions**:
The left-hand side can be combined as follows:
\[ \frac{(n+2) - (n-1)}{(n-1)(n+2)} = \frac{1}{60} \]
Simplifying the numerator:
\[ \frac{n + 2 - n + 1}{(n-1)(n+2)} = \frac{3}{(n-1)(n+2)} = \frac{1}{60} \]

5. **Cross-multiply to eliminate the fraction**:
\[ 3 \cdot 60 = (n-1)(n+2) \]
Simplifying gives:
\[ 180 = (n-1)(n+2) \]

6. **Expand the right-hand side**:
\[ 180 = n^2 + 2n - n - 2 \]
This simplifies to:
\[ 180 = n^2 + n - 2 \]

7. **Rearranging to form a quadratic equation**:
\[ n^2 + n - 182 = 0 \]

8. **Factor the quadratic equation**:
We need to factor \( n^2 + n - 182 \). We look for two numbers that multiply to \(-182\) and add to \(1\). The factors are \(14\) and \(-13\):
\[ (n + 14)(n - 13) = 0 \]

9. **Solve for \( n \)**:
Setting each factor to zero gives:
\[ n + 14 = 0 \quad \Rightarrow \quad n = -14 \quad (\text{not valid since } n \text{ must be positive}) \]
\[ n - 13 = 0 \quad \Rightarrow \quad n = 13 \]

Thus, the value of \( n \) is \( \boxed{13} \).