16. There are two regular polygons with the no of sides equal to $$ (n-1) $$ i $$ (n+2) $$. Theer extertor angles differ by $$ 6 \% $$. Find the ne of sides of the two polygons

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We are given two regular polygons with numbers of sides $$ n-1 $$ and $$ n+2 $$. Their exterior angles are

$$
\frac{360}{n-1} \quad \text{and} \quad \frac{360}{n+2},
$$

respectively. Since the polygon with fewer sides has a larger exterior angle, a 6% difference means that

$$
\frac{360}{n-1} - \frac{360}{n+2} = 0.06 \cdot \frac{360}{n+2}.
$$

Dividing both sides of the equation by 360 we obtain

$$
\frac{1}{n-1} - \frac{1}{n+2} = 0.06 \cdot \frac{1}{n+2}.
$$

Now, add $$\frac{1}{n+2}$$ to both sides:

$$
\frac{1}{n-1} = \left(0.06 + 1\right) \cdot \frac{1}{n+2} = 1.06 \cdot \frac{1}{n+2}.
$$

Multiplying both sides by $$(n-1)(n+2)$$ gives

$$
n+2 = 1.06 \, (n-1).
$$

Expanding the right side,

$$
n+2 = 1.06n - 1.06.
$$

Subtract $$ n $$ from both sides,

$$
2 = 0.06n - 1.06.
$$

Add $$ 1.06 $$ to both sides:

$$
3.06 = 0.06n.
$$

Dividing both sides by $$ 0.06 $$:

$$
n = \frac{3.06}{0.06} = 51.
$$

Thus, the two polygons have

$$
n-1 = 50 \quad \text{and} \quad n+2 = 53 \text{ sides}.
$$

To verify, the exterior angles are

$$
\frac{360}{50} = 7.2^\circ \quad \text{and} \quad \frac{360}{53} \approx 6.7925^\circ.
$$

Their difference is

$$
7.2^\circ - 6.7925^\circ \approx 0.4075^\circ,
$$

and this is $$ 6\% $$ of $$ 6.7925^\circ $$:

$$
0.06 \times 6.7925^\circ \approx 0.4075^\circ.
$$

Thus, the answer is confirmed.
The two polygons have 50 and 53 sides.

There are two regular polygons with the no of sides
equal to i . Theer extertor angles differ
by . Find the ne of sides of the two polygons

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There are two regular polygons with the no of sides equal to i . Theer extertor angles differ by . Find the ne of sides of the two polygons