The size of an interior angle of a n-sided polygon is 14 times of its exterior angle.Find the value of n

The interior angle is given by : (n-2) (180)/ n

And the exterior angle is given by :  360/n

And we have that

(n - 2) (180) / n  =  14* 360/ n      [multiply both sides by n  and simplify ]

180n -360  =  5040    add 360 to both sides

180n = 5400     divide both sides by 180

n = 30

The size of an interior angle of a n-sided polygon is 14 times of its exterior angle.Find the value of n

HOW TO SOLVE?

The Exterior Angles of a Polygon add up to $360^{\circ} \qquad sum_e = 360^{\circ}$
Sum of Interior Angles $= (n-2) × 180^{\circ}\qquad sum_i = (n-2) \cdot 180^{\circ}$

$\begin{array}{rcll} sum_i &=& 14 \cdot sum_e \\ (n-2) \cdot 180^{\circ} &=& 14 \cdot 360^{\circ} \quad & | \quad :180 \\ n-2 &=& 14 \cdot \frac{ 360^{\circ} } { 180^{\circ} } \\ n-2 &=& 14 \cdot 2 \\ n-2 &=& 28 \quad & | \quad +2 \\ \mathbf{ n } & \mathbf{=} & \mathbf{30} \\ \end{array}$