Solution

(a) Each interior angle \(=156^\circ\)

\[ \begin{aligned} \text{Let the polygon have } & n \text{ sides}\\[4pt] \text{Each interior angle} &= 156^\circ\\[4pt] \frac{(n-2)\times 180^\circ}{n} &= 156^\circ \\[4pt] 180n - 360 &= 156n \\[4pt] 180n - 156n &= 360 \\[4pt] 24n &= 360 \\[6pt] n &= \frac{360}{24} \\[6pt] &= 15 \end{aligned} \] So the regular polygon has \(\color{green}{15}\) sides.

(b) Each interior angle \(=108^\circ\)

\[ \begin{aligned} \text{Let the polygon have } & n \text{ sides}\\[4pt] \text{Each interior angle} &= 108^\circ\\[4pt] \frac{(n-2)\times 180^\circ}{n} &= 108^\circ \\[4pt] 180n - 360 &= 108n \\[4pt] 180n - 108n &= 360 \\[4pt] 72n &= 360 \\[6pt] n &= \frac{360}{72} \\[6pt] &= 5 \end{aligned} \] So the regular polygon has \(\color{green}{5}\) sides (a regular pentagon).

Answer (a) \(\color{red}{15\text{ sides }}\); (b) \(\color{red}{5\text{ sides}}\)

Solution

\[ \begin{aligned} x &= \frac{4}{3}y \\[6pt] x &= \frac{4}{3} \times \frac{3}{8}z \\[6pt] \implies x &= \frac{1}{2}z \\[6pt] \end{aligned} \]\[ \begin{aligned} AB \parallel CD & \text{ and } BC \text{ is a transversal} \\[4pt] \angle ABC + \angle BCD &= 180^\circ \ \color{magenta}\text{(co-interior angles are supplementary)} \\ x + y + z &= 180^\circ \end{aligned} \] \[ \begin{aligned} \frac{1}{2}z + \frac{3}{8}z + z &= 180^\circ \\[4pt] \frac{4z + 3z + 8z}{8} &= 180^\circ \\[4pt] \frac{15z}{8} &= 180^\circ \\[6pt] z &= \frac{8 \times \cancel{180}^{12}}{\cancel{15}_1} \\[6pt] z &= 12 \times 8 \\[6pt] z &= 96^\circ \\[6pt] x& = \frac{1}{2}z \\[6pt] x&= \frac{1}{2}\times 96^\circ \\[6pt] x &= 48^\circ \end{aligned} \]

Answer \(x = \color{red}{48^\circ}\)