Długość łuku obliczamy ze wzoru:
$x=\frac{\alpha }{{360}^o}\cdot 2\pi r$  

Pole wycinka koła obliczamy ze wzoru:
$P_w=\frac{\alpha }{{360}^o}\cdot \pi r^2$  

$a)x=\frac{{90}^o}{{360}^o}\cdot 2\pi \cdot 12=\frac{1}{4}\cdot 2\pi \cdot 12=\frac{24}{4}\pi =6\pi$  

$P_w=\frac{{90}^o}{{360}^o}\cdot \pi \cdot {12}^2=\frac{1}{4}\cdot \pi \cdot 144=\frac{144}{4}\pi =36\pi$   
$\underline{\underline{}}$  

$b)x=\frac{{60}^o}{{360}^o}\cdot 2\pi \cdot 15=\frac{1}{6}\cdot 2\pi \cdot 15=\frac{30}{6}\pi =5\pi$  

$P_w=\frac{{60}^o}{{360}^o}\cdot \pi \cdot {15}^2=\frac{1}{6}\cdot \pi \cdot 225=\frac{225}{6}\pi =37,5\pi$   
$\underline{\underline{}}$  

$c)x=\frac{{135}^o}{{360}^o}\cdot 2\pi \cdot 20=\frac{3}{8}\cdot 2\pi \cdot 20=\frac{120}{8}\pi =15\pi$   

$P_w=\frac{{135}^o}{{360}^o}\cdot \pi \cdot {20}^2=\frac{3}{8}\cdot \pi \cdot 400=\frac{1200}{8}\pi =150\pi$   
$\underline{\underline{}}$  

$d)x=\frac{{360}^o-{120}^o}{{360}^o}\cdot 2\pi \cdot 12=\frac{{240}^o}{{360}^o}\cdot 2\pi \cdot 12=\frac{2}{3}\cdot 2\pi \cdot 12=\frac{48}{3}\pi =16\pi$    

$P_w=\frac{{360}^o-{120}^o}{{360}^o}\cdot \pi \cdot {12}^2=\frac{{240}^o}{{360}^o}\cdot \pi \cdot 144=\frac{2}{3}\cdot \pi \cdot 144=\frac{288}{3}\pi =96\pi$   
$\underline{\underline{}}$  

$e)x=\frac{{360}^o-{200}^o}{{360}^o}\cdot 2\pi \cdot 15=\frac{{160}^o}{{360}^o}\cdot 2\pi \cdot 15=\frac{4}{9}\cdot 2\pi \cdot 15=\frac{120}{9}\pi =13\frac{1}{3}\pi$  

$P_w=\frac{{360}^o-{200}^o}{{360}^o}\cdot \pi \cdot {15}^2=\frac{{160}^o}{{360}^o}\cdot \pi \cdot 225=\frac{4}{9}\cdot \pi \cdot 225=\frac{900}{9}\pi =100\pi$