A circle with center O and radius 14cm. If the arc of the circle is 15cm, subtends to the center O with AB the chord of the circle, determine: a) the angle subtended to the circle, b) area of the minor sector, c) area of the minor segment, d) area of the major segment.

The formula for arc length is $s = r\theta$, where $\theta$ is in radians.

$$15 = 14\theta$$ $$\theta = \frac{15}{14} \text{ radians}$$

$$\theta_{degrees} = \frac{15}{14} \times \frac{180^\circ}{\pi}$$ $$\theta_{degrees} = \frac{2700}{14\pi}$$ $$\theta_{degrees} \approx \frac{2700}{14 \times 3.14159}$$ $$\theta_{degrees} \approx \frac{2700}{43.98226}$$ $$\theta_{degrees} \approx 61.388^\circ$$

Rounding to one decimal place:

$$\theta \approx 61.4^\circ$$

The angle subtended at the center is $\boxed{61.4^\circ}$.

The area of a sector $A_{sector}$ is given by $A_{sector} = \frac{1}{2}r^2\theta$, where $\theta$ is in radians.

$$A_{sector} = \frac{1}{2} (14)^2 \left(\frac{15}{14}\right)$$ $$A_{sector} = \frac{1}{2} \times 196 \times \frac{15}{14}$$ $$A_{sector} = 98 \times \frac{15}{14}$$ $$A_{sector} = 7 \times 15$$ $$A_{sector} = 105 \text{ cm}^2$$

The area of the minor sector is $\boxed{105.0 \text{ cm}^2}$.

The area of the triangle $A_{triangle}$ is given by $A_{triangle} = \frac{1}{2} r^2 \sin \theta$, where $\theta$ is in radians.

$$A_{triangle} = \frac{1}{2} (14)^2 \sin\left(\frac{15}{14}\right)$$ $$A_{triangle} = \frac{1}{2} \times 196 \times \sin(1.0714 \text{ rad})$$ $$A_{triangle} = 98 \times 0.878006$$ $$A_{triangle} \approx 86.0446 \text{ cm}^2$$

The area of the minor segment $A_{segment}$ is $A_{segment} = A_{sector} - A_{triangle}$.

$$A_{segment} = 105 - 86.0446$$ $$A_{segment} \approx 18.9554 \text{ cm}^2$$

Rounding to one decimal place:

$$A_{segment} \approx 19.0 \text{ cm}^2$$

The area of the minor segment is $\boxed{19.0 \text{ cm}^2}$.

The area of the circle $A_{circle}$ is given by $A_{circle} = \pi r^2$.

$$A_{circle} = \pi (14)^2$$ $$A_{circle} = 196\pi$$ $$A_{circle} \approx 196 \times 3.14159$$ $$A_{circle} \approx 615.752 \text{ cm}^2$$

The area of the major segment $A_{major\_segment}$ is $A_{major\_segment} = A_{circle} - A_{minor\_segment}$.

$$A_{major\_segment} \approx 615.752 - 18.9554$$ $$A_{major\_segment} \approx 596.7966 \text{ cm}^2$$

Rounding to one decimal place:

$$A_{major\_segment} \approx 596.8 \text{ cm}^2$$

The area of the major segment is $\boxed{596.8 \text{ cm}^2}$.

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