Question: A sector has radius $8$ cm and angle $1.1$ radians at the centre. Find the arc length and the area of the sector.

Step 1. The angle is in radians, so use the radian formulas directly — no $\frac{\theta}{360}$ fraction needed:

$$s = r\theta \qquad \text{and} \qquad A = \dfrac{1}{2}r^2\theta$$

Step 2. Arc length:

$$s = 8 \times 1.1 = 8.8 \text{ cm (exact)}$$

Step 3. Sector area:

$$A = \dfrac{1}{2}(8)^2(1.1) = \dfrac{1}{2}(64)(1.1) = 35.2 \text{ cm}^2 \text{ (exact)}$$

Step 4. If the perimeter of the sector is wanted, add the two radii:

$$\text{Perimeter} = 8.8 + 8 + 8 = 24.8 \text{ cm}$$

Check. $1.1$ rad is about $63°$, roughly $\frac{1}{6}$ of a circle. Full area $\pi(8)^2 \approx 201$, and $\frac{1}{6}$ of that $\approx 34$ — close to $35.2$. ✓

⚠ Watch out: These short formulas are radians-only. If the angle arrives in degrees, either convert it first or use the $\frac{\theta}{360°}$ versions — writing $\frac{1}{2}r^2(70)$ for a $70°$ sector is the classic error. Set the calculator to radian mode whenever the angle is in radians.

Question: $AOB$ is a sector of a circle, centre $O$ and radius $8$ cm, with $\angle AOB = 0.75$ radians. Find the length of the arc $AB$ and the area of the shaded segment (the region between the chord $AB$ and the arc).

[diagram in original sheet: sector AOB with the chord AB drawn, segment shaded]

Step 1. Arc length with the radian formula:

$$s = r\theta = 8(0.75) = 6 \text{ cm}$$

Step 2. See the segment as a subtraction:

$$\text{Segment} = \text{sector } AOB - \text{triangle } AOB$$

Step 3. Each piece, using $\theta$ in radians:

$$\text{Sector} = \dfrac{1}{2}r^2\theta = \dfrac{1}{2}(8)^2(0.75) = 24 \text{ cm}^2$$ $$\text{Triangle} = \dfrac{1}{2}r^2\sin\theta = \dfrac{1}{2}(8)^2\sin 0.75 = 21.812\ldots \text{ cm}^2$$

Step 4. Subtract:

$$\text{Segment} = 24 - 21.812\ldots = 2.187\ldots \approx 2.19 \text{ cm}^2 \quad (3 \text{ s.f.})$$

Check. $0.75$ rad is a fairly slim angle, so the chord hugs the arc and the segment should be small compared with the $24$ cm$^2$ sector. ✓

⚠ Watch out: The calculator must be in radian mode for $\sin 0.75$ — in degree mode you get $\sin 0.75° = 0.0131$, and the "segment" comes out absurdly large. The triangle term is $\frac{1}{2}r^2\sin\theta$, not $\frac{1}{2} \times \text{chord} \times r$.

Question: $TA$ and $TB$ are tangents from $T$ to a circle, centre $O$, radius $6$ cm, each of length $10$ cm. Find $\angle AOB$, then the perimeter and area of the shaded region between the tangents and the minor arc $AB$.

[diagram in original sheet: kite OATB — two radii, two tangents, minor arc AB shaded region outside the circle]

Step 1. A tangent meets its radius at $90°$, so $\triangle OAT$ is right-angled at $A$:

$$\tan\angle AOT = \dfrac{10}{6} \;\Rightarrow\; \angle AOT = 59.036\ldots°$$

Step 2. By symmetry of the kite (equal tangents from an external point):

$$\angle AOB = 2 \times 59.036\ldots° = 118.07\ldots° \approx 118.1° \quad (1 \text{ d.p.})$$

Step 3. Perimeter = arc $AB$ + the two tangents:

$$\text{Arc } AB = \dfrac{118.07\ldots}{360} \times 2\pi(6) = 12.364\ldots$$ $$\text{Perimeter} = 12.364\ldots + 10 + 10 \approx 32.4 \text{ cm} \quad (3 \text{ s.f.})$$

Step 4. Area = kite − sector = two triangles − sector:

$$2 \times \dfrac{1}{2}(6)(10) - \dfrac{118.07\ldots}{360} \times \pi(6)^2 = 60 - 37.093\ldots \approx 22.9 \text{ cm}^2$$

⚠ Watch out: Carry the unrounded $118.07\ldots°$ into the arc and sector — using the rounded $118.1°$ shifts the last digit. In $\tan\angle AOT$, the tangent length $10$ is opposite the centre angle: $\frac{10}{6}$, not $\frac{6}{10}$.

Question: A sector has radius $10$ cm and its arc is $22$ cm long. Find the angle at the centre in radians, and convert it to degrees.

Step 1. With the angle in radians, arc length has the short formula:

$$s = r\theta$$

Step 2. Substitute and solve for $\theta$:

$$22 = 10\,\theta \;\Rightarrow\; \theta = \dfrac{22}{10} = 2.2 \text{ rad}$$

Step 3. Convert using $\pi$ rad $= 180°$, so $1$ rad $= \dfrac{180°}{\pi}$:

$$\theta = 2.2 \times \dfrac{180°}{\pi} = 126.05\ldots° \approx 126.1° \quad (1 \text{ d.p.})$$

Check. $2.2$ rad is a bit more than $2$ rad $\approx 114.6°$, and a bit less than $\frac{3\pi}{4} \approx 2.36$ rad $= 135°$ — so $126.1°$ sits sensibly between. ✓

Going the other way, multiply by the reciprocal: to turn degrees into radians, use $1° = \dfrac{\pi}{180}$ rad.

⚠ Watch out: $s = r\theta$ (and $\frac{1}{2}r^2\theta$) only work with $\theta$ in radians — plugging in $126.1$ would be wrong by a factor of nearly $60$. Keep the radian value exact ($2.2$) for any later working; convert to degrees only when the question asks.