International Baccalaureate · DP 2: Applications and Interpretation SL

Revision: Geometry and Trigonometry

⌚ ~27 min ▢ 3 blocks ✓ 13 exercises ➣ Prerequisites : Voronoi Diagrams, Non-Right-Angled Triangle Trigonometry

I Toolbox: Arcs and Sectors in Degrees

Recappreviously studied — click to reread

Method — Finding the Length of an Arc

The length of an arc is a fraction of the full circumference.

Find the fraction of the circle. This is the arc’s central angle (in degrees) divided by $360^\circ$: $$ \text{Fraction} = \frac{\text{central angle } (\theta)}{360^\circ}. $$
Multiply the fraction by the full circumference. The circumference of a circle with radius $r$ is $C = 2 \pi r$: $$ \text{Arc Length} = \text{Fraction} \times (2 \pi r). $$

So, if the central angle is $\theta$ (in degrees) and the radius is $r$, $$ \text{Arc Length} = \frac{\theta}{360^\circ} \times 2\pi r. $$

Proposition — Area of a Sector

The area of a sector of radius $r$ and angle $\theta$ in degrees is the fraction $\dfrac{\theta}{360^\circ}$ of the area of the disc: $A = \dfrac{\theta}{360^\circ} \times \pi r^2$.

Example

Find the area of a sector of radius $6$ cm and angle $50^\circ$, exactly and to $0.1 \text{ cm}^2$.

Answer

$$\begin{aligned}[t]A &= \frac{\theta}{360^\circ} \times \pi r^2 && \text{(area of a sector)}\\ &= \frac{50^\circ}{360^\circ} \times \pi \times 6^2 && (\theta = 50^\circ,\ r = 6)\\ &= \frac{50 \times 36}{360}\,\pi && \text{(grouping the numbers)}\\ &= 5\pi \text{ cm}^2 && (1800 \div 360 = 5)\\ &\approx 15.7 \text{ cm}^2.\end{aligned}$$

Skills to practice

Finding an Angle or a Radius from an Arc or a Sector

II Toolbox: North and East Components

Recappreviously studied — click to reread

Method — Finding the North and East Components of a Journey

Unless it runs due North, East, South or West, a journey of length $d$ on a bearing $\theta$ is the hypotenuse of a right-angled triangle whose two other sides are its North–South part and its East–West part.

Draw the North–South line at the starting point, the journey, and the East–West segment from the end of the journey to the North–South line: these three segments form the right-angled triangle.
Find the acute angle $\alpha$ between the journey and the North–South line: $\alpha = \theta$ for a bearing between $000^\circ$ and $090^\circ$, $\alpha = 180^\circ - \theta$ between $090^\circ$ and $180^\circ$, $\alpha = \theta - 180^\circ$ between $180^\circ$ and $270^\circ$, $\alpha = 360^\circ - \theta$ between $270^\circ$ and $360^\circ$.
Compute the two parts in the right-angled triangle: the North–South part is $d\cos\alpha$ (the side adjacent to $\alpha$), the East–West part is $d\sin\alpha$ (the side opposite $\alpha$).
State the directions: north for a bearing below $090^\circ$ or above $270^\circ$, south for a bearing between $090^\circ$ and $270^\circ$; east for a bearing between $000^\circ$ and $180^\circ$, west for a bearing between $180^\circ$ and $360^\circ$.

A bearing of $000^\circ$, $090^\circ$, $180^\circ$ or $270^\circ$ points along the North–South or the East–West line: there is no triangle, one part is $d$ and the other is $0$.

Example

A boat sails $15$ km from a harbour $H$ on a bearing of $125^\circ$. How far south and how far east of $H$ is it (to $0.1$ km)?

Answer

The bearing lies between $090^\circ$ and $180^\circ$: the boat is south and east of $H$, and the journey makes the angle $\alpha = 180^\circ - 125^\circ = 55^\circ$ with the South line. $$\begin{aligned}[t] \text{south part} &= 15\cos 55^\circ \approx 8.6\ \text{km} && {\scriptstyle(\text{adjacent side})}\\ \text{east part} &= 15\sin 55^\circ \approx 12.3\ \text{km} && {\scriptstyle(\text{opposite side})} \end{aligned}$$ The boat is about $\mathbf{8.6}$ km south and $\mathbf{12.3}$ km east of the harbour. Check: $8.6^2 + 12.3^2 \approx 225 = 15^2$ (Pythagoras).

Skills to practice

Finding North and East Components

III Exam-Style Questions

Skills to practice

Working with Technology: Geometry and Trigonometry