International Baccalaureate · DP 1: Applications and Interpretation SL

Non-Right-Angled Triangle Trigonometry

⌚ ~98 min ▢ 10 blocks ✓ 45 exercises ➣ Prerequisites : Right-Angled Triangle Trigonometry

I The Unit Circle and the Area of a Triangle

I.1 Sine and Cosine of Any Angle

Definition — Unit circle

The unit circle is a circle with a radius of 1 centred at the origin. Travelled anticlockwise, it is called the trigonometric circle.

Proposition — Relationship between Angle and Coordinates

For an angle $\theta$, measured anticlockwise from the positive $x$-axis, $M(x,y)$ is the corresponding point on the unit circle.

The $x$-coordinate is the cosine of the angle: $\cos \theta = x$
The $y$-coordinate is the sine of the angle: $\sin \theta = y$

For an acute angle, these equalities are proved with the right-angled triangle formed by $O$, $M$ and the foot of the perpendicular from $M$ to the $x$-axis. For any other angle, these coordinates define $\cos\theta$ and $\sin\theta$.

Proof

For an acute angle $\theta$, using the trigonometric ratios in the right-angled triangle $OHM$: $$\begin{aligned} \cos \theta&=\frac{\text{adjacent}}{\text{hypotenuse}}\\ &=\frac{OH}{OM} \\ &=\frac{x}{1} \\ &=x \\ \end{aligned}$$ and $$\begin{aligned} \sin \theta&=\frac{\text{opposite}}{\text{hypotenuse}}\\ &=\frac{HM}{OM} \\ &=\frac{y}{1} \\ &=y \\ \end{aligned}$$

Skills to practice

Finding Angles from a Sine or a Cosine

I.2 The Area Formula

Recappreviously studied — click to reread

Discover

How wide is a lake you cannot walk across? Stand at a point $C$ on the shore from which both ends $A$ and $B$ are visible. You measure $CA=200\,\text{m}$, $CB=150\,\text{m}$, and the angle between them, $\Angle{ACB}=70^\circ$. There is no right angle anywhere — so Pythagoras' theorem and the ratios of the right-angled triangle do not apply — yet the distance $AB$ is fixed by what you measured.

Triangle calculation extends to any triangle: the area formula, the law of sines and the law of cosines compute a missing side or angle even when no angle is a right angle (here $AB=\sqrt{200^2+150^2-2\cdot200\cdot150\cos 70^\circ}$).

Everything here builds on two known tools: the ratios of the right-angled triangle — sine, cosine, tangent — and Pythagoras' theorem. No right angle is given, so the proofs manufacture one: each law is obtained by dropping a well-chosen altitude and applying those two tools.

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While SOH CAH TOA only applies to right-angled triangles, many real-world problems involve triangles that are not right-angled (also called oblique triangles). To solve these, we use two key results: the Law of Sines and the Law of Cosines.
Convention for labelling triangles: In any triangle $ABC$, the angles at the vertices are denoted by capital letters ($A, B, C$) and the sides opposite them are denoted by the corresponding lowercase letters ($a, b, c$).

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When two sides of a triangle and the angle between them (the included angle) are known, the area can be calculated using a formula involving the sine of this angle. This is particularly useful for non-right-angled triangles, where the height is not given directly.

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Proposition — Area Formula

The area of any triangle is half the product of two sides and the sine of their included angle: $$\text{Area} = \frac{1}{2} \times \textcolor{colordef}{\text{side}} \times \textcolor{colorprop}{\text{side}} \times \sin(\textcolor{olive}{\text{included angle}})$$ For example, if the two sides have lengths $b$ and $c$ and the included angle is $A$, then $\text{Area} = \dfrac12 bc \sin A$.

Proof

Consider triangle $ABC$ with base $a = BC$. Drop a perpendicular height $h$ from vertex $A$ to the base $a$, meeting $BC$ at point $H$.

The area of a triangle is given by the standard formula: $$ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2}ah. $$ In the right-angled triangle $AHC$, the sine ratio gives: $$ \sin(C) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{h}{b} \implies h = b\sin(C). $$ Substituting this expression for $h$ into the area formula gives: $$ \text{Area} = \frac{1}{2}a(b\sin C) = \frac{1}{2}ab\sin C. $$ (The figure places the foot of the altitude inside the triangle — the acute case; if the angle at $C$ is obtuse, the foot $H$ falls outside, beyond $C$; the right-angled triangle $AHC$ then has the angle $180^\circ - C$ at $C$, so the height is $h = b\sin(180^\circ - C) = b\sin C$: the formula is unchanged.)

The angles of a triangle may be obtuse. For an obtuse angle $\theta$, we use the values $\sin\theta = \sin(180^\circ - \theta)$ and $\cos\theta = -\cos(180^\circ - \theta)$, which a calculator applies automatically; these extended definitions come from the unit circle.

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Example

For the triangle

, find the area.

Answer

$$\begin{aligned} \text{Area} &= \frac{1}{2} \times \textcolor{colorprop}{10} \times \textcolor{colordef}{12} \times \sin(\textcolor{olive}{50^\circ})\\ & \approx 46 \, \text{cm}^2 \end{aligned}$$

Skills to practice

Finding the Area of a Triangle

II The Law of Cosines

II.1 The Law of Cosines

Recappreviously studied — click to reread

The Law of Cosines (also called Al-Kashi's theorem) relates all three sides of a triangle to one of its angles. It generalises Pythagoras' theorem and is used to solve triangles when given:

Three sides (SSS).
Two sides and their included angle (SAS).

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Proposition — Law of Cosines

For any triangle with sides $\textcolor{colordef}{a}$, $\textcolor{colorprop}{b}$, $\textcolor{olive}{c}$ and opposite angles $\textcolor{colordef}{A}$, $\textcolor{colorprop}{B}$, $\textcolor{olive}{C}$: $$\textcolor{olive}{c}^2 = \textcolor{colordef}{a}^2 + \textcolor{colorprop}{b}^2 - 2\textcolor{colordef}{a}\textcolor{colorprop}{b} \cos \textcolor{olive}{C}$$ This can be rearranged to find an angle: $$\cos \textcolor{olive}{C} = \dfrac{\textcolor{colordef}{a}^2 + \textcolor{colorprop}{b}^2 -\textcolor{olive}{c}^2}{2\textcolor{colordef}{a}\textcolor{colorprop}{b}}$$ Note that if $C = 90^\circ$, then $\cos C = 0$ and the formula reduces to $c^2 = a^2 + b^2$, Pythagoras' theorem.

Proof

Drop a perpendicular height $h$ from vertex $A$ to the side $a$ (line $\Line{BC}$), meeting it at point $D$.

In the right-angled triangle $ADC$:

By Pythagoras' theorem, $h^2 + x^2 = b^2 \implies h^2 = b^2 - x^2$.
$\cos(C) = x/b \implies x = b\cos(C)$.

In the right-angled triangle $ADB$, by Pythagoras' theorem: $$ c^2 = h^2 + (a-x)^2. $$ Substitute the expressions for $h^2$ and $x$ into this equation: $$ \begin{aligned} c^2 &= (b^2 - x^2) + (a^2 - 2ax + x^2) \\ &= a^2 + b^2 - 2ax \\ &= a^2 + b^2 - 2a(b\cos C) \\ &= a^2 + b^2 - 2ab\cos C. \end{aligned} $$ The figure places the foot $D$ on the side $\Segment{BC}$, between $B$ and $C$: this is the case when the angles $B$ and $C$ are both acute, even if the angle $A$ is obtuse. If the angle $B$ or the angle $C$ is obtuse, the foot $D$ falls outside the side $\Segment{BC}$. For $B$ obtuse, $D$ lies beyond $B$: $BD = x - a$ and $BD^2 = (a - x)^2$, so the computation above is unchanged. For $C$ obtuse, $D$ lies beyond $C$, and the right-angled triangle $ADC$ has the angle $180^\circ - C$ at $C$: $CD = b\cos(180^\circ - C) = -b\cos C$, $BD = a + CD = a - b\cos C$ and $h^2 = b^2 - CD^2 = b^2 - b^2\cos^2 C$. Then $$\begin{aligned}[t] c^2 &= h^2 + BD^2 && {\scriptstyle(\text{Pythagoras' theorem in } ADB)}\\ &= b^2 - b^2\cos^2 C + a^2 - 2ab\cos C + b^2\cos^2 C\\ &= a^2 + b^2 - 2ab\cos C, \end{aligned}$$ the same formula.

Example

For the triangle

, find side $c$.

Answer

Using the Law of Cosines: $$\begin{aligned} \textcolor{olive}{c}^2 &= \textcolor{colordef}{8}^2 + \textcolor{colorprop}{7}^2 - 2 \times\textcolor{colordef}{8} \times\textcolor{colorprop}{7} \times \cos \textcolor{olive}{50^\circ}\\ \textcolor{olive}{c} &= \sqrt{\textcolor{colordef}{8}^2 + \textcolor{colorprop}{7}^2 - 2 \times\textcolor{colordef}{8} \times\textcolor{colorprop}{7} \times \cos \textcolor{olive}{50^\circ}}\\ \textcolor{olive}{c} &\approx 6.4 \, \text{cm}\\ \end{aligned}$$

Skills to practice

Finding Sides and Angles with the Law of Cosines

III The Law of Sines

III.1 The Law of Sines

Recappreviously studied — click to reread

The Law of Sines gives a relationship between the sides of a triangle and the sines of their opposite angles. It solves a triangle as soon as at least one side and its opposite angle are both known, for example:

Two angles and any side (AAS or ASA).
Two sides and a non-included angle (SSA), which can lead to the ambiguous case (0, 1, or 2 possible triangles).

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Proposition — Law of Sines

Proof

The area of the triangle can be expressed in three different ways using the area formula: $$ \text{Area} = \frac{1}{2} bc \sin A = \frac{1}{2} ac \sin B = \frac{1}{2} ab \sin C. $$ Dividing each part of the equality by $\frac{1}{2}abc$ gives: $$ \frac{bc \sin A}{abc} = \frac{ac \sin B}{abc}= \frac{ab \sin C}{abc}. $$ So $$ \frac{\sin A}{a} = \frac{\sin B}{b}= \frac{\sin C}{c}. $$ Taking the reciprocal of each term gives the Law of Sines: $$ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}. $$

Example

For the triangle

, find side $a$.

Answer

First, find angle $A$: $A = 180^\circ - 40^\circ - 60^\circ = 80^\circ$.
Using the Law of Sines with the pair $(a, A)$ and the pair $(c, C)$: $$\begin{aligned} \textcolor{colordef}{\frac{a}{\sin 80^\circ}} &= \textcolor{olive}{\frac{5}{\sin 60^\circ} }\\ a &= 5 \cdot \frac{\sin 80^\circ}{\sin 60^\circ}\\ a &\approx 5.69 \, \text{cm} \end{aligned}$$

Skills to practice

Finding Sides and Angles with the Law of Sines

IV Solving Problems with the Laws of Sines and Cosines

IV.1 Solving Real-World Problems

Recappreviously studied — click to reread

The Laws of Sines and Cosines are powerful tools for solving a wide range of problems involving non-right-angled triangles, especially in real-world contexts. To solve these problems effectively, follow the structured steps below.

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Method — Solving Real-World Problems Using Sine and Cosine Laws

Draw a clear diagram representing the situation described in the problem.
Label the known and unknown sides and angles in the triangle.
Identify the configuration (for example, two sides and the included angle for the Law of Cosines, or two angles and a side for the Law of Sines).
Choose the appropriate law (Law of Sines for ratios of sides to sines of angles, Law of Cosines for relating sides and the cosine of an angle).
Write and solve the equation, then check that the answer is reasonable (size, units, and shape of the triangle).

Skills to practice

Solving Contextual Triangle Problems
Choosing and Justifying the Law

IV.2 Solving Bearing Problems

Recappreviously studied — click to reread

Discover

A ship leaves port $P$ on a bearing of $070^\circ$ and sails $40\,\text{km}$, then turns onto a bearing of $150^\circ$ and sails $30\,\text{km}$ to $R$. How far is the ship from port, and on what bearing must it steer to return? You cannot stretch a line across open water — but the two legs and the turn between them form a triangle, and a triangle is something you can calculate.

Bearings (three figures, clockwise from North) turn a navigation route into a triangle, which the sine and cosine laws solve — finding the ship's position without ever measuring across the sea.

A bearing problem is a triangle problem in disguise: the legs of the journey and the turn between them form a triangle, and the law of sines and the law of cosines solve it.

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Method — Solving a Bearing Problem

Sketch a Diagram: This is the most important step. Draw a clear diagram with a North line at every point involved. Label all given distances and bearings.
Find an Interior Angle: Use parallel line properties (for example, alternate interior or co-interior angles) to find at least one angle inside the triangle formed by the journey.
Apply the Law of Sines or the Law of Cosines: Use the Law of Cosines when two sides and the included angle (SAS) or all three sides (SSS) are known. Use the Law of Sines when two angles and a side (AAS) or two sides and a non-included angle (SSA) are known.
Calculate and Conclude: Solve for the required distance or bearing. Check that every bearing in the final answer is a three-figure value between $000^\circ$ and $360^\circ$, measured clockwise from North.

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

Calculating North and East Components
Calculating Distances from Bearings

V Going further

Skills to practice

Reasoning and Problem-Solving
Solving Exam-Style Problems
Quiz