International Baccalaureate · MYP 5

Non-Right-Angled Trigonometry

⌚ ~138 min ▢ 15 blocks ✓ 81 exercises ➣ Prerequisites : Right-Angled Trigonometry and Bearings, Quadratic Equations

I Finding the Sine and Cosine of an Obtuse Angle

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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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 $$\begin{aligned} \sin \theta&=\frac{\text{opposite}}{\text{hypotenuse}}\\ &=\frac{HM}{OM} \\ &=\frac{y}{1} \\ &=y \\ \end{aligned}$$

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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Proposition — Sine and Cosine of Supplementary Angles

For every angle $\theta$ with $0^\circ < \theta < 180^\circ$:

$\sin(180^\circ - \theta) = \sin\theta$
$\cos(180^\circ - \theta) = -\cos\theta$

On the unit circle, call $M$ the point of the angle $\theta$ and $M'$ the point of the angle $180^\circ - \theta$. The radius $OM'$ makes the angle $180^\circ - (180^\circ - \theta) = \theta$ with the negative $x$-axis, just as the radius $OM$ makes the angle $\theta$ with the positive $x$-axis: the reflection in the vertical axis, which swaps these two half-axes, sends $M$ to $M'$. So $M$ and $M'$ are symmetric about the vertical axis: they have the same $y$-coordinate and opposite $x$-coordinates. The coordinates of the point of an angle are its cosine and its sine, which gives the two equalities.

Skills to practice

Reading Cosine and Sine at an Angle in Degrees
Recognising Supplementary Angles
Using Supplementary Angles

II Calculating the Area of a Triangle

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.)

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
Finding Exact Areas

III Using the Law of Sines

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 $(c, C)$: $$\begin{aligned} \textcolor{colordef}{\frac{a}{\sin 80^\circ}} &= \textcolor{olive}{\frac{5}{\sin 60^\circ} }\\ a &= 5 \cdot \frac{\sin 80^\circ}{\sin 60^\circ}\\ a &\approx 5.69 \, \text{cm} \end{aligned}$$

When the Law of Sines is used with two sides and a non-included angle (SSA), the data may not pin the triangle down: swinging the known opposite side around its vertex can close the triangle in two ways, one way, or not at all. This is the ambiguous case.

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Proposition — The Ambiguous Case (SSA)

Suppose the angle $A$ and the sides $a$ (opposite $A$) and $b$ are known. The Law of Sines gives$$\sin B = \dfrac{b \sin A}{a}.$$

If $\dfrac{b \sin A}{a} > 1$: no angle $B$ exists — no triangle.
If $\dfrac{b \sin A}{a} = 1$: $B = 90^\circ$ — exactly one (right-angled) triangle when $A$ is acute; if $A$ is right or obtuse, $A + 90^\circ \geq 180^\circ$ and there is no triangle.
If $\dfrac{b \sin A}{a} < 1$: two candidate angles, $B_1 = \sin^{-1}\!\left(\dfrac{b\sin A}{a}\right)$ and $B_2 = 180^\circ - B_1$. Each candidate with $A + B < 180^\circ$ yields a triangle — zero, one or two triangles (if $A$ is obtuse, both candidates can fail the test).

\small The side $a$ swings around $C$: the circle of radius $a$ can cut the ray from $A$ twice ($B_1$, $B_2$), once, or not at all.

Proof

A triangle gives an admissible angle. Suppose that a triangle with the angle $A$ and the sides $a$ (opposite $A$) and $b$ exists, and call $B$ its angle opposite $b$. The Law of Sines gives $\dfrac{a}{\sin A} = \dfrac{b}{\sin B}$, that is $\sin B = \dfrac{b \sin A}{a}$, and the angle sum of the triangle gives $A + B < 180^\circ$. An admissible angle gives a triangle. Conversely, let $B$ be an angle with $0^\circ < B < 180^\circ$, $\sin B = \dfrac{b \sin A}{a}$ and $A + B < 180^\circ$. Then $C = 180^\circ - A - B$ is a positive angle, and there is a triangle, unique up to congruence, with a side $\Segment{BC}$ of length $a$ and the angles $B$ and $C$ at its ends (the case “a side and the two angles at its ends”); its third angle is $180^\circ - B - C = A$. By the Law of Sines, its side opposite $B$ measures $$\begin{aligned}[t] CA &= \frac{a \sin B}{\sin A} && \text{(Law of Sines)}\\ &= \frac{a}{\sin A} \times \frac{b \sin A}{a} && \text{(value of } \sin B\text{)}\\ &= b. \end{aligned}$$ So the triangles with the data $A$, $a$, $b$ correspond exactly to the angles $B$ with $\sin B = \dfrac{b \sin A}{a}$ and $A + B < 180^\circ$: one triangle for each such angle. It remains to count these angles, with the number $s = \dfrac{b \sin A}{a} > 0$.

If $s > 1$: the sine of an angle is at most $1$, so no angle $B$ exists — no triangle.
If $s = 1$: on the unit circle, the only point of height $1$ is the point for $90^\circ$, so $B = 90^\circ$ is the only candidate. It passes the test $A + 90^\circ < 180^\circ$ exactly when $A$ is acute: then exactly one triangle, right-angled at $B$; otherwise none.
If $s < 1$: the horizontal line of height $s$ cuts the upper half of the unit circle at two points, symmetric about the vertical axis, so the angles between $0^\circ$ and $180^\circ$ whose sine is $s$ are the acute angle $B_1 = \sin^{-1}(s)$ and its supplement $B_2 = 180^\circ - B_1$ (indeed $\sin(180^\circ - B_1) = \sin B_1$). Each of them gives a triangle exactly when it passes the test $A + B < 180^\circ$: zero, one or two triangles.

Method — Solving an SSA Triangle

Compute $\sin B = \dfrac{b\sin A}{a}$. If this exceeds $1$, stop: no triangle.
Find both candidates: $B_1 = \sin^{-1}\!\left(\dfrac{b\sin A}{a}\right)$ and $B_2 = 180^\circ - B_1$. If $\sin B = 1$, the two candidates coincide ($B_1 = B_2 = 90^\circ$): there is a single angle to test.
Test each candidate: keep $B$ whenever $A + B < 180^\circ$; complete each kept triangle ($C = 180^\circ - A - B$, then the Law of Sines for $c$).

Example

In triangle $ABC$, $A = 30^\circ$, $a = 6$ cm and $b = 10$ cm. Find all possible triangles (give the angles to $0.1^\circ$ and $c$ to $0.1$ cm).

Answer

$$\begin{aligned}[t] \sin B &= \dfrac{b \sin A}{a} = \dfrac{10 \sin 30^\circ}{6} = \dfrac{5}{6} && \text{(Law of Sines; } \tfrac{5}{6} < 1\text{: solutions exist)} \\ B_1 &\approx 56.4^\circ, \qquad B_2 = 180^\circ - B_1 \approx 123.6^\circ && \text{(the two candidates)} \end{aligned}$$ Both candidates are admissible: $30^\circ + 56.4^\circ < 180^\circ$ and $30^\circ + 123.6^\circ < 180^\circ$. Two triangles exist.
First triangle ($B_1 \approx 56.4^\circ$):$$C_1 = 180^\circ - 30^\circ - 56.4^\circ \approx 93.6^\circ, \qquad c_1 = \dfrac{a \sin C_1}{\sin A} \approx \dfrac{6 \times 0.998}{0.5} \approx 12.0 \text{ cm.}$$ Second triangle ($B_2 \approx 123.6^\circ$):$$C_2 = 180^\circ - 30^\circ - 123.6^\circ \approx 26.4^\circ, \qquad c_2 = \dfrac{a \sin C_2}{\sin A} \approx \dfrac{6 \times 0.445}{0.5} \approx 5.3 \text{ cm.}$$

Skills to practice

Finding Sides and Angles with the Law of Sines
Using the Law of Sines with Exact Values
Counting the Triangles in the Ambiguous Case

IV Using the Law of Cosines

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
Using the Law of Cosines with Exact Values

V Choosing the Law

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

VI Solving Bearing Problems

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.

Example

A ship leaves port $P$ on a bearing of $070^\circ$ and sails $40$ km to a point $Q$, then turns onto a bearing of $150^\circ$ and sails $30$ km to a point $R$.

Find the interior angle $\Angle{PQR}$ of the triangle $PQR$.
Find the distance $PR$ (to $0.1$ km).
Find the bearing of $R$ from $P$, then the bearing on which the ship must steer to return straight to $P$ (to the nearest degree).

Answer

The North lines at $P$ and $Q$ are parallel. The angle between $\Segment{QP}$ and the South line at $Q$ equals the angle of $70^\circ$ between the North line at $P$ and $\Segment{PQ}$ (alternate angles). The angle between the South line at $Q$ and $\Segment{QR}$ is $180^\circ - 150^\circ = 30^\circ$. Hence $\Angle{PQR} = 70^\circ + 30^\circ = 100^\circ$.
Two sides and the included angle are known (SAS): $$\begin{aligned}[t] PR^2 &= PQ^2 + QR^2 - 2 \times PQ \times QR \times \cos \Angle{PQR} && {\scriptstyle(\text{law of cosines})}\\ &= 40^2 + 30^2 - 2 \times 40 \times 30 \times \cos 100^\circ\\ &\approx 2\,916.8, \end{aligned}$$ so $PR \approx \mathbf{54.0}$ km. Since $\cos 100^\circ < 0$, the obtuse angle makes $PR$ longer than $\sqrt{40^2 + 30^2} = 50$ km.
With the unrounded $PR$: $$\begin{aligned}[t] \sin \Angle{QPR} &= \frac{QR \sin \Angle{PQR}}{PR} = \frac{30 \sin 100^\circ}{PR} && {\scriptstyle(\text{law of sines})}\\ &\approx 0.547, \end{aligned}$$ so $\Angle{QPR} \approx 33.2^\circ$. The angle is acute: it faces $QR = 30$ km, and the longest side $\Segment{PR}$ faces the obtuse angle. The ship turned clockwise at $Q$, so seen from $P$ the point $R$ lies clockwise from the first leg: the bearing of $R$ from $P$ is $70^\circ + 33.2^\circ = 103.2^\circ$, that is $\mathbf{103^\circ}$. To return, the ship follows the back bearing $103.2^\circ + 180^\circ = 283.2^\circ$, that is $\mathbf{283^\circ}$.

Check: the bearing $103^\circ$ lies between the bearings $070^\circ$ and $150^\circ$ of the two legs, as the figure shows.

Skills to practice

Solving Bearing Problems with the Two Laws

VII Going further

Skills to practice

Reasoning and Problem-Solving
Investigating Patterns in Triangles
Quiz