Right-angled triangles

Naming the sides

Trigonometry links the angles of a right-angled triangle to its sides. Before you can use it, you must name the three sides, and the names depend on which angle you are working with.

• The hypotenuse is opposite the right angle. It is the longest side, and its name never changes.
• The opposite side is across the triangle from your angle. It does not touch the angle.
• The adjacent side is next to your angle, and is not the hypotenuse.

The angle you are using is often marked $\theta$ (the Greek letter theta) or $x$, or is given in degrees. If you switch to the other acute angle, the opposite and adjacent sides swap over. So always label the sides from the angle in the question, and do it on the diagram, every time.

Labelled from the angle $\theta$ at A: the opposite side does not touch $\theta$, and the adjacent side runs from $\theta$ to the right angle.

The three ratios and why they work

For an acute angle $\theta$ in a right-angled triangle:

$$\sin \theta =\frac{\text{opposite}}{\text{hypotenuse}}\cos \theta =\frac{\text{adjacent}}{\text{hypotenuse}}\tan \theta =\frac{\text{opposite}}{\text{adjacent}}$$

Many students remember them as SOH CAH TOA: Sine is Opposite over Hypotenuse, Cosine is Adjacent over Hypotenuse, Tangent is Opposite over Adjacent.

Why a ratio depends only on the angle. Take any two right-angled triangles that both have an angle of ${35}^{°}$. Their third angles are both ${55}^{°}$, so the triangles have the same angles and are similar: one is an enlargement of the other. An enlargement multiplies every side by the same scale factor, so the ratio of two sides does not change. That is why a calculator can store a single value, $\sin {35}^{°}=0.5735\dots$, that works for every right-angled triangle with a ${35}^{°}$ angle, large or small.

Two useful facts follow. The hypotenuse is the longest side, so for an acute angle $\sin \theta$ and $\cos \theta$ are always between 0 and 1. And $\tan {45}^{°}=1$, because a right-angled triangle with a ${45}^{°}$ angle is isosceles, so its opposite and adjacent sides are equal.

Calculator check: your calculator must be in degree mode. Type $\sin 30$: it should give 0.5. If it gives $-0.988\dots$, it is in radians, and every answer will be wrong.

Finding a side

1. Label the sides opposite, adjacent and hypotenuse from the angle you know.
2. Mark the side you know and the side you want.
3. Choose the ratio that uses exactly those two sides: O and H means sin, A and H means cos, O and A means tan.
4. Write the equation, for example $\sin {38}^{°}=\frac{x}{14}$.
5. Solve it.

• If the unknown is on top of the fraction, multiply: $x=14\sin {38}^{°}$.
• If the unknown is on the bottom, it swaps with the ratio: from $\tan {52}^{°}=\frac{9.5}{y}$ you get $y=\frac{9.5}{\tan {52}^{°}}$.

The swap works because $\tan {52}^{°}=\frac{9.5}{y}$ means $y\times \tan {52}^{°}=9.5$; dividing both sides by $\tan {52}^{°}$ leaves $y$ on its own.

Check your answer: the opposite and adjacent sides must both be shorter than the hypotenuse.

Finding an angle

When you know two sides and want an angle, use the ratio backwards. The inverse functions ${\sin }^{-1}$, ${\cos }^{-1}$ and ${\tan }^{-1}$ (usually SHIFT then sin, cos or tan) take a ratio and give the angle.

1. Label the sides from the angle you want.
2. Choose the ratio that uses the two sides you know.
3. Write the ratio as a fraction, for example $\tan x=\frac{7.3}{11.6}$.
4. Use the inverse function on the whole fraction: $x={\tan }^{-1}\frac{7.3}{11.6}$.

Angles are given in degrees, and an answer that is not exact is written correct to 1 decimal place. Do not round the fraction first: type the whole fraction into the calculator, because rounding $\frac{7.3}{11.6}$ to $0.63$ moves the angle.

Once you have one acute angle, the other is ${90}^{°}$ minus it, because the angles of the triangle add up to ${180}^{°}$.

Pythagoras or trigonometry?

Read what you know and what you want:

• Two sides known, third side wanted, no angle involved: Pythagoras' theorem.
• An angle is known or wanted: trigonometry.
• One side and one angle known: trigonometry finds a second side; then either method finds the third.

In a long question you will often use both: Pythagoras to find a missing side, then trigonometry to find an angle, or the other way round. If you can, find a value from the given numbers rather than from your own earlier answers, so an early slip does not spread.

Bearings in right-angled triangles

A bearing is an angle measured clockwise from north, written with three figures: ${062}^{°}$, ${111.4}^{°}$, ${290}^{°}$. North, east, south and west are at right angles to each other, so a journey due east and then due north makes a right-angled triangle.

To find a bearing with trigonometry:

1. Draw a north line at the point you are measuring the bearing from. "The bearing of Q from H" means stand at H.
2. Find an angle inside the right-angled triangle with ${\tan }^{-1}$, ${\sin }^{-1}$ or ${\cos }^{-1}$.
3. Turn it into an angle clockwise from north. Sketch it: is the direction between north and east (bearing ${000}^{°}$ to ${090}^{°}$), between east and south (${090}^{°}$ to ${180}^{°}$), and so on?

The return direction differs by ${180}^{°}$: if the bearing of Q from H is ${058.6}^{°}$, the bearing of H from Q is ${058.6}^{°}+{180}^{°}={238.6}^{°}$. This works because the north lines at H and Q are parallel.

A distance travelled on a bearing splits into an east part and a north part. A ship that sails $d$ km on a bearing of $\theta$ (between ${000}^{°}$ and ${090}^{°}$) goes $d\sin \theta$ km east and $d\cos \theta$ km north, because the bearing angle is at the start, between the north line and the path.

Problems in two dimensions

Harder questions combine several right-angled triangles, or hide them inside other shapes. The method is always the same: find a right-angled triangle in which you know two things, solve it, and use what you found in the next triangle.

• Two triangles sharing a side: the shared side links them. Find it in one triangle, then use it in the other.
• Isosceles triangles: split down the line of symmetry into two right-angled triangles.
• Trapeziums and other quadrilaterals: drop a perpendicular to make a right-angled triangle, and work out its sides from the lengths given.

Keep every intermediate value at full accuracy: leave it on the calculator display, or store it in the memory. Rounding a length to 3 significant figures and then using it again can change the final answer, and that costs the accuracy mark. Round only the answer you write on the answer line.

Elevation: looking up from the horizontal. Depression: looking down from the horizontal.