Use the trigonometric ratios sine, cosine and tangent to find unknown sides and angles in right-angled triangles, including the exact ratios of 30, 45 and 60 degrees, and apply them to angles of elevation and depression and to compass and true bearings

What this dot point is asking

This is the start of trigonometry in the Advanced course, and almost everything later, the unit circle, the sine and cosine rules, the wave graphs, grows out of what is on this page: the three ratios in a right-angled triangle. NESA wants you to use sine, cosine and tangent to find an unknown side or an unknown angle, to know the exact values of $30\degree$, $45\degree$ and $60\degree$ without a calculator, and to apply all of this to the two contexts the HSC asks about every year, angles of elevation and depression, and compass and true bearings.

Very few of the marks are for memorising a formula. They are for three decisions and one habit. The first decision is which ratio fits the two sides involved (the SOH CAH TOA choice). The second is whether you are solving for a side or for an angle, because that decides whether you multiply, divide, or press an inverse-trig button. The third, in word problems, is which right-angled triangle the words describe and which side is opposite, adjacent or the hypotenuse. The habit is keeping the calculator in degrees and carrying full precision to the last line. Get the ratio, the triangle and the calculator mode right and the arithmetic is a single button press.

The answer

Every ratio on this page refers to the same picture: a right-angled triangle, one acute angle $\theta$, and its three sides named relative to $\theta$. The hypotenuse is the longest side, always opposite the right angle. The opposite is the side across the triangle from $\theta$. The adjacent is the remaining side, the one that touches $\theta$ but is not the hypotenuse. The names of two of the sides depend on which acute angle you call $\theta$, so fix $\theta$ first, then label.

The three ratios: SOH CAH TOA

Because any two right-angled triangles with the same acute angle $\theta$ are similar (the AA test), the ratio of any two sides depends only on $\theta$, not on the size of the triangle. Those fixed ratios are the trigonometric functions:

The mnemonic SOH CAH TOA stores all three: Sine is Opposite over Hypotenuse, Cosine is Adjacent over Hypotenuse, Tangent is Opposite over Adjacent. The deeper fact worth carrying is that sine and cosine are the two ways of comparing a leg to the hypotenuse, while tangent compares the two legs directly and so never mentions the hypotenuse at all. That is exactly why tangent is the natural ratio for elevation, depression and bearings, where the hypotenuse (the line of sight) is usually the thing you neither know nor want.

Finding an unknown side

When the unknown is a side, choose the ratio that pairs the unknown with a known side, write it with the unknown on top, then make the unknown the subject in one step. If the unknown is on top, the last step is a multiplication; if the unknown sits in the denominator (the known side is the opposite or adjacent and you want the hypotenuse), it is a division. Keeping the unknown on top from the start avoids the cross-multiplication slip that loses easy marks.

Finding an unknown angle

When two sides are known and the unknown is the angle, work out from those two sides which ratio is fixed, write it, then apply the matching inverse function. For example, if the opposite and adjacent are known, $\tan\theta$ is fixed, so $\theta = \tan^{-1}\!\left(\dfrac{O}{A}\right)$. The inverse-trig keys ($\sin^{-1}$, $\cos^{-1}$, $\tan^{-1}$) return the acute angle directly, which is all a right-angled triangle ever needs.

The exact ratios of 30, 45 and 60 degrees

For three special angles the ratios are exact surds, and NESA expects them without a calculator (questions that say "find the exact value" forbid a decimal). They come from two triangles you can reconstruct in seconds.

Cut a unit square along its diagonal: the two legs are $1$ and $1$, the diagonal is $\sqrt{1^2 + 1^2} = \sqrt{2}$, and the angles are both $45\degree$. Cut an equilateral triangle of side $2$ down its altitude: the base halves to $1$, the altitude is $\sqrt{2^2 - 1^2} = \sqrt{3}$, and the angles are $30\degree$ and $60\degree$. Reading SOH CAH TOA off these two triangles gives the table.

Two patterns make the table self-checking. Down the sine column the values rise $\frac{1}{2}, \frac{1}{\sqrt2}, \frac{\sqrt3}{2}$, and the cosine column is the same list reversed, because the cosine of an angle is the sine of its complement ($\cos 60\degree = \sin 30\degree$). And $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$, so the tangent column is just the sine entry divided by the cosine entry. Some answers are left with a surd in the denominator; rationalising is optional, so $\tan 30\degree$ may be written $\dfrac{1}{\sqrt{3}}$ or equivalently $\dfrac{\sqrt{3}}{3}$.

Angles of elevation and depression

An angle of elevation is measured upward from the horizontal to a line of sight; an angle of depression is measured downward from the horizontal. Both are always taken from the horizontal, never the vertical, and both are acute. They are linked by a fact that saves a step in almost every question: the angle of depression from a high point down to a low point equals the angle of elevation from the low point back up to the high point, because the two horizontal lines are parallel and the line of sight is a transversal, making the two angles alternate angles.

In the single right-angled triangle this makes, the elevation or depression angle $\theta$, the horizontal distance $x$ and the vertical height $H$ are tied together by $\tan\theta = \dfrac{H}{x}$. That one relation, rearranged as $x = \dfrac{H}{\tan\theta}$ or $H = x\tan\theta$, solves the great majority of these problems. When two sightings are taken from the same height, compute each horizontal distance separately and subtract.

Compass and true bearings

Directions in the HSC are given in one of two systems. A true bearing is the angle measured clockwise from north, written with three digits from $000\degree$ to $360\degree$. The leading zeros are not decoration; they mark the number as a bearing and stop $075$ being misread. A compass bearing names a primary direction (N or S), an acute angle, then E or W, for example N$30\degree$E or S$45\degree$W.

To convert a compass bearing to a true bearing, walk clockwise from north: N$30\degree$E is $030\degree$; S$45\degree$W is past south ($180\degree$) by a further $45\degree$, so $225\degree$; N$50\degree$W is short of a full turn by $50\degree$, so $360\degree - 50\degree = 310\degree$. Two more facts close out almost every Year 11 bearings problem:

• Each point has its own north. A bearing is read from the local north where you are standing, so a diagram needs a north arrow at every point you take a bearing from. The north lines are all parallel.
• A back-bearing reverses by $180\degree$. The bearing of $A$ from $B$ is the bearing of $B$ from $A$ plus or minus $180\degree$ (add if the original is under $180\degree$, subtract if over), because those two north arrows are parallel and the line between the points is a transversal.

In Year 11 the bearings questions are built so that the triangle is right-angled, usually because one leg runs due north or south and the other due east or west, which meet at $90\degree$. You then find the unknown leg or the unknown angle with SOH CAH TOA (or Pythagoras), and translate the angle back into a three-figure bearing. The non-right-angled bearings problems, which need the sine and cosine rules, come later in the sine rule, cosine rule and area and in triangle trigonometry and applications.

How exam questions ask about right-angled trigonometry

The wording tells you which ratio and which step to reach for:

• "Find the length of ... / find $x$" with one angle and one side given. A side problem: pick the ratio pairing the unknown with the known side (SOH CAH TOA), unknown on top, then one step.
• "Find the size of the angle ... / find $\theta$" with two sides given. An angle problem: choose the ratio fixed by those two sides, then take the inverse trig function.
• "Find the exact value ..." A forbid-the-calculator signal: use the $30\degree$, $45\degree$, $60\degree$ special triangles and leave the answer as a surd.
• "Correct to the nearest degree / minute / two decimal places" sets the rounding for the final line only; keep full precision until then.
• "The angle of elevation / depression of ... is ..." Draw the right-angled triangle, mark $\tan\theta = \dfrac{\text{height}}{\text{horizontal}}$, and use the alternate-angle link if the angle is quoted at the other end of the line of sight.
• "... on a bearing of $\,\dots\,\degree$T", "find the bearing of ... from ...", "how far ..." Sketch a north arrow at the relevant point, build the right-angled triangle (a north or south leg and an east or west leg), solve it, then convert the angle into a three-figure bearing; reverse by $180\degree$ for a back-bearing.
• "Show that ... = [given value]" A scaffold step that hands you the next part's number, so set out the working that lands exactly on it.

Building a bearings problem stage by stage

The two-leg, right-angled bearings problem is the archetype the HSC reuses. The reliable method is to draw it up one leg at a time, close the triangle, then read off the distance and the bearing. Here is the build for: a yacht sails $9$ km due north from harbour $H$ to a buoy $A$, then $5$ km due east to a marker $B$; find the distance $HB$ and the bearing of $B$ from $H$.

Stage 1, anchor the starting point and its north line. Mark the harbour $H$ and draw a north arrow there. Every bearing in the problem is read clockwise from this north line, so it is the reference for the final answer.

Stage 2, draw the first leg. From $H$, the first leg runs $9$ km due north (straight up the north line) to the buoy $A$.

Stage 3, add the second leg and the right angle. From $A$, the second leg runs $5$ km due east to the marker $B$. North and east are perpendicular, so the angle at $A$ is exactly $90\degree$, which is what makes this a right-angled triangle.

Stage 4, close the triangle and read off the answers. Join $H$ to $B$. By Pythagoras, $HB = \sqrt{9^2 + 5^2} = \sqrt{106} \approx 10.30$ km. The bearing of $B$ from $H$ is the angle east of north at $H$: $\tan\theta = \dfrac{5}{9}$, so $\theta = \tan^{-1}\!\left(\dfrac{5}{9}\right) \approx 29.05\degree$, which is the three-figure bearing $029\degree$.

So $HB \approx 10.30$ km and $B$ is on a bearing of $029\degree$ from $H$. As a final reading check, the back-bearing of $H$ from $B$ is $029\degree + 180\degree = 209\degree$.