Triangle Trigonometry and Bearings

Updated 7 October 2026

Two triangles with the same angles have sides in the same proportion, whatever their size. So once an angle is fixed, the ratio of any two sides is fixed too, and that ratio can be looked up instead of measured. Sine, cosine and tangent are the names of three such ratios.

What do sine, cosine and tangent actually mean?

In a right triangle, sine, cosine and tangent are ratios of sides, named from one acute angle $\theta$θ. The hypotenuse faces the right angle, the opposite side faces $\theta$θ, and the adjacent side lies beside $\theta$θ. Then $\sin \theta$sin θ is opposite over hypotenuse, $\cos \theta$cos θ is adjacent over hypotenuse, and $\tan \theta$tan θ is opposite over adjacent. In a 3-4-5 triangle, with $\theta$θ facing the side of 3, $\sin \theta =\frac{3}{5}$sin θ = 3/5, $\cos \theta =\frac{4}{5}$cos θ = 4/5 and $\tan \theta =\frac{3}{4}$tan θ = 3/4. The ratios depend only on the angle. Any right triangle with the same $\theta$θ is an enlargement of this one, so doubling the sides to 6, 8 and 10 leaves $\frac{6}{10}=\frac{3}{5}$6/10 = 3/5. The labels move when the angle does: for the other acute angle, the side of 4 is opposite, so that angle's sine is $\frac{4}{5}$4/5, the cosine of $\theta$θ. Check the three values together, since $\tan \theta$tan θ must equal $\sin \theta /\cos \theta$sin θ / cos θ: here $\frac{3}{5}÷\frac{4}{5}=\frac{3}{4}$3/5 ÷ 4/5 = 3/4.

The memory aid SOHCAHTOA lists the three ratios in order. See Sine, cosine and tangent.

The two acute angles of a right triangle add to 90°, so in general $\sin \theta =\cos \left(90^{\circ }-\theta \right)$sin θ = cos(90° − θ). See Sine and cosine of complementary angles.

How do you find a missing side or angle?

To find a missing side, label the sides from the given angle, choose the ratio that links the side you know to the side you want, and solve. With a hypotenuse of 10 and an angle of 30°, the opposite side needs sine: it is 10 × sin 30°, and $10\times \frac{1}{2}=5$10 × 1/2 = 5. The unknown may land in the denominator. With an adjacent side of 12 and an angle of 60°, cos 60° = 12/hypotenuse, so the hypotenuse is $12/\cos 60^{\circ }$12 / cos 60°, which is $12÷\frac{1}{2}=24$12 ÷ 1/2 = 24. This works because each ratio is fixed for a given angle, so one known side sets the scale. When two sides are known and the angle is missing, apply the inverse function to their ratio: with opposite 5 and hypotenuse 10, $\theta ={\sin }^{-1}\left(\frac{1}{2}\right)=30^{\circ }$θ = sin⁻¹(1/2) = 30°. Check with Pythagoras: the third side is $5\sqrt{3}$5√3, whose square is 75, and 25 + 75 = 100, the square of 10.

1. Mark the hypotenuse. It always faces the right angle.
2. Mark opposite and adjacent from the angle in the question.
3. Write the ratio that links the side you know to the side you want, then solve.

See Solving right triangles.

${\sin }^{-1}x$sin⁻¹ x is not $1/\sin x$1 / sin x; it is the angle whose sine is x. See Finding an angle from a ratio.

An angle of elevation is measured up from the horizontal, and an angle of depression down from it. Because the two horizontals are parallel, the two angles are equal. See Elevation and depression.

Which values should you know without a calculator?

Learn sine, cosine and tangent at 30°, 45° and 60°, because two triangles give all nine values. Cut a square of side 1 along its diagonal, which is $\sqrt{2}$√2: $\sin 45^{\circ }=\cos 45^{\circ }=\frac{1}{\sqrt{2}}$sin 45° = cos 45° = 1/√2, or $\frac{\sqrt{2}}{2}$√2/2, and tan 45° = 1. Cut an equilateral triangle of side 2 in half: each piece has hypotenuse 2, short side 1 and height $\sqrt{3}$√3, since 4 − 1 = 3. So $\sin 30^{\circ }=\frac{1}{2}$sin 30° = 1/2, $\cos 30^{\circ }=\frac{\sqrt{3}}{2}$cos 30° = √3/2 and $\tan 30^{\circ }=\frac{1}{\sqrt{3}}$tan 30° = 1/√3, and 60° swaps the sine and cosine: $\sin 60^{\circ }=\frac{\sqrt{3}}{2}$sin 60° = √3/2, $\cos 60^{\circ }=\frac{1}{2}$cos 60° = 1/2, $\tan 60^{\circ }=\sqrt{3}$tan 60° = √3. They are exact because the lengths are. From the unit circle, sin 0° = 0 and sin 90° = 1, while tan 90° is undefined. In radians the three angles are $\frac{\pi }{6}$π/6, $\frac{\pi }{4}$π/4 and $\frac{\pi }{3}$π/3. Check a pair with ${\sin }^2\theta +{\cos }^2\theta =1$sin²θ + cos²θ = 1: at 60°, the squares are $\frac{3}{4}$3/4 and $\frac{1}{4}$1/4, and $\frac{3}{4}+\frac{1}{4}=1$3/4 + 1/4 = 1.

Its height, $\sqrt{3}$√3, is where every $\sqrt{3}$√3 in the table comes from.

Keep these in surd form: $\frac{\sqrt{3}}{2}$√3/2 is exact, and a rounded decimal is not. See Exact sine, cosine and tangent at 30, 45 and 60°.

What happens past 90 degrees?

Past 90°, sine and cosine are defined on the unit circle, radius 1, centered at the origin. Turn a radius counterclockwise through $\theta$θ from the positive x-axis to reach the point $\left(\cos \theta ,\sin \theta \right)$(cos θ, sin θ), and $\tan \theta =\sin \theta /\cos \theta$tan θ = sin θ / cos θ. At 150° the point is at the same height as the point for 30°, mirrored across the y-axis, so $\sin 150^{\circ }=\frac{1}{2}$sin 150° = 1/2 and $\cos 150^{\circ }=-\frac{\sqrt{3}}{2}$cos 150° = −√3/2. This matches the triangle definitions below 90°, where the point, its foot on the x-axis and the origin form a right triangle with hypotenuse 1. The signs change by quadrant: all three ratios are positive up to 90°, only sine up to 180°, only tangent up to 270°, and only cosine up to 360°. For sin 210°, take the reference angle, 210 − 180 = 30, then the sign: $\sin 210^{\circ }=-\frac{1}{2}$sin 210° = −1/2. Check the 150° values: their squares are $\frac{1}{4}$1/4 and $\frac{3}{4}$3/4, and $\frac{1}{4}+\frac{3}{4}=1$1/4 + 3/4 = 1.

In general an obtuse angle and its supplement have the same sine and opposite cosines. See Sine and cosine of angles past 90 degrees.

What do the trigonometric graphs look like?

The sine graph is a wave: the height of the point on the unit circle plotted against the angle turned. It starts at 0, rises to 1 at 90°, returns to 0 at 180°, falls to −1 at 270°, reaches 0 again at 360°, and repeats. See The graphs of sine and cosine. Cosine is the same wave shifted 90° to the left, so it starts at its peak of 1. The wave repeats every 360° because a full turn brings the point back, and it stays between −1 and 1 because the radius is 1. A stretch changes the numbers but not the shape: y = 3 sin 2x runs between −3 and 3, with period 360 ÷ 2 = 180 degrees, because 2x completes a turn while x completes half of one. Check the shift: cos 60° should equal the sine of 150°, and 60 + 90 = 150, where the sine is $\frac{1}{2}$1/2, as is cos 60°.

Tangent is $\sin \theta /\cos \theta$sin θ / cos θ. At 90° and 270°, $\cos \theta =0$cos θ = 0, so the graph has a vertical asymptote. Tangent repeats every 180° and takes every value. See The graph of tangent.

In y = a sin(b(x − c)) + d, the amplitude a is the distance from the principal axis to the crest, the period is 360° ÷ b, the phase shift c slides the wave sideways, and d is the height of the principal axis. See Amplitude and the principal axis and The period and phase shift of a wave.

How do you solve a trigonometric equation?

Rearrange the equation to sin x, cos x or tan x equal to a number, take the inverse function for one solution, and use the graph's symmetry for the rest. For $\sin x=\frac{1}{2}$sin x = 1/2 on $0^{\circ }\le x\le 360^{\circ }$0° ≤ x ≤ 360°, ${\sin }^{-1}$sin⁻¹ gives 30°, and since the sine curve is symmetric about 90°, the other solution is 180 − 30 = 150 degrees. See Solving simple trigonometric equations. The symmetry holds because the unit-circle points at x and 180° − x are at the same height. Cosine pairs x with 360° − x, and tangent repeats every 180°, so its second solution is x + 180°. With $\sin x=-\frac{1}{2}$sin x = −1/2 the start changes: the inverse gives −30°, outside the range, so add 360° for 330°, and symmetry gives 210°. Check: 210° is 30° past a half turn and 330° is 30° short of a full one, as 330 + 30 = 360 shows, so both have sine $-\frac{1}{2}$−1/2.

$2{\sin }^{2}x-\sin x-1=0$2 sin²x − sin x − 1 = 0 is a quadratic in sin x. Write s = sin x, factor, and solve sin x = each root separately, rejecting any root outside −1 to 1. See Quadratic equations in sine, cosine or tangent.

If the equation is in 2x, widen the interval before solving. For $0^{\circ }\le x\le 360^{\circ }$0° ≤ x ≤ 360°, 2x runs from 0° to 720°, so collect every solution for 2x in that range and then halve each one. See Trigonometric equations with a multiple angle.

The depth of water in a harbor rises and falls on a fixed cycle, so it fits the same equation. See Modeling tides and daylight with a sine function.

What do you do when the triangle has no right angle?

Without a right angle, use the sine rule or the cosine rule. The sine rule says $a/\sin A=b/\sin B=c/\sin C$a / sin A = b / sin B = c / sin C: each side over the sine of its opposite angle. Use it when a side, its opposite angle and one more part are known. For A = 30°, a = 5 and B = 45°, the common ratio is $5/\sin 30^{\circ }$5 / sin 30°, which is $5÷\frac{1}{2}=10$5 ÷ 1/2 = 10, so $b=10\sin 45^{\circ }=5\sqrt{2}$b = 10 sin 45° = 5√2. See The sine rule. It holds because a height can be written with either base angle. With the angle between two known sides, or three sides known, no pair matches, so the cosine rule, $a^2=b^2+c^2-2\text{b}\text{c}\cos A$a² = b² + c² − 2 b c cos A, takes over. For sides 3 and 5 with 60° between them, $a^2$a² is 9 + 25 − 15 = 19. Check: C is 105°, the largest angle, so c must be the longest side.

At A = 90°, cos A = 0, the correction term in the cosine rule vanishes, and Pythagoras is left. See The cosine rule.

With three sides known, rearrange for the cosine: $\cos A=\left(b^2+c^2-a^2\right)/2$cos A = (b² + c² − a²) / 2bc. See Finding angles with the cosine rule and Finding angles with the sine rule.

The area of any triangle is ½ab sin C, because the height onto a is b sin C. See Area with sine.

Why does the sine rule sometimes give two triangles?

The sine rule can give two triangles because an angle and its supplement have the same sine, and the inverse sine returns only the acute one. It happens when the known angle is not between the two known sides. Take A = 30°, a = 6 and b = 10: then $\sin B=10\times \left(\frac{1}{2}\right)/6=\frac{5}{6}$sin B = 10 × (1/2) / 6 = 5/6, so B is acute or 180° minus that acute angle. Picture side a swinging from its corner: it is longer than the height b sin A, which is $10\times \frac{1}{2}=5$10 × 1/2 = 5, but shorter than b, so it meets the base line twice. The test is whether the obtuse B leaves room for a third angle. As $\frac{5}{6}$5/6 lies between $\frac{1}{2}$1/2 and $\frac{\sqrt{3}}{2}$√3/2, the obtuse B is between 120° and 150°, so A + B stays under 180° and both triangles exist. Check the boundary: at a = 5, sin B = 1 and the two triangles merge into one.

Take A = 20° and $\sin B=\frac{1}{2}$sin B = 1/2. With B = 30° the third angle is 130°; with B = 150° it is 10°. Both fit. With A = 75° the obtuse candidate leaves no third angle, so only B = 30° fits. See The ambiguous case of the sine rule.

The cosine rule never has this problem. Cosine is negative for an obtuse angle and positive for an acute one, so the sign of the answer says which kind it is.

How do bearings work?

A bearing is a direction given as an angle measured clockwise from north, always with three figures. North is 000°, east 090°, south 180° and west 270°, and a bearing of 47° is written 047°. See Three-figure bearings. To find one, stand at the starting point, face north, and turn clockwise until you face the other point. If B is southeast of A, the turn passes east and stops halfway to south, so the bearing is 90 + 45 = 135 degrees. Three figures and one fixed starting direction leave no doubt about how the angle was measured. The return journey changes the steps. The bearing of A from B is measured from north at B, and the two north lines are parallel, so it differs by 180°: 135° + 180° = 315°. For an outward bearing over 180°, subtract 180° instead. Check against the compass: 315° lies between west and north, so the return from a southeast trip points northwest.

The eight-point compass names north, east, south, west and the four directions between them. Northeast is 045° and southwest is 225°. See The eight-point compass.

The return direction is called the back bearing. See Back bearings.

A navigation problem is usually a sine-rule or cosine-rule triangle. Convert the bearings to angles inside the triangle first. See Bearings with the sine and cosine rules.

How does trigonometry work in three dimensions?

In three dimensions, find a right triangle lying flat inside the solid and work in that triangle alone. In a box 3 by 4 by 12, find the floor diagonal first: 9 + 16 = 25, so it is 5. The space diagonal forms a right triangle with that diagonal and the height of 12, so its length is 13, since 25 + 144 = 169. Its angle to the floor has tangent $\frac{12}{5}$12/5, height over floor diagonal. This turns one problem in space into two flat ones, each in a single plane where the triangle rules apply. The choice of angle is the step that changes: the angle between a line and a plane is measured to the line's projection on the plane, its shadow straight down, not to any other line in the plane. Check with the box formula $\sqrt{a^2+b^2+c^2}$√(a² + b² + c²): 9, 16 and 144 add to 169, and 13 × 13 = 169.

That angle lives in the same flat triangle. See Trigonometry in three dimensions, or solve the same problems with vectors.

The mistakes worth naming

• Labeling opposite and adjacent before choosing the angle. Only the hypotenuse is fixed.
• Reading ${\sin }^{-1}x$sin⁻¹ x as $1/\sin x$1 / sin x. It is the angle whose sine is x.
• Giving one solution when the interval holds several. Use the symmetry of the graph to find the rest.
• Using the sine rule for an angle that may be obtuse. Check the supplement, or use the cosine rule.
• Halving a multiple-angle interval too early. Widen the interval, solve, then halve.

The right-angle groundwork is in lines, angles and Pythagoras, the same problems in space are in vectors, and identities and radian measure are in radians and trigonometric identities.