Geometry and measures · KS3

Trigonometry: sin, cos and tan

Pythagoras links the three sides of a right-angled triangle. Trigonometry goes further: it links the sides to the angles. Know one angle and one side, and you can work out any other side. Know two sides, and you can work out the angles. It's how surveyors measure buildings they never climb.

National curriculum KS3: use Pythagoras' theorem and trigonometric ratios in similar triangles to solve problems involving right-angled triangles

1 · Naming the sides from an angle

Everything on this page happens in a right-angled triangle: a triangle with one 90° corner, marked with a little square. We pick one of the other two angles to work with and call it θ (the Greek letter "theta", just a name for an angle, like x is a name for a number).

Then we give each side a name depending on where it sits compared with θ:

Drag the top corner (or use the slider) to change the shape. Then move θ to the other corner and watch what happens to the labels.

θ is at:
Hypotenuse
Opposite
Adjacent

Common mistake: labelling from the wrong angle. "Opposite" and "adjacent" only make sense once you know which angle you're talking about. Swap θ to the other corner and they swap over. The hypotenuse is the only side that keeps its name. Always find H first (opposite the square), then O (across from θ), and the one left over is A.

2 · Practise: label the sides

The triangle can be turned any way round, so don't rely on "the bottom is always adjacent". Use the rules: H is opposite the right angle, O is across from θ, A touches θ. Tap the side you're asked for.

3 · Same angle, same ratios

Here's the big idea that makes trigonometry work. Take a right-angled triangle with a 35° angle and divide one side by another, say opposite ÷ hypotenuse. A division of two lengths like this is called a ratio. Now make the triangle twice as big. Both sides double, so the ratio doesn't change: 2 × O ÷ 2 × H is still O ÷ H.

Triangles that are the same shape but different sizes are called similar (it's the same idea as enlarging by a scale factor on the scale factors page). Every right-angled triangle with a 35° angle is similar to every other one, because the angles fix the shape. So the ratios belong to the angle, not to the triangle. Grow the triangle with the size slider and watch the ratios stay put. Then change the angle and watch them change.

O ÷ H
A ÷ H
O ÷ A

Look at the table: the lengths are all different, but each ratio column is the same all the way down. That's why it's worth giving these ratios names.

4 · sin, cos and tan (SOH CAH TOA)

Each of the three ratios gets a name. They're short for sine, cosine and tangent, and we write them with the angle after them:

SOHsin θ = O ÷ H
CAHcos θ = A ÷ H
TOAtan θ = O ÷ A

Remember them as SOH CAH TOA ("so-ca-toe-a"). Each chunk reads: the function, then the top of the fraction, then the bottom. So "SOH" means Sin = Opposite over Hypotenuse. Here the hypotenuse is fixed at 10 so you can see the numbers. Slide θ and check each division yourself.

You don't need to draw a triangle every time to find these numbers: your calculator has them stored. Type the function, then the angle: sin 35 gives 0.5736 (to 4 d.p.). That means every right-angled triangle with a 35° angle has an opposite that is 0.5736 times its hypotenuse.

Using your calculator: degree mode

Calculators can measure angles in two systems: degrees (what you use) and radians (used at A level). Look for a small D or DEG at the top of the screen. If you see R or RAD, every answer will be wrong. Try it on this pretend calculator.

mode:

Quick test for degree mode: work out sin 30. In degrees it's exactly 0.5. If you get −0.988…, your calculator is in radians. Change the mode in the setup menu before you start.

5 · Finding a missing side

Now we can use the ratios. You know one angle and one side, and you want another side. The method is always the same:

  1. Label the sides H, O, A from the angle you know.
  2. Decide which side you know and which you want. Ignore the third side.
  3. Choose the ratio that uses exactly those two letters: O and H → SOH, A and H → CAH, O and A → TOA.
  4. Write the equation, put the numbers in, and solve it.

Solving has two cases. If the side you want is on the top of the fraction, you multiply. If it's on the bottom, you divide. The solver below shows every line. Tick what you know and what you want, then press "Next step".

I know: = I want:

Common mistake: dividing the wrong way. If cos 40° = 8 ÷ x, the answer is x = 8 ÷ cos 40° = 10.4, not 8 × cos 40° = 6.1. Sense check: the hypotenuse must be the longest side, so it can't come out smaller than the 8 you started with.

6 · Finding a missing angle

Now turn it round. If you know two sides, you can work out the ratio, and then ask "which angle has this ratio?". Your calculator answers with the inverse functions, written sin⁻¹, cos⁻¹ and tan⁻¹ (say "inverse sine" and so on). They undo sin, cos and tan: since sin 30° = 0.5, sin⁻¹(0.5) = 30°. On most calculators you press SHIFT then sin.

The "⁻¹" is a label, not a power. sin⁻¹ does not mean 1 ÷ sin.

Common mistake: using inverse for a side. Use sin⁻¹, cos⁻¹, tan⁻¹ only when you're finding an angle. When you're finding a side, use plain sin, cos or tan. A quick check: the answer to an inverse button is an angle in degrees, so for a right-angled triangle it must be between 0° and 90°. Also, O ÷ H and A ÷ H can never be more than 1 (the hypotenuse is the longest side), so if you get sin⁻¹(1.2) your calculator will say "Math ERROR": you've put the sides in upside down.

7 · Real-world problems

Real problems hide the triangle in a picture. Find the right angle (usually where something meets flat ground), mark the angle you know, label O, A, H, then carry on exactly as before. A few words you'll meet:

Surveying and building

Surveyors measure a distance along the ground and an angle up, then use tan to find heights of buildings, trees and hills without climbing them.

Navigation

Ships and planes split their journey into north and east parts using sin and cos of their bearing. GPS and maps rely on the same ideas.

Games and animation

To move a character at an angle, a game works out how far to step across (cos) and how far to step up (sin) each frame.

8 · Exact values (extension)

For three special angles you can find sin, cos and tan exactly, without a calculator, from two simple triangles. Cut a square with side 1 along its diagonal: you get a 45° triangle with sides 1, 1 and √2 (by Pythagoras). Cut an equilateral triangle with side 2 in half: you get a 30°–60° triangle with sides 1, √3 and 2. Choose an angle and read the ratios off the sides. Notice that 30° and 60° use the same triangle, with θ in different corners, so opposite and adjacent swap.

θ =
30°45°60°
sin1/21/√2 = √2/2√3/2
cos√3/21/√2 = √2/21/2
tan1/√3 = √3/31√3

Worked examples

Find the opposite: θ = 35°, hypotenuse = 12 cm
Label: H = 12 is known, O = x is wanted. A isn't involved.
Choose: O and H → SOH, so use sin.
Write: sin 35° = x ÷ 12
Solve: x is on top, so multiply both sides by 12: x = 12 × sin 35°
Calculate: x = 12 × 0.5735… = 6.8829… so x = 6.9 cm to 1 d.p.
Check: 6.9 is less than the hypotenuse 12 ✓
Find the hypotenuse: θ = 40°, adjacent = 8 cm
Label: A = 8 is known, H = x is wanted.
Choose: A and H → CAH, so use cos.
Write: cos 40° = 8 ÷ x
Solve: x is on the bottom. Multiply both sides by x: x × cos 40° = 8. Divide both sides by cos 40°: x = 8 ÷ cos 40°
Calculate: x = 8 ÷ 0.7660… = 10.4432… so x = 10.4 cm to 1 d.p.
Check: the hypotenuse (10.4) is longer than the adjacent (8) ✓. If you'd multiplied you'd get 6.1, which is too short.
Find an angle: opposite = 5 cm, adjacent = 9 cm
Choose: O and A → TOA, so use tan.
Write: tan θ = 5 ÷ 9 = 0.5555…
Inverse: θ = tan⁻¹(5 ÷ 9). Type SHIFT tan ( 5 ÷ 9 ).
Calculate: θ = 29.0546…° so θ = 29.1° to 1 d.p.
Check: O is smaller than A, so θ should be less than 45° ✓ (at 45°, O and A are equal).
Ladder: a 6 m ladder leans against a wall at 70° to the ground. How high up the wall does it reach?
Picture: the wall meets the ground at a right angle. The ladder is the hypotenuse, H = 6. The 70° is at the foot of the ladder.
Label: the height up the wall is across from the 70°, so it's O = x.
Choose and write: O and H → sin. sin 70° = x ÷ 6
Solve: x = 6 × sin 70° = 6 × 0.9396… = 5.6381… so 5.6 m to 1 d.p.
Bonus: the foot of the ladder is 6 × cos 70° = 2.0521… ≈ 2.1 m from the wall.
Angle of elevation: from 30 m away, the top of a tree is at 28° above eye level. Eye level is 1.5 m above the ground. How tall is the tree?
Picture: the right angle is where the horizontal line from your eye meets the tree. The distance 30 m is next to the 28° angle: A = 30. The height of the tree above eye level is across from it: O = x.
Choose and write: O and A → tan. tan 28° = x ÷ 30
Solve: x = 30 × tan 28° = 30 × 0.5317… = 15.9512… m
Don't forget eye height: tree = 15.9512… + 1.5 = 17.4512… so 17.5 m to 1 d.p.
Ramp: a ramp rises 0.5 m over a horizontal distance of 6 m. What angle does it make with the ground?
Label: from the angle at the bottom of the ramp, the rise is O = 0.5 and the horizontal distance is A = 6.
Choose: O and A → tan. tan θ = 0.5 ÷ 6 = 0.08333…
Inverse: θ = tan⁻¹(0.08333…) = 4.7636…° so θ = 4.8° to 1 d.p.

Quick check

What you've learned