Sine, cosine and tangent in a right triangle
What the four ratios mean, the two triangles that give every special angle, and the square-root trick for remembering the table.
First, the rules
Four ratios, two triangles, one trick
Everything in this lesson follows from these. The small grey text says which worked example uses each rule; ★ marks the ones used most.
The ratios
Special angles
| 30° | 45° | 60° | |
|---|---|---|---|
| sin | 12 | √22 | √32 |
| cos | √32 | √22 | 12 |
| tan | √33 | 1 | √3 |
| cot | √3 | 1 | √33 |
See it
Move the angle, watch the ratios
Drag the slider or press a special angle. The sides are named from the highlighted angle; switch to the other angle and watch "opposite" and "adjacent" trade places while the hypotenuse stays put.
The two triangles to memorise
Example: sin 30° = opposite / hypotenuse = 1 / 2. cos 45° = 1 / √2 = √2 / 2. tan 60° = √3 / 1.
Reading a ratio from a triple
If a question gives tan α = 43, draw a right triangle with opposite 4k and adjacent 3k. Pythagoras gives the hypotenuse 5k, and now every ratio is available: sin α = 45, cos α = 35. The k cancels in every ratio, so you may simply write 3, 4, 5. More on triples in Pythagoras and the triples.
Step by step
Worked examples
Try each one before opening the solution. Red lines in a figure are the length you are looking for.
Find sin x from a ratio of sines and cosines
ratio → tangent → trianglesin x + cos x2 sin x − cos x = 34 and x is acute. Find sin x.
Ask yourself: What is the first move in any equation with a fraction? (Cross-multiply.)
Show the solution
- Cross-multiply: 4(sin x + cos x) = 3(2 sin x − cos x), so 4 sin x + 4 cos x = 6 sin x − 3 cos x.
- Sines to one side, cosines to the other: 7 cos x = 2 sin x.
- Divide both sides by 2 cos x: tan x = 72.
- Draw a right triangle with opposite 7 and adjacent 2. Pythagoras: hypotenuse √49 + 4 = √53.
- sin x = 7√53; rationalise the denominator by multiplying top and bottom by √53.
Answer sin x = 7√5353
cos β from a tangent and an equal mark
triple · PythagorasIn the figure tan α = 43 and |AD| = |DC|. Find cos β.
Ask yourself: tan α = 4/3 — how do you write the sides of triangle ABD with a k? (A 3-4-5 triangle.)
Show the solution
- In ABD, opposite to α is BD and adjacent is AB: |BD| = 4k, |AB| = 3k. It is a 3-4-5 triangle, so |AD| = 5k.
- The mark says |DC| = |AD| = 5k, so |BC| = 4k + 5k = 9k.
- Pythagoras in ABC: |AC|² = (3k)² + (9k)² = 90k², so |AC| = 3√10k.
- cos β = |BC||AC| = 9k3√10k = 3√10
Answer cos β = 3√1010
Height of a building
tangent in daily lifeAn instrument 1.5 m above the ground, 36 m from a building, measures the angle to the top as 30°. How tall is the building?
Ask yourself: You want the side opposite the angle and you know the adjacent side. Which ratio links them? (tan.)
Show the solution
- Call the part of the building above the instrument h. Then tan 30° = h36 and tan 30° = 1√3.
- h = 36√3 = 36√33 = 12√3.
- Add the height of the instrument: 12√3 + 1.5 ≈ 20.78 + 1.5. The most common slip is forgetting this 1.5 m.
Answer 12√3 + 1.5 m ≈ 22.3 m
Practice: ratios from a triangle
0 / 0A right triangle with two known sides, or one ratio given — find the one asked for. Answers are exact fractions. Your best score is kept in this browser only.
About this lesson
Learn sine, cosine, tangent and cotangent as ratios of the sides of a right triangle: opposite, adjacent and hypotenuse relative to an angle. Move the angle and watch the ratios change, read every special-angle value from the 30-60-90 and 45-45-90 triangles, use the square-root trick for the table, and practise finding one ratio from another with the 3-4-5 method.
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