Fix the angle and the side ratios lock: sin, cos, tan.
A carpenter sets a ramp, a surveyor sights a distant peak, a sailor reads a bearing — and every one of them is secretly doing the same thing: turning an angle into a ratio of lengths. The magic fact, and the whole engine of this lesson, is that inside a right triangle the angle alone decides the proportions of the sides. Make the triangle twice as big and every side doubles, yet "opposite over hypotenuse" never budges. So we give those locked ratios names — sine, cosine, tangent — and from then on a single angle hands us a number, and a single number lets us measure heights and distances we could never reach with a tape. By the end you will read the special-angle table off two school set-squares, prove sin2θ + cos2θ = 1 with nothing but the Pythagorean theorem, and solve a whole triangle from one side and one angle.
You already met slope with linear functions: a line's steepness is riserun — how far it climbs for each step across. A wheelchair ramp that rises 1 m over a 12 m run has slope 112 ≈ 0.083. That number answers "how steep?", but it does not directly answer "at what angle?" — and a builder's protractor speaks in degrees.
Here is the bridge. Draw the ramp as a right triangle: the run is the horizontal leg, the rise is the vertical leg, and the angle θ the ramp makes with the ground sits at the bottom. The slope riserun is exactly the ratio "opposite over adjacent" for that angle — and that ratio is what we will name the tangent of θ. So slope and angle are two languages for one thing: slope = tan θ. A 1-in-12 ramp sits at tan−1(0.083) ≈ 4.8°, gentle enough for a wheelchair.
Steepness is a ratio of two lengths, and that ratio is fixed by the angle. Trigonometry is the dictionary that translates between the angle and the ratio.
Pick one acute angle θ of a right triangle. The three sides now have roles relative to that angle: the hypotenuse is the long side facing the right angle (it never changes role); the opposite leg is the one across the room from θ; the adjacent leg is the one touching θ. From these we build exactly three ratios:
| name | ratio | memory |
|---|---|---|
| sin θ | oppositehypotenuse | SOH |
| cos θ | adjacenthypotenuse | CAH |
| tan θ | oppositeadjacent | TOA |
Say it aloud once — SOH-CAH-TOA — and you own the definitions for life. But why is each ratio a property of the angle and not of the particular triangle you drew? Because of similar triangles. Any two right triangles that share the acute angle θ are similar (they agree in all three angles), and similar triangles have proportional sides. Scale the triangle up by a factor of 3 and the opposite and the hypotenuse both triple, so their ratio is unchanged. The size washes out; only the angle survives. That is what makes "sin θ" a well-defined number.
"Opposite" and "adjacent" are relative to the angle you chose. Pick the other acute angle of the same triangle and the two legs swap roles, so sin and cos trade places. Sine is not "the top side" or "the tall side" — it is opposite-over-hypotenuse for your angle. Always name the angle first.
Two triangles from any geometry set — the school set-squares — hand us exact values, no calculator needed. Reuse the Pythagorean theorem on each.
The 45° set-square is a right isosceles triangle: two legs of length 1, so the hypotenuse is √(12+12) = √2. Therefore sin 45° = cos 45° = 1√2 = √22 ≈ 0.707, and tan 45° = 11 = 1.
The 30°–60° set-square is half of an equilateral triangle of side 2. Cutting it down the middle gives a right triangle with hypotenuse 2, short leg 1 (half the base, opposite the 30°), and long leg √(22−12) = √3 (opposite the 60°). So sin 30° = 12, cos 30° = √32, and reading the 60° angle from the same picture, sin 60° = √32, cos 60° = 12.
tan 60° = sin 60°cos 60° = √3 / 21 / 2 = √3 ≈ 1.732 — the long leg is √3 times the short one, just as the picture shows.
The three ratios for one angle are not independent — two beautiful relations bind them. Imagine the triangle scaled so the hypotenuse is exactly 1. Then the opposite leg is sin θ and the adjacent leg is cos θ. Drop those two legs into the Pythagorean theorem:
sin2θ + cos2θ = 1
This is the most-used identity in all of trigonometry, and it is just Pythagoras wearing a costume. (You will meet it again on the unit circle in §25.3, where the hypotenuse-of-1 becomes the circle's radius.) The second relation comes straight from the definitions:
tan θ = sin θcos θ
because opp/hypadj/hyp = oppadj, the hypotenuses cancelling. Check it on 45°: tan 45° = (√2/2) ÷ (√2/2) = 1. ✓
"Solving" a triangle means finding every unknown side and angle. In a right triangle one angle is already 90°, so one side plus one acute angle is enough to find everything else. The recipe: name your angle, decide which ratio links the side you know to the side you want, then solve.
A 6 m ladder leans against a wall at 70° to the ground. How high up the wall does it reach? The ladder is the hypotenuse, the height is opposite the 70° angle — that calls for sine:
height = 6 · sin 70° = 6 · 0.940 ≈ 5.64 m.
And the foot sits 6 · cos 70° ≈ 2.05 m from the wall.
Choosing the ratio is the whole skill: known and wanted are both legs → tangent; one of them is the hypotenuse → sine or cosine depending on whether the other is opposite or adjacent. When you need an angle instead, run the ratio backward with the inverse keys (sin−1, cos−1, tan−1).
Now the payoff. An angle of elevation is measured up from the horizontal to a sight-line; an angle of depression is measured down from the horizontal (the two are equal when you look back, being alternate angles across the parallel horizontals). A bearing is a direction given as an angle clockwise from north — "N 40° E" or simply "070°". In each case you build a right triangle from the sight-line and read a real distance off a ratio.
From 50 m away, the angle of elevation to the top of a tower is 38°. The tower height is opposite, the 50 m ground distance is adjacent, so tangent links them:
height = 50 · tan 38° = 50 · 0.781 ≈ 39.1 m.
The next widget makes this live: dial the sun's elevation and watch a tower's height fall out of its shadow.
A higher sun makes a shorter shadow for a fixed tower — but here we fix the shadow and dial the angle, so a steeper angle means a taller tower casting that same shadow. Always sketch the triangle and label which side is opposite the angle before reaching for a button.
Fix an acute angle of a right triangle and three side-ratios lock: sin θ = opp/hyp, cos θ = adj/hyp, tan θ = opp/adj (SOH-CAH-TOA). They depend on the angle alone because similar triangles share their proportions. Memorize the special-angle table and the two same-angle identities:
| θ | 30° | 45° | 60° |
|---|---|---|---|
| sin θ | 12 | √22 | √32 |
| cos θ | √32 | √22 | 12 |
| tan θ | √33 | 1 | √3 |
sin2θ + cos2θ = 1 · tan θ = sin θ / cos θ
Next, in §25.2, we free the angle from the right-triangle cage by letting a ray spin — and measure the turn in radians.
Six questions to lock it in. Tap the answer you think is right.
This lesson covers the right-triangle definitions of sine, cosine, and tangent and their use in applied problems — aligned with the Common Core high-school geometry trigonometry cluster HSG-SRT.C.6 (ratios in similar right triangles define the trig functions), HSG-SRT.C.7 (the relationship between sine and cosine of complementary angles, here as sin2θ + cos2θ = 1 and the role-swap when the angle changes), and HSG-SRT.C.8 (using trig ratios and the Pythagorean theorem to solve right triangles in applied problems, including angles of elevation, depression, and bearing). Similar-triangle reasoning from Stage 17 is the justification students should be able to give for why the ratios are well defined.