Special Right Triangles and the Unit Circle

The angle decides the ratios

The same right-triangle ratios extend to unit-circle coordinates and to geometric contexts such as angles of elevation.

Earlier work with right triangles built the fluency: because two right triangles sharing an acute angle are always similar, the angle alone fixes every ratio of the sides. That fact, the angle decides the ratios, is worth carrying forward as this topic pushes it further. Sine, cosine, and tangent are simply the three ways to pair those sides.

Figure 1. Relative to the marked angle, the sides earn names: opposite, adjacent, and the hypotenuse. The angle alone fixes their ratios.

As a quick refresher: from angle $\theta$, one leg is opposite, one is adjacent, and the hypotenuse faces the right angle.

$\sin\theta = \frac{\text{opp}}{\text{hyp}}$, $\quad \cos\theta = \frac{\text{adj}}{\text{hyp}}$, $\quad \tan\theta = \frac{\text{opp}}{\text{adj}}$

SOH-CAH-TOA is that pairing compressed into six letters, nothing more.

The 45-45-90 and 30-60-90 triangles

Two right triangles are worth recognizing on sight, with no calculator needed. The half-square, a square cut along its diagonal, gives the $45$-$45$-$90$ ratio $1 : 1 : \sqrt2$.

Figure 2. The $45$-$45$-$90$ triangle: equal legs, hypotenuse $\sqrt2$ times a leg. Its angles force the side ratio $1 : 1 : \sqrt2$.

The half-equilateral, an equilateral triangle cut down the middle, gives the $30$-$60$-$90$ ratio $1 : \sqrt3 : 2$.

Figure 3. The $30$-$60$-$90$ triangle: side ratio $1 : \sqrt3 : 2$, with the shortest side opposite the smallest angle.

Reading ratios by picture rather than by memorized table is worth the habit: reach for the triangle before the calculator, since it hands back the cleaner exact form, $\frac{\sqrt3}{2}$, rather than a decimal like $0.866\ldots$. In a $30$-$60$-$90$ triangle, the side opposite $30^\circ$ is the short leg and the hypotenuse is exactly twice it. If the side opposite $30^\circ$ is $7$, the hypotenuse is $14$ and the other leg is $7\sqrt3$. A common slip is to hand back $7\sqrt3$, the long leg, when the hypotenuse was asked for.

Worked: a radical perimeter hides a special triangle. A square of side $s$ is cut along its diagonal. One of the resulting right triangles has perimeter $10 + 5\sqrt2$. What is $s$? Cutting a square along its diagonal is exactly the half-square construction: legs $s$ and $s$, hypotenuse $s\sqrt2$. The perimeter is their sum: $s + s + s\sqrt2 = 10 + 5\sqrt2$, so $s(2 + \sqrt2) = 5(2 + \sqrt2)$ and $s = 5$. The radical is not a separate quantity to isolate; it is part of the same factor multiplying $s$ on both sides. Recognizing $10 + 5\sqrt2$ as $5(2 + \sqrt2)$, the constant matched to the constant and the radical term to the radical term, turns the equation into a one-step division instead of a fight with an unfamiliar radical equation.

Not every right triangle is special. Legs of $8$ and $15$ give a hypotenuse of $\sqrt{64 + 225} = \sqrt{289} = 17$. The smaller acute angle sits opposite the shorter leg, so its sine is $\frac{8}{17}$ and its cosine is $\frac{15}{17}$. Always identify which leg is opposite the angle in the question before choosing a ratio.

Complementary angles: sine and cosine trade places

One more relationship is worth a refresher: a right triangle's two acute angles are complementary, since the third angle is $90^\circ$. If one is $\theta$, the other is $90^\circ - \theta$, and the same leg is opposite one angle and adjacent to the other.

Figure 4. The same leg is "opposite" to $\theta$ and "adjacent" to $90^\circ - \theta$. That single fact is why sine and cosine trade places across complementary angles.

Which means $\sin\theta = \cos(90^\circ - \theta)$ and $\cos\theta = \sin(90^\circ - \theta)$.

This is why a question that states $\cos(x^\circ) = \sin(58^\circ)$ is not asking for a calculator at all. It is asking whether you know the two angles are partners. If both are acute, then $x$ and $58$ must add to $90$, so $x = 32$. The whole item is just the complementary relationship in disguise. The same idea handles an equation like $\sin(5x^\circ) = \cos((x + 12)^\circ)$ with acute angles: set $5x + (x + 12) = 90$ and solve.

Radians and the unit circle

The Circles unit built the full radian system: the conversion $\pi = 180^\circ$, and arc length as $r\theta$. The one addition here is where radians and the trig ratios meet, which the unit circle makes visual. Place a circle of radius $1$ at the origin and walk a point counterclockwise from the positive $x$-axis through an angle $\theta$. Drop that point straight down to the $x$-axis, and you have built a right triangle whose hypotenuse is the radius $1$.

Figure 5. A point on the unit circle sits at $(\cos\theta, \sin\theta)$: the same opposite-and-adjacent legs, now read as coordinates.

The horizontal leg is the adjacent side, so it has length $\cos\theta$; the vertical leg is opposite, so it has length $\sin\theta$. The point's coordinates are therefore $(\cos\theta, \sin\theta)$. Nothing new was defined; the same two legs you already named are simply being read as a horizontal and a vertical distance. This is the quiet doorway between the trigonometry of triangles and the trigonometry of the whole plane. It also means every point on the unit circle satisfies $x^2 + y^2 = 1$. If $x = -\frac35$, then $y^2 = 1 - \frac{9}{25} = \frac{16}{25}$, and the quadrant decides whether $y$ is $\frac45$ or $-\frac45$.

That is only the first quadrant of the story. Walk the point past $90^\circ$ and it keeps producing coordinates, $(\cos\theta, \sin\theta)$, the whole way around the circle; each quadrant simply changes which way the two legs point. A leg pointing left of the origin is negative; one pointing below it is negative too. Track only that, and every quadrant's signs fall out for free.

• Quadrant I, $0^\circ$ to $90^\circ$: $\cos\theta$ and $\sin\theta$ both positive.
• Quadrant II, $90^\circ$ to $180^\circ$: $\sin\theta$ positive, $\cos\theta$ negative.
• Quadrant III, $180^\circ$ to $270^\circ$: $\cos\theta$ and $\sin\theta$ both negative.
• Quadrant IV, $270^\circ$ to $360^\circ$: $\cos\theta$ positive, $\sin\theta$ negative.

Every angle outside the first quadrant has a twin inside it: its reference angle, the acute angle between its terminal side and the $x$-axis. The reference angle sets the size of $\cos\theta$ and $\sin\theta$; the quadrant sets the sign. The reference angle of $150^\circ$ is $30^\circ$; of $210^\circ$, also $30^\circ$; of $300^\circ$, $60^\circ$. Pair that reference angle with the special-triangle ratios already in hand, attach the quadrant's sign, and an angle anywhere on the circle reduces to arithmetic you have already done.

Worked: find $\cos 120^\circ$. $120^\circ$ sits in Quadrant II, $30^\circ$ past the $90^\circ$ mark, so its reference angle is $180^\circ - 120^\circ = 60^\circ$. The reference value is $\cos 60^\circ = \frac12$. Quadrant II makes cosine negative, so $\cos 120^\circ = -\frac12$.

The same move runs in radians without a fresh rule: $120^\circ$ is $\frac{2\pi}{3}$, and $\pi - \frac{2\pi}{3} = \frac{\pi}{3}$ is the reference angle in radian form, $60^\circ$ again. For $\frac{7\pi}{6}$, which is $210^\circ$ in Quadrant III, the reference angle is $30^\circ$, so the point is $\left(-\frac{\sqrt3}{2}, -\frac12\right)$. Desmos can confirm any of this, but the angle mode must match the angle you typed: switch Desmos to degree mode before graphing $y = \cos(x)$ and reading a degree value, or convert to radians first and stay in radian mode. On a question with no graphing step offered, the quadrant and the reference angle are the route there regardless.

Trig ratios from a point's coordinates

The unit circle used a radius of $1$ to make the coordinates equal $(\cos\theta, \sin\theta)$ exactly, but the same right triangle forms for a point at any distance from the origin. For a point $(x, y)$ on the terminal side of an angle $\theta$ in standard position, $\tan\theta$ is simply $\frac{y}{x}$, the vertical distance over the horizontal one, the same ratio that defines slope.

Worked. Point $(3, 4)$ lies on the terminal side of angle $\theta$, measured from the positive $x$-axis. What is $\tan\theta$? $\tan\theta = \frac{y}{x} = \frac43$. No radius, hypotenuse, or triangle needs to be drawn: $\tan\theta$ is already the slope of the segment from the origin to $(3, 4)$. Finding $\sin\theta$ or $\cos\theta$ from the same point does require the distance to the origin, $\sqrt{3^2 + 4^2} = 5$, giving $\sin\theta = \frac45$ and $\cos\theta = \frac35$; tangent alone skips that step because it is a ratio of the two coordinates to each other, not to the radius.

If the point is in another quadrant, the signs of the coordinates carry through. For $(-5, 12)$, $\tan\theta = \frac{12}{-5} = -\frac{12}{5}$, and the distance to the origin is $13$, so $\sin\theta = \frac{12}{13}$ and $\cos\theta = -\frac{5}{13}$.

Mixed problems: angles of elevation and depression

Most word-problem trigonometry arrives through a line of sight. You look up at the top of something, and the angle your gaze makes with the level ground is the angle of elevation; look down, and it is the angle of depression. Either way, the line of sight, the ground, and the object's height frame a right triangle, and one angle plus one side unlocks the rest.

Figure 6. An angle of elevation builds a right triangle from the line of sight, the level ground, and the height. One angle and one side determine the others.

Suppose you stand $40$ feet from the base of a flagpole and the angle of elevation to its top is $30^\circ$. The height is the side opposite the angle, and the $40$ feet is the side adjacent, so the ratio that pairs them is tangent: $\tan 30^\circ = \frac{h}{40}$, so $h = 40\tan 30^\circ = \frac{40}{\sqrt3} = \frac{40\sqrt3}{3} \approx 23.1$ ft. Notice the exact form $\frac{40\sqrt3}{3}$ carries the answer more honestly than the rounded $23.1$, and notice too that the whole setup was one decision: which ratio pairs the side I have with the side I want? That single question handles the great majority of these.

An angle of depression is measured from the horizontal line through the observer, looking down. By alternate interior angles, it equals the angle of elevation from the object back up to the observer, so you can draw the same right triangle either way.

Use the answer choices to check the target

Trigonometry questions tend to be engineered around a small set of predictable confusions, and the choices usually display them. When two options are reciprocals of each other, the question is testing opposite-against-adjacent: whether you reached for tangent or its flip. When a sine value and a cosine value both appear, the item is often about the complementary partner, $90^\circ - \theta$. When a clean radian value sits beside a degree value, the conversion $180^\circ = \pi$ is the real subject. When both $\theta$ and $90^\circ - \theta$ appear among the choices, check which angle the requested ratio actually uses. Read the spread of choices as a description of what the question is actually about, then answer that.

Summary

One acute angle fixes the shape of a right triangle, and a fixed shape fixes every ratio of its sides. Sine, cosine, and tangent are those three ratios named; the special triangles let you read them exactly; the complementary relationship is two corners sharing one leg; radians are the same angles measured by their arcs; and the unit circle reads the same legs as coordinates, with the reference angle setting the size and the quadrant setting the sign. Carry the one idea, the angle decides the ratios, and the rest stops being a list to memorize and becomes a picture you can rebuild whenever you need it.