Angles, radians and right-triangle ratios

An angle can record a rotation, including its direction

Imagine a ray turning about its endpoint. Its starting position is the initial ray and its final position is the terminal ray. A degree is 1/360 of a full turn, so a quarter turn is 90°, a half turn is 180° and one full turn is 360°. For an oriented rotation, counterclockwise is positive and clockwise is negative.

To compare rotations in a diagram, we often put the vertex at the origin and the initial ray along the positive horizontal axis. This is called standard position. A rotation of −90° turns clockwise through a quarter turn, whereas 270° turns counterclockwise through three quarters of a turn. Both finish on the same downward ray, but they describe different rotations.

Adding an integer number of full turns preserves the terminal ray: d° and (d + 360n)° finish together for every integer n. Such angles are called coterminal. A rotation of 450° contains a full turn followed by 90°; replacing it by 90° preserves its terminal ray while losing the extra turn. Always decide whether a question asks for the full rotation or only its final direction.

Radians compare an arc length with its positive radius

Picture a quarter-turn on a circle of radius 3 cm. Its arc is one quarter of the 6π cm circumference, so its length is 3π/2 cm. Dividing that arc length by the radius gives (3π/2)/3 = π/2. This quotient describes the angle without keeping a centimetre unit.

Take a circle of radius r > 0. For a nonnegative central angle tracing an arc of length s, define its radian measure by θ = s/r. The arc length and radius must be expressed in the same length unit. One radian is the angle that traces an arc whose length equals the radius. A larger circle gives proportionally larger arcs for the same angle, so their quotient is unchanged.

A full circumference has length 2πr. Consequently one full turn measures 2πr/r = 2π radians, a half turn measures π radians and a quarter turn measures π/2 radians. The radius cannot be zero in this definition: division by zero is unavailable, and a single point does not supply an ordinary circle or its arc.

The quotient s/r has no remaining length unit. Radians are therefore dimensionless, but the label “rad” tells us which angle measure a number represents. A bare number such as 60 must not be substituted into a formula requiring radians when the given angle is 60°. Keep exact multiples of π until a decimal approximation is actually needed.

Signed radian measure follows the same direction convention as degrees, and accumulated rotations may exceed 2π. For example, a clockwise quarter turn is −π/2. Its negative sign records direction; the physical length of the travelled quarter-circle arc is positive. An unsigned arc length and a signed rotation are different quantities.

Conversion changes the unit, not the rotation

Since 180° equals π radians, a degree measure d converts to θ = (d/180)π radians. Conversely, a radian measure θ converts to d = 180θ/π degrees. These formulas apply to positive, negative and multi-turn rotations. They preserve the sign and the number of turns; conversion is not a reduction to a preferred terminal direction.

For instance, 150° becomes (150/180)π = 5π/6 radians. A rotation of −90° becomes −π/2, while 450° becomes 5π/2. To find a coterminal angle afterwards, add or subtract integer multiples of 2π. Choosing the representative in 0 ≤ θ < 2π is a separate operation: it sends −π/2 to 3π/2 and 5π/2 to π/2.

The number of signed turns is d/360, equivalently θ/(2π). This gives a useful check on both conversions. The fraction −1/4 means one clockwise quarter turn, while 5/4 means one and a quarter counterclockwise turns. A terminal-ray diagram alone cannot show that distinction.

Use a fraction of one full circle for an ordinary sector

Use the same radius of 3 cm and a counterclockwise quarter-turn. The curved arc is a length, while the swept quarter-disc is a region with an area. Predict which result needs centimetres and which needs square centimetres. One quarter of the 9π cm² disc area is 9π/4 cm²; the arc remains 3π/2 cm.

For this lesson's ordinary once-traced sector, require r > 0 and a radian angle 0 ≤ θ ≤ 2π. The angle specifies a counterclockwise sweep from the initial radius, so it chooses which region is swept even when the two radii alone could enclose two different sectors. Interior angles in this interval may describe a minor or a major sector. The endpoints are included as the zero-sweep and full-disc cases.

The swept fraction of a full turn is θ/(2π). Multiplying this fraction by the circumference gives s = [θ/(2π)] × 2πr = rθ. Multiplying it by the disc area gives S = [θ/(2π)] × πr² = r²θ/2. These use the established proportionality of a central angle with its intercepted arc and once-covered sector area. The derivation works for every admissible r and θ, independently of any selected example.

The angle in those formulas must be in radians. If the given angle is d degrees, convert first or use s = (d/360) × 2πr and S = (d/360) × πr² with 0 ≤ d ≤ 360. With radius measured in centimetres, s is in centimetres and S is in square centimetres; a coincidental equality of their numerical values does not make them the same quantity.

At θ = 0, both the swept arc length and sector area are zero, although the supporting circle still has positive radius. At θ = 2π, they are the entire circumference and disc area. These two sweeps have the same terminal ray but different swept quantities. A negative or larger-than-2π rotation remains a valid angle; it lies outside this particular once-traced sector convention. Repeated winding does not make the area of a region covered once larger than the full disc.

Side names depend on the chosen acute angle

The right triangle with legs 3 cm and 4 cm and hypotenuse 5 cm gives a concrete starting point. Its right angle is where the two legs meet; choose one of its other angles before naming a leg opposite or adjacent. We use the established triangle angle sum, equal-side angle property, Pythagoras and its converse, and AA/SSS criteria in the proofs below.

Now choose an acute angle θ in a nondegenerate right triangle: 0 < θ < π/2, or 0° < θ < 90°. The hypotenuse h is opposite the right angle. Relative to θ, let o be the opposite leg and a the adjacent leg. “Adjacent” here means the leg meeting θ, not the hypotenuse, which also meets that vertex. Choosing the other acute angle exchanges o and a but leaves h unchanged.

In a convenient perpendicular coordinate frame with one common length unit, put A = (0, 0), B = (a, 0) and C = (0, o), with o > 0 and a > 0. The right angle is at A and the chosen angle θ is at B. Thus AC is opposite θ, AB is its adjacent leg and BC is the hypotenuse. These coordinates are justified by the known right angle; imposing them on an arbitrary triangle would assume something not yet proved.

When side data are supplied as a proposed right triangle, first convert every length to the same unit. Require o > 0, a > 0, h > 0 and o² + a² = h². Pythagoras and its converse justify the right-triangle classification. Positive o and a give h > o and h > a and strict triangle inequalities, so the triangle is nondegenerate and its other angles are acute.

Data (o, a, h) = (0, 4, 4) pass the squared equality but have a zero side and no nondegenerate triangle. Data (3, 4, 6) have positive sides and satisfy the strict triangle inequalities, so an ordinary triangle exists. They fail 3² + 4² = 6², however, and do not define the proposed right triangle. Distinguish a failed right-angle hypothesis from a claim that no triangle is possible.

Sine, cosine and tangent are ratios of lengths

Choose the angle at B in a right triangle whose opposite leg is 3 cm, adjacent leg 4 cm and hypotenuse 5 cm. Three useful comparisons are 3/5, 4/5 and 3/4: opposite to hypotenuse, adjacent to hypotenuse, and opposite to adjacent. The next paragraph gives these comparisons their names.

For the chosen acute angle in the valid right triangle, define sin θ = o/h, cos θ = a/h and tan θ = o/a. The words sine, cosine and tangent name these three different ratios. The notation sin θ means the sine of θ; it is not a product of a variable called “sin” with θ. State the chosen angle before assigning its opposite and adjacent legs.

Each ratio divides one length by another expressed in the same unit, so the length unit cancels. The result is dimensionless: (3 m)/(5 m) = 3/5, with no metre remaining. All three ratios are positive in this acute-angle setting. Because h is longer than either leg, 0 < sin θ < 1 and 0 < cos θ < 1. Tangent can be less than, equal to or greater than 1, according to the two legs.

These definitions presently apply to acute angles in nondegenerate right triangles. They do not yet define the sine, cosine or tangent of 0°, 90°, negative angles or full-turn angles. Angles of those sizes are valid rotations from the earlier sections; extending trigonometric functions to them needs the unit-circle construction in the next lesson.

Similarity proves that the angle determines the ratios

Scale the 3, 4, 5 side lengths by 3/2. The new lengths are 9/2, 6 and 15/2 in the same unit. Before calculating, predict whether the ratios, perimeter and area stay the same. The paired drawing uses one coordinate frame so corresponding lengths can be compared directly; the general reason follows.

Why may we call these ratios values of an angle instead of values of one triangle? Take any two nondegenerate right triangles with the same acute angle θ. Both also have a right angle, so their third angles agree. The angle-angle similarity theorem gives a common positive scale factor k between corresponding side lengths: o′ = ko, a′ = ka and h′ = kh.

Cancelling this nonzero factor gives o′/h′ = ko/(kh) = o/h, a′/h′ = ka/(kh) = a/h and o′/a′ = ko/(ka) = o/a. This proves the three ratios agree for every such pair of triangles. A set of numerical examples may check the reasoning, but the arbitrary positive k and similarity theorem supply the general proof.

Changing the common length unit is another positive multiplication of all numerical side lengths, and therefore leaves the ratios unchanged. Mixed units must first be normalised. For lengths 3 m, 400 cm and 5 m, convert 400 cm to 4 m before checking Pythagoras or taking numerical ratios. The raw numbers 3, 400 and 5 do not express three lengths in one unit.

For side-length scaling, k > 0 is essential. A multiplier of zero collapses all sides, so the fractions would have zero denominators and the acute triangle disappears. A negative multiplier produces negative proposed physical lengths. Signed coordinate transformations use a different convention; they do not make a negative number an admissible physical side-length multiplier here.

The other acute angle exchanges sine and cosine

The two acute angles in a right triangle sum to π/2. If the chosen angle is θ, call the other φ = π/2 − θ. Both remain acute. The leg opposite θ is adjacent to φ, and the leg adjacent to θ is opposite φ. The hypotenuse stays the same.

Using the definitions with these exchanged roles gives sin φ = a/h = cos θ, cos φ = o/h = sin θ and tan φ = a/o = 1/(tan θ). Every denominator is positive under the stated hypotheses. This explains the complementary-angle relations directly from a triangle, without a rounded angle measurement.

Pythagoras also gives (sin θ)² + (cos θ)² = (o² + a²)/h² = 1. Dividing sine by cosine gives (sin θ)/(cos θ) = (o/h)/(a/h) = o/a = tan θ. Here the squares apply to the ratio values, and cosine is nonzero because θ is acute. These are general identities within the present acute-angle scope, not a definition of the functions for all rotations.

An isosceles right triangle gives exact 45° values

Start with a right triangle whose two legs are each 1 unit long. Equal legs make the acute angles equal; their sum is 90°, so each is 45°. Now keep the same construction at any size, with both legs of length ℓ > 0. Equal legs have equal opposite angles. The two acute angles sum to 90°, so each is 45°, or π/4 radians. Pythagoras gives h² = ℓ² + ℓ² = 2ℓ² and hence h = ℓ√2; the positive root is required for a length.

Relative to either acute angle, the opposite and adjacent legs are both ℓ. Thus sin 45° = cos 45° = ℓ/(ℓ√2) = 1/√2 = √2/2, and tan 45° = ℓ/ℓ = 1. Multiplying numerator and denominator of 1/√2 by √2 explains the equivalent exact form √2/2.

The proof uses arbitrary positive ℓ, so it describes every size of this special right triangle. Setting ℓ = 1 makes a convenient diagram, while similarity explains why changing the size cannot change the ratios. A decimal such as 0.707 is only an approximation to √2/2.

Halving an equilateral triangle gives exact 30° and 60° values

Begin with an equilateral triangle of side 2 units and mark the midpoint of its base, making two base halves of length 1. More generally, use side 2ℓ with ℓ > 0 and join the top vertex to the midpoint of the base. The resulting triangles have equal long sides, equal base halves and a shared middle segment. Side-side-side congruence makes their angles at the midpoint equal; those adjacent angles sum to 180°, so each is 90°. Their equal angles at the top split 60° into two 30° angles, while the base angles remain 60°.

Each right triangle has hypotenuse 2ℓ and short leg ℓ. If its other leg is t > 0, Pythagoras gives t² = (2ℓ)² − ℓ² = 3ℓ², so t = ℓ√3. Relative to 30°, the opposite leg is ℓ and the adjacent leg is ℓ√3. The ratios are sin 30° = 1/2, cos 30° = √3/2 and tan 30° = 1/√3 = √3/3.

Relative to the complementary 60° angle, the two legs exchange roles. Consequently sin 60° = √3/2, cos 60° = 1/2 and tan 60° = √3. These values and the 45° values come from exact geometry and algebra. A measured sketch cannot replace their proofs, and rounding the radicals would lose their exactness.

Compare exact cases, then connect each check to its general reason

The explorer has four scenes with eight choices each. Conversion keeps the complete signed rotations 0°, 30°, 45°, 60°, 90°, 180°, −90° and 450°. Compare each exact radian value and signed turn count with its separately reduced terminal direction. Equal terminal rays do not erase different directions of travel or whole turns. The circle in this scene illustrates rotation and radians; it does not introduce global trigonometric functions.

The ratio scene uses (opposite, adjacent, proposed hypotenuse) triples (3, 4, 5), (4, 3, 5), (5, 12, 13), (12, 5, 13), (8, 15, 17), (7, 24, 25), (0, 4, 4) and (3, 4, 6), with one common length unit. The article and tables both use o for opposite and a for adjacent. For admissible right-triangle data, its diagram uses A = (0, 0), B = (adjacent, 0), C = (0, opposite), with the chosen acute angle at B. Check positivity and the exact Pythagorean equality before accepting any ratio as a right-triangle sine, cosine or tangent. A zero leg is degenerate; the last triple is an ordinary non-right triangle rather than a valid right-triangle input. The invalid right-triangle choices retain their exact checks without a diagram asserting that right triangle.

The similarity scene multiplies the base lengths (3, 4, 5) by k = 1/2, 1, 2, 3/2, 3, 11/5, 0 or −1. Positive choices preserve the three ratios by the general cancellation proof. Zero and negative choices fail the physical side-length requirement, even though formal arithmetic can still be written. A handful of matching ratios illustrates the proof; it does not establish invariance for every positive scale.

The sector scene lists (radius, radian angle divided by π) as (2, 1/6), (3, 1/2), (1/2, 1), (4, 2), (2, 0), (0, 1/2), (2, −1/2) and (2, 5/2). Apply r > 0 and 0 ≤ θ ≤ 2π before drawing an ordinary once-traced sector or assigning its length and area. The positive-radius zero and full sweeps are valid endpoints. The last three choices fail this construction's guards; signed and multi-turn angle conversion remains valid. Read the exact formulas and proof beside each case, and keep all 32 selected checks separate from the general arguments above.

Worked examples

Try it yourself

1. Convert a counterclockwise rotation of 150° to radians exactly. What fraction of a full turn does it represent?

   Show the answer and reasoning

   150° = 5π/6 radians, representing 5/12 of a counterclockwise full turn.

   Multiply the degree measure by π/180: 150π/180 = 5π/6. The signed turn count is 150/360 = 5/12, also (5π/6)/(2π). Both calculations preserve the positive rotation rather than replacing it by another angle.

2. Convert −90° and 270° to radians. Starting from the same initial ray, do they have the same terminal ray and do they describe the same signed rotation?

   Show the answer and reasoning

   −90° = −π/2 and 270° = 3π/2 radians. They have the same terminal ray but different signed rotations, differing by one positive full turn of 2π.

   The conversions multiply each signed degree measure by π/180. Their difference is 270° − (−90°) = 360°, or 3π/2 − (−π/2) = 2π. One turns clockwise through a quarter turn and the other counterclockwise through three quarters; reducing the terminal direction must not erase that information.

3. A circle has radius 2 cm and a once-traced sector angle of 60°. Find its arc length and sector area exactly. Are they the same physical quantity?

   Show the answer and reasoning

   The arc length is 2π/3 cm and the sector area is 2π/3 cm². Their numerical coefficients agree, but they are different quantities with different units.

   Convert 60° to π/3 radians and check r > 0 and 0 ≤ π/3 ≤ 2π. Then s = 2 × π/3 = 2π/3 cm, while S = 2² × (π/3)/2 = 2π/3 cm². One measures a curve and the other a region, so the matching numbers do not identify the quantities.

4. A proposed right triangle has opposite leg 8 units, adjacent leg 15 units and hypotenuse 17 units relative to an acute angle θ. Verify the data and find its three ratios.

   Show the answer and reasoning

   8² + 15² = 17² verifies the positive right-triangle data. For the stated θ, sin θ = 8/17, cos θ = 15/17 and tan θ = 8/15.

   The three lengths are positive in a common unit, and 64 + 225 = 289 = 17². Use the supplied angle's side roles rather than exchanging the legs: sine divides the opposite leg by the hypotenuse, cosine divides the adjacent leg by the hypotenuse, and tangent divides the opposite leg by the adjacent leg. All units cancel.

5. In a 3–4–5 right triangle, θ is opposite the 3-unit leg. Let φ be the other acute angle. Express φ in terms of θ and find sin φ, cos φ and tan φ.

   Show the answer and reasoning

   φ = π/2 − θ. Its ratios are sin φ = 4/5, cos φ = 3/5 and tan φ = 4/3.

   The two acute angles are complementary. Relative to θ, the opposite and adjacent legs are 3 and 4; relative to φ they are 4 and 3. The hypotenuse stays 5. Thus sine and cosine exchange values and tangent becomes the reciprocal: 1/(3/4) = 4/3. Both angles are acute, so no denominator is zero.

6. Multiply the opposite leg, adjacent leg and hypotenuse of a 3–4–5 right triangle by k = 3/2. Give the new lengths and three ratios. Explain why every positive multiplier would preserve those ratios.

   Show the answer and reasoning

   The new lengths are 9/2, 6 and 15/2 units. The ratios remain 3/5, 4/5 and 3/4. For every k > 0, the common factor cancels from ko/(kh), ka/(kh) and ko/(ka).

   Multiply all three original lengths by the same positive factor. The new squared equality is (9/2)² + 6² = 81/4 + 144/4 = 225/4 = (15/2)². Their ratios are (9/2)/(15/2) = 3/5, 6/(15/2) = 4/5 and (9/2)/6 = 3/4. Positivity keeps a nondegenerate triangle; the arbitrary-factor cancellation and similarity argument explain the general claim, beyond this one check.

7. Two triples (opposite, adjacent, proposed hypotenuse), all in one length unit, are (0, 4, 4) and (3, 4, 6). Which may be used as a nondegenerate right triangle for acute-angle ratios? Does rejecting the second as right-triangle data mean no ordinary triangle exists?

   Show the answer and reasoning

   Neither is a valid nondegenerate right-triangle input. The first has a zero side; the second fails 3² + 4² = 6². The positive lengths 3, 4 and 6 do form an ordinary non-right triangle.

   For (0, 4, 4), the equality 0² + 4² = 4² does not overcome the missing positive leg or supply an acute angle. For (3, 4, 6), 9 + 16 = 25 ≠ 36, so Pythagoras fails. Nevertheless 3 + 4 > 6, 3 + 6 > 4 and 4 + 6 > 3 are all strict, which allows a nondegenerate ordinary triangle. Its side quotients cannot be labelled the acute right-triangle ratios from this lesson.

What to carry forward

• Degrees and radians describe the same rotation in different units: d° = (d/180)π radians. Conversion preserves signs and whole turns.
• Coterminal angles finish on the same ray after differing by an integer number of full turns; they need not describe the same signed rotation.
• For a positive-radius circle, radian measure compares common-unit arc length and radius. Ordinary once-traced sectors with 0 ≤ θ ≤ 2π use s = rθ and S = r²θ/2, with radians and distinct length and area units.
• Acute right-triangle ratios require positive common-unit sides and the Pythagorean equality. Opposite and adjacent depend on the chosen angle; the hypotenuse is always opposite the right angle.
• Sine is opposite/hypotenuse, cosine is adjacent/hypotenuse and tangent is opposite/adjacent. Similarity and positive-factor cancellation prove these dimensionless ratios depend on the acute angle, not triangle size.
• Complementary acute angles exchange opposite and adjacent legs, so sine and cosine exchange and tangent becomes its reciprocal. Pythagoras gives (sin θ)² + (cos θ)² = 1 in this scope.
• Isosceles-right and halved-equilateral triangle proofs give exact 45°, 30° and 60° ratios. Selected explorer cases illustrate general arguments; all-angle unit-circle trigonometric functions come next.