The six trig functions of an acute angle

Explain it like I am five

Keep the ladder: 17 feet long, bottom 8 feet from the wall, top 15 feet up, and θ the angle where it meets the floor. A trig function is a machine: an angle goes in and a ratio comes out. A ratio compares two lengths by dividing one by the other.

Sine of θ, written sin θ, is opposite over hypotenuse: $15 over 17$ ≈ 0.88 (≈ means about), so 88% of the ladder's length becomes height. Cosine, cos θ, is adjacent over hypotenuse: $8 over 17$ ≈ 0.47, the share that becomes floor distance. Tangent, tan θ, is opposite over adjacent: $15 over 8$ = 1.875, the steepness, 1.875 feet up for every foot across.

Turn each fraction upside down for the other three: cosecant, csc θ = $17 over 15$ ; secant, sec θ = $17 over 8$ ; cotangent, cot θ = $8 over 15$ . Why only the angle matters: a 34-foot ladder at the same tilt reaches 30 feet up, and $30 over 34$ = $15 over 17$ . Doubling every side doubles the top and the bottom, and the 2s cancel.

In plain words

Imagine measuring a ladder's shape without caring about its size. Compare two side lengths by dividing: that comparison is a ratio. An angle goes into a trig function and a ratio comes out, like a vending machine taking a coin and returning one item. Sine compares opposite with hypotenuse, cosine compares adjacent with hypotenuse, and tangent compares opposite with adjacent. Flip those fractions for cosecant, secant, and cotangent. Their abbreviations are sin, cos, tan, csc, sec and cot; sin θ is read sine of theta. SOH-CAH-TOA names sine's opposite and hypotenuse, cosine's adjacent and hypotenuse, and tangent's opposite and adjacent. Three sides give six choices of top and bottom. Each side can be divided by either of the other two. These side ratios use an acute angle.

Every ratio uses two of these three side names, chosen from θ.

Reminder

Read the names. sin θ means sine of theta, not sin times θ. α is read alpha and can name the angle too.
A reciprocal. The reciprocal of $3 over 5$ is $5 over 3$ , because their product is 1.
Reduce. $9 over 15$ = $3 over 5$ : divide top and bottom by the same nonzero factor 3.
Rationalize. $1 over square root of 5$ × $square root of 5 over square root of 5$ = $square root of 5 over 5$ ; the multiplier is 1.
Missing side. With legs 8 and 15, hyp = $square root of 8 squared + 15 squared$ = 17. Add before taking the root.

Why it works. Enlarge a 3, 4, 5 triangle on a copier until its sides are 6, 8 and 10. Its angles stay the same, and sine stays the same too: = . Every side ratio keeps its value because the same scale factor appears on top and bottom and cancels. Triangles with the same angles are similar triangles, meaning scaled copies, so any right triangle with this angle gives the same ratios. That is why the angle alone determines the output of each trig function.

RuleFor an acute angle in a right triangle: sin θ = , cos θ = , tan θ = .
csc θ = , sec θ = , cot θ = .

The same idea, five ways

Say it

Sine of theta is opposite divided by hypotenuse.

Write it

A trigonometric function takes an angle as input and returns the indicated ratio as output.

In math

sin θ = $\frac{o p p}{h y p}$
cos θ = $\frac{a d j}{h y p}$
tan θ = $\frac{o p p}{a d j}$
csc θ = $\frac{h y p}{o p p}$
sec θ = $\frac{h y p}{a d j}$
cot θ = $\frac{a d j}{o p p}$

A ladder's height per foot of ladder describes its tilt even after you buy a longer ladder.

See it

Every ratio uses two of these three side names, chosen from θ.

The same idea, other ways

The ladder measurements

Height divided by ladder length is sine. Floor distance divided by ladder length is cosine. Height divided by floor distance is tangent. Turning each comparison around gives cosecant, secant and cotangent.

At the ladder's foot, opposite is height and adjacent is floor distance.

Small numbers

For opp = 3, adj = 4 and hyp = 5, sine is 3 ÷ 5, cosine is 4 ÷ 5 and tangent is 3 ÷ 4. Their flips are 5 ÷ 3, 5 ÷ 4 and 4 ÷ 3.

Name the sides from the marked angle before making a fraction.

The memory words

Read SOH as sine, opposite, hypotenuse; CAH as cosine, adjacent, hypotenuse; TOA as tangent, opposite, adjacent. The last two letters name top then bottom. Flip the matching fraction for the other three functions.

SOH: sin θ =

CAH: cos θ =

TOA: tan θ =

Each word gives a function and its numerator and denominator.

Why size cannot matter

A copy machine doubles 3, 4, 5 to 6, 8, 10. $6 over 10$ reduces to $3 over 5$ , and $6 over 8$ reduces to $3 over 4$ . You changed the size, but division cancels the enlargement.

3, 4, 5 become 6, 8, 10

Equal scaling preserves every ratio.

Function	Ratio	Its flip	Ratio
sin θ	$\frac{o p p}{h y p}$	csc θ	$\frac{h y p}{o p p}$
cos θ	$\frac{a d j}{h y p}$	sec θ	$\frac{h y p}{a d j}$
tan θ	$\frac{o p p}{a d j}$	cot θ	$\frac{a d j}{o p p}$

.1Sine (sin θ)

Sine compares wall height with ladder length. It asks what share of the hypotenuse is the side facing your chosen angle.

sin θ = $\frac{o p p}{h y p}$
This side ratio applies to an acute angle; both lengths are positive.

Sine uses opp on top and hyp on bottom.

The same idea, five ways

Say it

Read sin θ as sine of theta.

Write it

The sine of an acute angle is its opposite leg divided by its hypotenuse.

In math

sin θ = $\frac{o p p}{h y p}$

Sine compares wall height with ladder length. It asks what share of the hypotenuse is the side facing your chosen angle.

See it

Sine uses opp on top and hyp on bottom.

Worked exampleRead sin from a small triangle

opp = 3, adj = 4 and hyp = 5. Find sin θ. In words, divide opp by hyp.

Sine uses opp on top and hyp on bottom.

Write sin θ = $\frac{o p p}{h y p}$ .The definition identifies the numerator and denominator.
Substitute sin θ = $3 over 5$ .Use the side labels from the chosen angle.
Keep $3 over 5$ in reduced exact form.The two numbers have no common factor greater than 1.

Answer

sin θ = .

Check Double the side lengths to 6, 8 and 10. The new ratio is = , confirming the same value from a larger similar triangle.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Reverse top and bottom but keep the same function name.

Reversing the fraction creates the reciprocal function.

✓ Instead: sin θ = .

Tips and tricks

Say the numerator side first, then the denominator side.

.2Cosine (cos θ)

Cosine compares floor distance with ladder length. It asks what share of the hypotenuse is the leg beside your chosen angle.

cos θ = $\frac{a d j}{h y p}$
This side ratio applies to an acute angle; both lengths are positive.

Cosine uses adj on top and hyp on bottom.

The same idea, five ways

Say it

Read cos θ as cosine of theta.

Write it

The cosine of an acute angle is its adjacent leg divided by its hypotenuse.

In math

cos θ = $\frac{a d j}{h y p}$

Cosine compares floor distance with ladder length. It asks what share of the hypotenuse is the leg beside your chosen angle.

See it

Cosine uses adj on top and hyp on bottom.

Worked exampleRead cos from a small triangle

opp = 3, adj = 4 and hyp = 5. Find cos θ. In words, divide adj by hyp.

Cosine uses adj on top and hyp on bottom.

Write cos θ = $\frac{a d j}{h y p}$ .The definition identifies the numerator and denominator.
Substitute cos θ = $4 over 5$ .Use the side labels from the chosen angle.
Keep $4 over 5$ in reduced exact form.The two numbers have no common factor greater than 1.

Answer

cos θ = .

Check Double the side lengths to 6, 8 and 10. The new ratio is = , confirming the same value from a larger similar triangle.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Reverse top and bottom but keep the same function name.

Reversing the fraction creates the reciprocal function.

✓ Instead: cos θ = .

Tips and tricks

Say the numerator side first, then the denominator side.

.3Tangent (tan θ)

Tangent compares wall height with floor distance, like a ramp's steepness. The hypotenuse is not involved.

tan θ = $\frac{o p p}{a d j}$
This side ratio applies to an acute angle; both lengths are positive.

Tangent uses opp on top and adj on bottom.

The same idea, five ways

Say it

Read tan θ as tangent of theta.

Write it

The tangent of an acute angle is its opposite leg divided by its adjacent leg.

In math

tan θ = $\frac{o p p}{a d j}$

Tangent compares wall height with floor distance, like a ramp's steepness. The hypotenuse is not involved.

See it

Tangent uses opp on top and adj on bottom.

Worked exampleRead tan from a small triangle

opp = 3, adj = 4 and hyp = 5. Find tan θ. In words, divide opp by adj.

Tangent uses opp on top and adj on bottom.

Write tan θ = $\frac{o p p}{a d j}$ .The definition identifies the numerator and denominator.
Substitute tan θ = $3 over 4$ .Use the side labels from the chosen angle.
Keep $3 over 4$ in reduced exact form.The two numbers have no common factor greater than 1.

Answer

tan θ = .

Check Double the side lengths to 6, 8 and 10. The new ratio is = , confirming the same value from a larger similar triangle.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Reverse top and bottom but keep the same function name.

Reversing the fraction creates the reciprocal function.

✓ Instead: tan θ = .

Tips and tricks

Say the numerator side first, then the denominator side.

.4Cosecant (csc θ)

Cosecant reverses sine's comparison: ladder length per unit of wall height. Divide hypotenuse by opposite.

csc θ = $\frac{h y p}{o p p}$
This side ratio applies to an acute angle; both lengths are positive.

Cosecant uses hyp on top and opp on bottom.

The same idea, five ways

Say it

Read csc θ as cosecant of theta.

Write it

The cosecant of an acute angle is its hypotenuse divided by its opposite leg.

In math

csc θ = $\frac{h y p}{o p p}$

Cosecant reverses sine's comparison: ladder length per unit of wall height. Divide hypotenuse by opposite.

See it

Cosecant uses hyp on top and opp on bottom.

Worked exampleRead csc from a small triangle

opp = 3, adj = 4 and hyp = 5. Find csc θ. In words, divide hyp by opp.

Cosecant uses hyp on top and opp on bottom.

Write csc θ = $\frac{h y p}{o p p}$ .The definition identifies the numerator and denominator.
Substitute csc θ = $5 over 3$ .Use the side labels from the chosen angle.
Keep $5 over 3$ in reduced exact form.The two numbers have no common factor greater than 1.

Answer

csc θ = .

Check Double the side lengths to 6, 8 and 10. The new ratio is = , confirming the same value from a larger similar triangle.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Reverse top and bottom but keep the same function name.

Reversing the fraction creates the reciprocal function.

✓ Instead: csc θ = .

Tips and tricks

Say the numerator side first, then the denominator side.

.5Secant (sec θ)

Secant reverses cosine's comparison: ladder length per unit of floor distance. Divide hypotenuse by adjacent.

sec θ = $\frac{h y p}{a d j}$
This side ratio applies to an acute angle; both lengths are positive.

Secant uses hyp on top and adj on bottom.

The same idea, five ways

Say it

Read sec θ as secant of theta.

Write it

The secant of an acute angle is its hypotenuse divided by its adjacent leg.

In math

sec θ = $\frac{h y p}{a d j}$

Secant reverses cosine's comparison: ladder length per unit of floor distance. Divide hypotenuse by adjacent.

See it

Secant uses hyp on top and adj on bottom.

Worked exampleRead sec from a small triangle

opp = 3, adj = 4 and hyp = 5. Find sec θ. In words, divide hyp by adj.

Secant uses hyp on top and adj on bottom.

Write sec θ = $\frac{h y p}{a d j}$ .The definition identifies the numerator and denominator.
Substitute sec θ = $5 over 4$ .Use the side labels from the chosen angle.
Keep $5 over 4$ in reduced exact form.The two numbers have no common factor greater than 1.

Answer

sec θ = .

Check Double the side lengths to 6, 8 and 10. The new ratio is = , confirming the same value from a larger similar triangle.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Reverse top and bottom but keep the same function name.

Reversing the fraction creates the reciprocal function.

✓ Instead: sec θ = .

Tips and tricks

Say the numerator side first, then the denominator side.

.6Cotangent (cot θ)

Cotangent reverses tangent's comparison: floor distance per unit of wall height. Divide adjacent by opposite.

cot θ = $\frac{a d j}{o p p}$
This side ratio applies to an acute angle; both lengths are positive.

Cotangent uses adj on top and opp on bottom.

The same idea, five ways

Say it

Read cot θ as cotangent of theta.

Write it

The cotangent of an acute angle is its adjacent leg divided by its opposite leg.

In math

cot θ = $\frac{a d j}{o p p}$

Cotangent reverses tangent's comparison: floor distance per unit of wall height. Divide adjacent by opposite.

See it

Cotangent uses adj on top and opp on bottom.

Worked exampleRead cot from a small triangle

opp = 3, adj = 4 and hyp = 5. Find cot θ. In words, divide adj by opp.

Cotangent uses adj on top and opp on bottom.

Write cot θ = $\frac{a d j}{o p p}$ .The definition identifies the numerator and denominator.
Substitute cot θ = $4 over 3$ .Use the side labels from the chosen angle.
Keep $4 over 3$ in reduced exact form.The two numbers have no common factor greater than 1.

Answer

cot θ = .

Check Double the side lengths to 6, 8 and 10. The new ratio is = , confirming the same value from a larger similar triangle.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Reverse top and bottom but keep the same function name.

Reversing the fraction creates the reciprocal function.

✓ Instead: cot θ = .

Tips and tricks

Say the numerator side first, then the denominator side.

Strategy: step by step

Name opp, adj and hyp from the angle asked about.
If a side is missing, find it with the Pythagorean theorem.
Write sin, cos and tan with SOH-CAH-TOA.
Flip each one: csc is sin upside down, sec is cos upside down, cot is tan upside down.
Reduce every fraction and rationalize any square root left on the bottom.

Strategy

Find all six values from a triangle

Is the chosen angle acute in a right triangle?

YesLabel opp, adj and hyp.

NoUse a later point or unit-circle definition.

↓

Is the hypotenuse missing?

YesAdd the squared legs and take the nonnegative root.

NoIf a leg is missing, subtract the known squared leg from hy and take the nonnegative root.

↓

Does an answer have a root on the bottom?

YesMultiply top and bottom by that root, then reduce.

NoReduce common numerical factors.

Name opp, adj and hyp from the angle asked about.
If a side is missing, find it with the Pythagorean theorem.
Write sin, cos and tan with SOH-CAH-TOA.
Flip each one: csc is sin upside down, sec is cos upside down, cot is tan upside down.
Reduce every fraction and rationalize any square root left on the bottom.

Worked exampleSix trig functions of an acute angle when a leg is missing

In right triangle KLM the right angle is at L. The leg KL is 3 units long and the hypotenuse KM is 7 units long. Find the exact values of all six trigonometric functions of angle K. Write each answer as a reduced fraction with no square root in the denominator.

Name the sides from angle K. The hypotenuse is KM = 7, the side across from the right angle at L. The adjacent side is KL = 3, the leg that touches K. The opposite side is LM, the leg across from K. Its length is not given.Opp and adj depend on which acute angle is asked about, but hyp is always across from the right angle. Naming the sides from angle K first keeps its ratios from being swapped with those of angle M.
Find the missing leg with the Pythagorean theorem: K $L^{2}$ + L $M^{2}$ = K $M^{2}$ , so $3 squared$ + L $M^{2}$ = $7 squared$ . Then 9 + L $M^{2}$ = 49, so L $M^{2}$ = 40 and LM = $square root of 40$ = $square root of 4 · 10$ = 2 $square root of 10$ .In a right triangle the squares of the legs add up to the square of the hypotenuse. A length is positive, so only the positive root counts. Taking the perfect square 4 out of 40 puts the root in simplest form.
Write the three basic ratios with SOH-CAH-TOA, using opp = 2 $square root of 10$ , adj = 3 and hyp = 7: sin K = $\frac{o p p}{h y p}$ = $2 square root of 10 over 7$ , cos K = $\frac{a d j}{h y p}$ = $3 over 7$ , tan K = $\frac{o p p}{a d j}$ = $2 square root of 10 over 3$ .SOH: sine is opposite over hypotenuse. CAH: cosine is adjacent over hypotenuse. TOA: tangent is opposite over adjacent.
Flip each ratio to get the other three: csc K = $7 over 2 square root of 10$ , sec K = $7 over 3$ , cot K = $3 over 2 square root of 10$ .Cosecant, secant and cotangent are the reciprocals of sine, cosine and tangent: csc K = $\frac{h y p}{o p p}$ , sec K = $\frac{h y p}{a d j}$ , cot K = $\frac{a d j}{o p p}$ .
Two answers have a root on the bottom, so multiply top and bottom of each by $square root of 10$ : csc K = $7 over 2 square root of 10$ · $square root of 10 over square root of 10$ = $7 square root of 10 over 20$ , and cot K = $3 over 2 square root of 10$ · $square root of 10 over square root of 10$ = $3 square root of 10 over 20$ . Then confirm that all six fractions are reduced. None of the pairs 2 and 7, 3 and 7, 2 and 3, 7 and 20, or 3 and 20 shares a common factor. $square root of 10 over square root of 10$ equals 1, so multiplying by it does not change the value. Since 2 $square root of 10$ · $square root of 10$ = 2 · 10 = 20, the root leaves the bottom. A fraction is in simplest form when the whole numbers on top and bottom share no factor greater than 1.

Answer

sin K = , cos K = , tan K = , csc K = , sec K = , cot K =

Check Pythagorean identity: K + K = ( + ( = + = = 1. Tangent from sine and cosine: tan K = = ÷ = . Each flipped pair multiplies to 1: sin K · csc K = · = = 1, cos K · sec K = · = 1, and tan K · cot K = · = = 1.

Work to write

opp = LM, adj = KL = 3, hyp = KM = 7
L $M^{2}$ = $7 squared$ − $3 squared$ = 49 − 9 = 40
LM = $square root of 40$ = 2 $square root of 10$
sin K = $2 square root of 10 over 7$
cos K = $3 over 7$
tan K = $2 square root of 10 over 3$
csc K = $7 over 2 square root of 10$ = $7 square root of 10 over 20$
sec K = $7 over 3$
cot K = $3 over 2 square root of 10$ = $3 square root of 10 over 20$

sin K = $2 square root of 10 over 7$ , cos K = $3 over 7$ , tan K = $2 square root of 10 over 3$ , csc K = $7 square root of 10 over 20$ , sec K = $7 over 3$ , cot K = $3 square root of 10 over 20$

Ladder: from easy to exam-hard. Press Try it first on any rung to hide its steps and use them as hints.

Rung 1Rung 1: all lengths given

opp = 3, adj = 4, hyp = 5. Find all six functions. In words, form the six ordered side ratios.

Name the sides from the marked angle before making a fraction.

sin θ = $3 over 5$ , cos θ = $4 over 5$ , tan θ = $3 over 4$ .SOH-CAH-TOA gives the first three ratios.
csc θ = $5 over 3$ , sec θ = $5 over 4$ , cot θ = $4 over 3$ .Flip the first three in their matching pairs.

Answer

sin θ = $3 over 5$
cos θ = $4 over 5$
tan θ = $3 over 4$
csc θ = $5 over 3$
sec θ = $5 over 4$
cot θ = $4 over 3$

Check + = , and each flip times its partner is 1.

Rung 2Rung 2: change the chosen angle

Choose the other acute corner of the same 3, 4, 5 triangle. Now opp = 4 and adj = 3. Find all six functions.

The same lengths are redrawn with the other corner at θ.

hyp stays 5; opp = 4 and adj = 3.Changing the chosen acute corner swaps the leg roles.
sin θ = $4 over 5$ , cos θ = $3 over 5$ , tan θ = $4 over 3$ .Substitute the new labels into SOH-CAH-TOA.
csc θ = $5 over 4$ , sec θ = $5 over 3$ , cot θ = $3 over 4$ .Flip each matching ratio.

Answer

sin θ = $4 over 5$
cos θ = $3 over 5$
tan θ = $4 over 3$
csc θ = $5 over 4$
sec θ = $5 over 3$
cot θ = $3 over 4$

Check Sine and cosine traded values, while tangent became its former reciprocal, matching the switched viewpoint.

Rung 3Rung 3: reduce enlarged ratios

opp = 9, adj = 12, hyp = 15. Find all six functions. In words, divide named lengths and reduce.

Name the sides from the marked angle before making a fraction.

sin θ = $9 over 15$ = $3 over 5$ , cos θ = $12 over 15$ = $4 over 5$ , tan θ = $9 over 12$ = $3 over 4$ .Divide each top and bottom by their common factor 3.
csc θ = $5 over 3$ , sec θ = $5 over 4$ , cot θ = $4 over 3$ .Flipping reduced ratios gives reduced reciprocals.

Answer

sin θ = $3 over 5$
cos θ = $4 over 5$
tan θ = $3 over 4$
csc θ = $5 over 3$
sec θ = $5 over 4$
cot θ = $4 over 3$

Check + = . Tripling every side of rung 1 kept the ratios unchanged.

Rung 4Rung 4: a missing hypotenuse

opp = 8 and adj = 15. Find hyp, then all six functions. In words, recover the missing side before making ratios.

Name the sides from the marked angle before making a fraction.

hyp = $square root of 8 squared + 15 squared$ = $square root of 64 + 225$ = $square root of 289$ = 17.Add the squared legs according to the Pythagorean theorem.
sin θ = $8 over 17$ , cos θ = $15 over 17$ , tan θ = $8 over 15$ .Use the newly known three sides in SOH-CAH-TOA.
csc θ = $17 over 8$ , sec θ = $17 over 15$ , cot θ = $15 over 8$ .Flip each matching fraction.

Answer

hyp = 17
sin θ = $8 over 17$
cos θ = $15 over 17$
tan θ = $8 over 15$
csc θ = $17 over 8$
sec θ = $17 over 15$
cot θ = $15 over 8$

Check = 289 = 64 + 225; hyp is longer than either leg.

Rung 5Rung 5: rationalize a missing hypotenuse

opp = 1 and adj = 2. Find hyp and all six functions. In words, expect roots on the bottoms of sine and cosine.

Name the sides from the marked angle before making a fraction.

hyp = $square root of 1 squared + 2 squared$ = $square root of 5$ .Add squared legs and take the nonnegative root.
sin θ = $1 over square root of 5$ = $1 × square root of 5 over square root of 5 × square root of 5$ = $square root of 5 over 5$ .Multiply by root five over itself, a fraction equal to 1.
cos θ = $2 over square root of 5$ = $2 square root of 5 over 5$ .The same multiplier removes the bottom root.
tan θ = $1 over 2$ , cot θ = 2, sec θ = $square root of 5 over 2$ , csc θ = $square root of 5$ .Use the leg ratios and the reversed hypotenuse ratios.

Answer

hyp = $square root of 5$
sin θ = $square root of 5 over 5$
cos θ = $2 square root of 5 over 5$
tan θ = $1 over 2$
cot θ = 2
sec θ = $square root of 5 over 2$
csc θ = $square root of 5$

Check + = 5 = (. × = 1 checks the sine pair.

Rung 6Rung 6: the missing side is a leg

hyp = 6 and adj = 4. Find opp, then all six functions. In words, subtract squares to recover the missing leg.

Name the sides from the marked angle before making a fraction.

op $p^{2}$ = $6 squared$ − $4 squared$ = 36 − 16 = 20, so opp = $square root of 20$ = 2 $square root of 5$ .Subtract ad $j^{2}$ from both sides of the Pythagorean theorem; $square root of 20$ = $square root of 4 × 5$ .
sin θ = $2 square root of 5 over 6$ = $square root of 5 over 3$ , cos θ = $4 over 6$ = $2 over 3$ .Use SOH and CAH, then reduce by 2.
tan θ = $2 square root of 5 over 4$ = $square root of 5 over 2$ .Use opposite over adjacent, then reduce by 2.
csc θ = $3 over square root of 5$ = $3 square root of 5 over 5$ , sec θ = $3 over 2$ , cot θ = $2 over square root of 5$ = $2 square root of 5 over 5$ .Flip the reduced ratios, then rationalize the two remaining root denominators.

Answer

opp = 2 $square root of 5$
sin θ = $square root of 5 over 3$
cos θ = $2 over 3$
tan θ = $square root of 5 over 2$
csc θ = $3 square root of 5 over 5$
sec θ = $3 over 2$
cot θ = $2 square root of 5 over 5$

Check (2 + = 20 + 16 = 36 = , checking the recovered leg.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: sin θ = .

SOH uses the opposite leg.

✓ Instead: sin θ = ; is cosine.

✗ Not this: The flip of sine is secant.

Sine pairs with cosecant; cosine pairs with secant.

✓ Instead: Write sin pairs with csc, cos pairs with sec, tan pairs with cot.

✗ Not this: Doubling a triangle doubles its sine.

Both top and bottom double, so the common factor cancels.

✓ Instead: = , not 2 × .

✗ Not this: Use SOH-CAH-TOA for a 120° corner in a right triangle.

A right triangle's two remaining corners must be acute.

✓ Instead: The later point definition handles other angles.

Tips and tricks

SOH-CAH-TOA gives the first three ratios. Remember the flip pairs: sin with csc, cos with sec, tan with cot.
All six acute-angle values are positive; a negative answer signals an error.
Sine and cosine are below 1 because each leg is shorter than hyp. Tangent may exceed 1 when opposite is longer than adjacent.
Rebuild three flip ratios instead of memorizing six independent fractions.
Read reference tables by columns: keep the input label and its output in the same vertical column.

Trap. Naming sides by where they sit on the page. Figures put the angle in different corners, so the leg on the left is not always adjacent. Always name the sides from the angle's point of view.

Keep in mind

Name the sides from the angle's point of view, not from where they sit on the page: at the top of the ladder the 8-foot floor leg is opposite, so sin = $8 over 17$ there.
Each flip pair multiplies to 1, a quick check on csc, sec and cot: $15 over 17$ × $17 over 15$ = 1.
The size of the triangle never matters, only the angle: a 3, 4, 5 triangle and a 6, 8, 10 triangle give the same sine, $3 over 5$ = $6 over 10$ .
SOH-CAH-TOA works only inside a right triangle, since only a right triangle has a hypotenuse: in a triangle with angles 30°, 30° and 120°, the side across from 30° divided by the longest side is about 0.577, not sin 30° = 0.5.

Memory hookSOH-CAH-TOA: Sine is Opposite over Hypotenuse, Cosine is Adjacent over Hypotenuse, Tangent is Opposite over Adjacent. Flip each for csc, sec, cot: every flip pair holds exactly one "co".

Flash cards: say the answer out loud, then flip

What does SOH-CAH-TOA stand for?

sin = $\frac{o p p}{h y p}$
cos = $\frac{a d j}{h y p}$
tan = $\frac{o p p}{a d j}$

What is cosecant?

The flip of sine: csc θ = = .

Legs 8 and 15, hypotenuse 17. Angle A faces the leg 8. Find sin A, cos A and tan A.

sin A = $8 over 17$
cos A = $15 over 17$
tan A = $8 over 15$

Legs 9 and 40, hypotenuse 41. Angle B faces the leg 9. Find csc B and cot B.

csc B = $41 over 9$
cot B = $40 over 9$

One ladder has sin θ = 0.7. A ladder twice as long leans at the same angle. What is its sine?

0.7. The ratio depends only on the angle.

Is tan θ = ?

No. tan θ = . is sin θ.