What does SOH-CAH-TOA stand for?

It is a memory aid for the three trig ratios in a right triangle: Sine = Opposite/Hypotenuse, Cosine = Adjacent/Hypotenuse, and Tangent = Opposite/Adjacent. Pick an acute angle, label the sides relative to it, and read off the ratio you need.

How do you find an angle from two sides on the ACT?

Form the appropriate ratio, then apply the inverse trig function. For example, if the opposite and adjacent legs are 5 and 12, then tan θ = 5/12 and θ = tan⁻¹(5/12) ≈ 22.6°. Use the sin⁻¹, cos⁻¹, or tan⁻¹ buttons in degree mode.

Does SOH-CAH-TOA work for any triangle?

No. The sine, cosine, and tangent ratios are defined only for right triangles. For a triangle with no right angle, use the law of sines or the law of cosines instead.

Should the calculator be in degrees or radians for ACT trig?

Use degree mode. ACT trigonometry problems state angles in degrees, so a radian setting will produce wrong answers. A quick check is that sin 30° should read exactly 0.5.

CramClub • Trigonometry: SOH-CAH-TOA

$\sin\theta = \dfrac{\text{opposite}}{\text{hypotenuse}}$

$\cos\theta = \dfrac{\text{adjacent}}{\text{hypotenuse}}$

$\tan\theta = \dfrac{\text{opposite}}{\text{adjacent}}$

$\theta = \tan^{-1}\!\left(\dfrac{\text{opp}}{\text{adj}}\right)$

$\sin\theta = \cos(90^\circ - \theta)$

Right-Triangle Trigonometry on the ACT

Every ACT Math section includes a handful of trigonometry questions, and the overwhelming majority are built on one right-triangle idea: the three trigonometric ratios — sine, cosine, and tangent. Learn the mnemonic SOH-CAH-TOA and you can answer most of them without memorizing anything else.

Trigonometry connects the angles of a right triangle to the lengths of its sides. Pick one of the two non-right angles — call it $\theta$ — and label the three sides from that angle's point of view:

the hypotenuse is the longest side, always across from the right angle;
the opposite side is the leg directly across from $\theta$ ;
the adjacent side is the leg touching $\theta$ that is not the hypotenuse.

Because "opposite" and "adjacent" depend on which angle you pick, the very first move on any trig problem is to mark your angle and relabel the legs.

The Three Ratios

$\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}, \quad \cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}}, \quad \tan\theta = \frac{\text{opposite}}{\text{adjacent}}$

The mnemonic spells it out: Sin = Opp/Hyp, Cos = Adj/Hyp, Tan = Opp/Adj. For a right triangle with legs $3$ and $4$ and hypotenuse $5$ , the angle across from the side of length $3$ has $\sin\theta = \tfrac{3}{5}$ , $\cos\theta = \tfrac{4}{5}$ , and $\tan\theta = \tfrac{3}{4}$ .

Finding a Side

If you know one acute angle and one side, you can find any other side. Choose the ratio that links the side you have to the side you want, then solve.

A guy-wire runs from the top of a pole to the ground at a $40^\circ$ angle, and the wire is $30$ feet long. How high is the attachment point? The wire is the hypotenuse and the height is opposite the $40^\circ$ angle, so use sine:

$\sin 40^\circ = \frac{h}{30} \quad\Rightarrow\quad h = 30\sin 40^\circ \approx 19.3 \text{ ft}$

Set up the ratio first with symbols, then reach for the calculator.

Finding an Angle — Inverse Trig

When you know two sides and want the angle, use an inverse trig function, written $\sin^{-1}$ , $\cos^{-1}$ , or $\tan^{-1}$ (also called arcsin, arccos, arctan). These "undo" the ratio:

$\theta = \tan^{-1}\!\left(\frac{\text{opposite}}{\text{adjacent}}\right)$

If a ramp rises $5$ feet over a horizontal run of $12$ feet, its angle of elevation is $\theta = \tan^{-1}\!\left(\tfrac{5}{12}\right) \approx 22.6^\circ$ . The inverse buttons on your calculator (often labeled with the little $-1$ ) are how you get from a ratio back to a degree measure.

Cofunctions: Sine and Cosine Are Partners

In any right triangle the two acute angles add to $90^\circ$ — they are complementary. The side opposite one angle is adjacent to the other, which produces the cofunction identity:

$\sin\theta = \cos(90^\circ - \theta) \qquad \cos\theta = \sin(90^\circ - \theta)$

So $\sin 25^\circ = \cos 65^\circ$ . The ACT loves to hide this in algebra: if $\sin(3x + 5)^\circ = \cos(2x)^\circ$ , the angles must be complementary, so $(3x + 5) + 2x = 90$ , giving $x = 17$ .

Special Right Triangles

Two right triangles have exact, memorizable ratios that skip the calculator:

Triangle	Angles	Side ratio
45-45-90	$45^\circ,45^\circ,90^\circ$	$1 : 1 : \sqrt{2}$
30-60-90	$30^\circ,60^\circ,90^\circ$	$1 : \sqrt{3} : 2$

From these, $\sin 30^\circ = \tfrac{1}{2}$ , $\cos 30^\circ = \tfrac{\sqrt{3}}{2}$ , and $\sin 45^\circ = \tfrac{\sqrt{2}}{2}$ — worth knowing cold.

A Reliable Game Plan

Label first: mark $\theta$ , then tag each side as opposite, adjacent, or hypotenuse.
Have an angle + a side, need a side? Pick the ratio that uses both, then solve.
Have two sides, need the angle? Use the matching inverse function.
See " $\sin(\ldots) = \cos(\ldots)$ "? Set the two arguments to sum to $90$ .
Set your calculator to DEGREE mode — ACT trig is in degrees, not radians.

Common Mistakes to Avoid

Swapping opposite and adjacent because you forgot to fix an angle first.
Leaving the calculator in radian mode and getting a nonsense decimal.
Using $\sin$ when the question asks for the angle (you need $\sin^{-1}$ ).
Trying SOH-CAH-TOA on a triangle that has no right angle — that requires the law of sines or cosines instead.

A right triangle has legs $6$ and $8$ and hypotenuse $10$ . For the acute angle $\theta$ opposite the leg of length $6$ , find $\sin\theta$ , $\cos\theta$ , and $\tan\theta$ .

In a right triangle, an acute angle measures $30^\circ$ and the hypotenuse is $12$ . What is the length of the side opposite the $30^\circ$ angle?

A right triangle has an acute angle of $37^\circ$ whose adjacent leg is $15$ . Find the length of the opposite leg, rounded to the nearest tenth.

A loading ramp is $24$ feet long and meets the ground at an angle of elevation of $18^\circ$ . How high, to the nearest tenth of a foot, is the top of the ramp above the ground?

A right triangle has legs of length $9$ and $12$ . What is the measure, to the nearest degree, of the angle opposite the leg of length $9$ ?

If $\sin(3x + 15)^\circ = \cos(x + 35)^\circ$ and both angles are acute, what is the value of $x$ ?

Labeling "opposite" and "adjacent" before choosing an angle, so the two legs get swapped.

Always fix the reference angle $\theta$ first; the opposite leg is across from it and the adjacent leg touches it. The hypotenuse never changes.

Leaving the calculator in radian mode, which turns $\sin 30^\circ$ into a strange decimal instead of $0.5$ .

Switch the calculator to DEGREE mode for ACT trig. Sanity-check: $\sin 30^\circ$ should read $0.5$ .

Using $\sin$ , $\cos$ , or $\tan$ when the question wants the angle, e.g. computing $\tan(0.75)$ to "find" $\theta$ .

Going from a side ratio to an angle requires the inverse: $\theta = \tan^{-1}(0.75)$ , not $\tan(0.75)$ .

Applying SOH-CAH-TOA to a triangle with no right angle.

The three ratios only describe right triangles. For an oblique triangle, switch to the law of sines or the law of cosines.

In a right triangle, the side opposite an acute angle is $8$ and the hypotenuse is $17$ . What is the sine of that angle?

A right triangle has legs of length $5$ and $12$ . What is the tangent of the angle opposite the leg of length $5$ ?

In a right triangle an acute angle measures $28^\circ$ , and its adjacent leg is $20$ . To the nearest tenth, how long is the leg opposite the $28^\circ$ angle?

A $32$ -foot guy-wire runs from the top of a pole to the ground, making a $65^\circ$ angle with the ground. To the nearest tenth, how high up the pole is the wire attached?

A right triangle has legs of length $7$ and $24$ . To the nearest degree, what is the measure of the angle opposite the leg of length $7$ ?

If $\sin(2x + 20)^\circ = \cos(4x + 10)^\circ$ and both angles are acute, what is the value of $x$ ?