SOH-CAH-TOA is a mnemonic for the three trigonometric ratios in a right-angled triangle. SOH means sine equals opposite over hypotenuse. CAH means cosine equals adjacent over hypotenuse. TOA means tangent equals opposite over adjacent.

What does sin squared plus cos squared equal?

Sin squared theta plus cos squared theta equals 1. This is the Pythagorean identity, and it holds for every angle theta. It comes directly from Pythagoras' theorem applied to a right triangle inscribed in a unit circle.

SOH-CAH-TOA is a mnemonic for the three trigonometric ratios in a right-angled triangle. SOH means sine equals opposite over hypotenuse. CAH means cosine equals adjacent over hypotenuse. TOA means tangent equals opposite over adjacent.

What does sin squared plus cos squared equal?

Sin squared theta plus cos squared theta equals 1. This is the Pythagorean identity, and it holds for every angle theta. It comes directly from Pythagoras' theorem applied to a right triangle inscribed in a unit circle.

Year 11 General Applications Of Trigonometry

Review Of Basic Trigonometry

Q: How do I know which side is opposite and which is adjacent?

The opposite side is directly across from the angle theta. The adjacent side is next to theta, not counting the hypotenuse. The hypotenuse is always the longest side, opposite the right angle, and never changes regardless of which acute angle you pick.

Q: What are the exact values of sin, cos and tan for 30, 45 and 60 degrees?

For 30 degrees: sin is one half, cos is root three over two, tan is one over root three. For 45 degrees: sin and cos are both root two over two, tan is 1. For 60 degrees: sin is root three over two, cos is one half, tan is root three.

Q: Is sin of 90 minus theta the same as cos theta?

Yes. Sin of (90 minus theta) equals cos theta, and cos of (90 minus theta) equals sin theta. These are the complementary angle identities. In a right triangle the two acute angles add to 90 degrees, so the opposite side for one angle is the adjacent side for the other.

12 practice questions 2 video lessons Theory + worked examples

Theory

Trigonometry relates the angles of a right-angled triangle to the lengths of its sides. The three sides — opposite, adjacent and hypotenuse — are named relative to a chosen angle $\theta$, and the three ratios sin, cos and tan are remembered with SOH-CAH-TOA.

In a right-angled triangle, the three sides are named relative to a chosen acute angle $\theta$:

Hypotenuse — the longest side, opposite the right angle. Never changes regardless of which acute angle you pick.
Opposite — the side directly across from $\theta$.
Adjacent — the side next to $\theta$ (the other one, not the hypotenuse).

The three trig ratios connect these sides to the angle. The mnemonic SOH-CAH-TOA tells you which side goes on top:

SOH — $\sin\theta = \dfrac{\text{Opposite}}{\text{Hypotenuse}}$
CAH — $\cos\theta = \dfrac{\text{Adjacent}}{\text{Hypotenuse}}$
TOA — $\tan\theta = \dfrac{\text{Opposite}}{\text{Adjacent}}$

Sides named relative to angle $\theta$

Pick the other acute angle $\varphi$ and opposite/adjacent swap

The three trig ratios (SOH-CAH-TOA)

\[\sin\theta = \dfrac{\text{Opposite}}{\text{Hypotenuse}}\]

\sin θ = \frac{Opposite}{Hypotenuse}

\[\cos\theta = \dfrac{\text{Adjacent}}{\text{Hypotenuse}}\]

\cos θ = \frac{Adjacent}{Hypotenuse}

\[\tan\theta = \dfrac{\text{Opposite}}{\text{Adjacent}}\]

\tan θ = \frac{Opposite}{Adjacent}

Exact values for special angles

$\theta$	$\sin\theta$	$\cos\theta$	$\tan\theta$
$30^\circ$	$\dfrac{1}{2}$	$\dfrac{\sqrt{3}}{2}$	$\dfrac{1}{\sqrt{3}}$
$45^\circ$	$\dfrac{\sqrt{2}}{2}$	$\dfrac{\sqrt{2}}{2}$	$1$
$60^\circ$	$\dfrac{\sqrt{3}}{2}$	$\dfrac{1}{2}$	$\sqrt{3}$

Useful identities

\[\sin^2\theta + \cos^2\theta = 1\]

\sin^{2} θ + \cos^{2} θ = 1

\[\dfrac{\sin\theta}{\cos\theta} = \tan\theta\]

\frac{\sin θ}{\cos θ} = \tan θ

          Complementary angles: \(\sin(90^\circ - \theta) = \cos\theta\) and \(\cos(90^\circ - \theta) = \sin\theta\). In a right triangle the two acute angles add to \(90^\circ\), so swapping which angle you pick swaps the opposite and adjacent sides.
        

How to apply SOH-CAH-TOA

Mark the chosen angle $\theta$ on your triangle (the angle that is given or asked about — never the right angle).
Label the three sides relative to $\theta$: the hypotenuse (opposite the right angle), the opposite (across from $\theta$), and the adjacent (next to $\theta$).
Choose the ratio that uses the two sides you know or want: sin for Opp/Hyp, cos for Adj/Hyp, tan for Opp/Adj.
Solve for the unknown. If you know the angle and one side, multiply or divide. If you know two sides and want the angle, take $\sin^{-1}$, $\cos^{-1}$ or $\tan^{-1}$.

Example 1 — Identify the three ratios

A right-angled triangle has $AB = 8$, $BC = 15$, $AC = 17$, with the right angle at $B$ and $\theta$ at $A$. Find $\sin\theta$, $\cos\theta$ and $\tan\theta$.

Solution

Sketch the triangle and label the sides relative to $\theta = \angle A$:

Read off each ratio:

$\sin\theta$	$=$	$\dfrac{\text{opp}}{\text{hyp}} = \dfrac{15}{17}$
$\cos\theta$	$=$	$\dfrac{\text{adj}}{\text{hyp}} = \dfrac{8}{17}$
$\tan\theta$	$=$	$\dfrac{\text{opp}}{\text{adj}} = \dfrac{15}{8}$

\sin θ = \frac{15}{17}, \cos θ = \frac{8}{17}, \tan θ = \frac{15}{8}

Example 2 — Find others from one ratio

For an acute angle, $\sin\theta = \dfrac{3}{5}$. Find $\cos\theta$ and $\tan\theta$.

Solution

Sketch a triangle with opposite $=3$ and hypotenuse $=5$, then use Pythagoras to find the adjacent side $a$:

$a^2$	$=$	$5^2 - 3^2 = 16$
$a$	$=$	$4$

So adjacent $=4$. Then:

$\cos\theta$	$=$	$\dfrac{4}{5}$
$\tan\theta$	$=$	$\dfrac{3}{4}$

\cos θ = \frac{4}{5}, \tan θ = \frac{3}{4}

Example 3 — Exact values

Find the exact value of $\sin 60^\circ + \cos 30^\circ$.

Solution

Both terms have exact value $\dfrac{\sqrt{3}}{2}$:

$\sin 60^\circ + \cos 30^\circ$	$=$	$\dfrac{\sqrt{3}}{2} + \dfrac{\sqrt{3}}{2}$
	$=$	$\sqrt{3}$

\sin 60 ° + \cos 30 ° = \sqrt{3}

Example 4 — Complementary angles

Triangle $PQR$ has the right angle at $R$, with $PR = 12$, $QR = 5$, $PQ = 13$. Let $\theta = \angle P$. Find $\tan(90^\circ - \theta)$.

Solution

Sketch the triangle. At $P$: opposite $= QR = 5$, adjacent $= PR = 12$. At $Q$ (which is $90^\circ - \theta$) the opposite and adjacent swap:

$\tan\theta$	$=$	$\dfrac{5}{12}$
$\tan(90^\circ - \theta)$	$=$	$\dfrac{12}{5}$

\tan (90 ° - θ) = \frac{12}{5}

Common pitfalls

Opposite and adjacent depend on the chosen angle. In the same triangle, picking the other acute angle swaps which side is opposite and which is adjacent. The hypotenuse never changes.

The hypotenuse is always opposite the right angle. It's the longest side and never sits next to the right angle. Don't accidentally call a non-hypotenuse side the "hypotenuse" just because it looks long.

Use Pythagoras to find a missing side from one ratio. If

\sin θ = \frac{3}{5}

, opposite

= 3

and hypotenuse

= 5

, but adjacent isn't given. Use

a^{2} + b^{2} = c^{2}

to find it before computing

\cos θ

\tan θ

Exact values are not decimals. If a question asks for an exact value, leave answers as fractions and surds (e.g.

\frac{\sqrt{3}}{2}

), not

0.866

Frequently asked questions

What is SOH-CAH-TOA?

SOH-CAH-TOA is a mnemonic for the three trigonometric ratios. SOH: $\sin θ = \frac{Opp}{Hyp}$ . CAH: $\cos θ = \frac{Adj}{Hyp}$ . TOA: $\tan θ = \frac{Opp}{Adj}$ .

How do I know which side is opposite and which is adjacent?

The opposite side is directly across from $θ$ . The adjacent side is next to $θ$ , not counting the hypotenuse. The hypotenuse is always the longest side, opposite the right angle, and never changes regardless of which acute angle you pick.

What are the exact values of sin, cos and tan for 30°, 45° and 60°?

At $30^{\circ}$ : $\sin = \frac{1}{2}$ , $\cos = \frac{\sqrt{3}}{2}$ , $\tan = \frac{1}{\sqrt{3}}$ . At $45^{\circ}$ : $\sin = \cos = \frac{\sqrt{2}}{2}$ , $\tan = 1$ . At $60^{\circ}$ : $\sin = \frac{\sqrt{3}}{2}$ , $\cos = \frac{1}{2}$ , $\tan = \sqrt{3}$ .

How do I find cos and tan when I only know sin?

Treat the sin ratio as fixing the opposite side and hypotenuse. For example, $\sin θ = \frac{3}{5}$ means opposite $= 3$ and hypotenuse $= 5$ . Use Pythagoras ( $a^{2} + b^{2} = c^{2}$ ) to find the missing adjacent side, then read off $\cos θ$ and $\tan θ$ from the three sides.

What does

\sin^{2} θ + \cos^{2} θ

equal?

It equals $1$ for every angle $θ$ . This is the Pythagorean identity, and comes directly from Pythagoras' theorem applied to a right triangle inscribed in a unit circle.

\sin (90^{\circ} - θ)

the same as

\cos θ

Yes. $\sin (90^{\circ} - θ) = \cos θ$ , and $\cos (90^{\circ} - θ) = \sin θ$ . These are the complementary angle identities. In a right triangle the two acute angles add to $90^{\circ}$ , so the opposite side for one angle is the adjacent side for the other — which swaps sin and cos.

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Finding An Unknown Side In A Right-Angled Triangle

\(\theta\)	\(\sin\theta\)	\(\cos\theta\)	\(\tan\theta\)
\(30^\circ\)	\(\dfrac{1}{2}\)	\(\dfrac{\sqrt{3}}{2}\)	\(\dfrac{1}{\sqrt{3}}\)
\(45^\circ\)	\(\dfrac{\sqrt{2}}{2}\)	\(\dfrac{\sqrt{2}}{2}\)	\(1\)
\(60^\circ\)	\(\dfrac{\sqrt{3}}{2}\)	\(\dfrac{1}{2}\)	\(\sqrt{3}\)

\(\sin\theta\)	\(=\)	\(\dfrac{\text{opp}}{\text{hyp}} = \dfrac{15}{17}\)
\(\cos\theta\)	\(=\)	\(\dfrac{\text{adj}}{\text{hyp}} = \dfrac{8}{17}\)
\(\tan\theta\)	\(=\)	\(\dfrac{\text{opp}}{\text{adj}} = \dfrac{15}{8}\)

\(a^2\)	\(=\)	\(5^2 - 3^2 = 16\)
\(a\)	\(=\)	\(4\)

\(\cos\theta\)	\(=\)	\(\dfrac{4}{5}\)
\(\tan\theta\)	\(=\)	\(\dfrac{3}{4}\)

\(\sin 60^\circ + \cos 30^\circ\)	\(=\)	\(\dfrac{\sqrt{3}}{2} + \dfrac{\sqrt{3}}{2}\)
	\(=\)	\(\sqrt{3}\)

\(\tan\theta\)	\(=\)	\(\dfrac{5}{12}\)
\(\tan(90^\circ - \theta)\)	\(=\)	\(\dfrac{12}{5}\)

Review Of Basic Trigonometry

📖 Theory

The three trig ratios (SOH-CAH-TOA)

Exact values for special angles

Useful identities

How to apply SOH-CAH-TOA

Common pitfalls

Frequently asked questions

🎬 Video Lessons

✏️ Practice Questions

Theory

Video Lessons

Practice Questions