The Primary Trigonometric Ratios

In a right triangle, the angles and the sides are linked: if you know one acute angle, the ratios of the sides are fixed. The three primary trigonometric ratios, sine, cosine and tangent, give names to those ratios. With them (and the Pythagorean theorem) you can find every missing side and angle of a right triangle from just two pieces of information. All angles on this page are in degrees.

Key ideas

Naming the sides

In a right triangle, the hypotenuse is the longest side, across from the right angle. The other two sides are named relative to the acute angle you’re working with:

• the opposite side is across from the angle
• the adjacent side is next to the angle (and isn’t the hypotenuse)

The same side can be “opposite” for one angle and “adjacent” for the other, so always ask: opposite or adjacent to which angle?

The hypotenuse never changes, but “opposite” and “adjacent” swap when you switch angles.

Why the ratios depend only on the angle

Take any two right triangles that both have a $37^\circ$ angle. Each also has a $90^\circ$ angle, so by AA they are similar. Corresponding sides of similar triangles are proportional, so

$$\frac{\text{opposite}}{\text{hypotenuse}}$$

has the same value in both triangles, no matter how big they are. For example, a $3$-$4$-$5$ triangle and a $6$-$8$-$10$ triangle have the same shape, and in both the side across from the smaller acute angle divided by the hypotenuse is $\dfrac{3}{5} = \dfrac{6}{10} = 0.6$. The ratio depends only on the angle, so it makes sense to give it a name.

Sine, cosine and tangent

For an acute angle $\theta$ in a right triangle:

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

A memory aid: SOH CAH TOA.

	Stands for
SOH	Sine is Opposite over Hypotenuse
CAH	Cosine is Adjacent over Hypotenuse
TOA	Tangent is Opposite over Adjacent

Using your calculator

Your calculator must be in degree mode. Test it: $\sin 30^\circ$ should give exactly $0.5$. If you get $-0.988\ldots$, you’re in radian mode.

• To find a ratio from an angle, press the ratio key: $\sin 40^\circ \approx 0.6428$.
• To find an angle from a ratio, use the inverse keys $\sin^{-1}$, $\cos^{-1}$, $\tan^{-1}$ (often 2nd or shift then the ratio key). If $\sin \theta = 0.8$, then $\theta = \sin^{-1}(0.8) \approx 53^\circ$.

The $-1$ in $\sin^{-1}$ means “the angle whose sine is”, not a reciprocal.

Unless a question says otherwise, round lengths to one decimal place and angles to the nearest degree. Keep full calculator values until the end, and round only the final answer.

Choosing the ratio

1. Mark the angle you know (or want).
2. Label the sides you know and want as opposite, adjacent or hypotenuse.
3. Pick the ratio that uses exactly those two sides: SOH, CAH or TOA.
4. Write the equation and solve.

Solving a right triangle

To solve a triangle means to find all its missing sides and angles. In a right triangle, use:

• the trigonometric ratios, for sides and angles
• the Pythagorean theorem, $a^2 + b^2 = c^2$, when you know two sides
• the angle sum: the two acute angles add to $90^\circ$

Worked examples

Example 1: Writing the ratios

In right $\triangle ABC$, $\angle C = 90^\circ$, $BC = 8$, $AC = 15$ and $AB = 17$. Write the three primary trigonometric ratios for $\angle A$ and for $\angle B$.

Solution. The hypotenuse is $AB = 17$ (across from the right angle).

For $\angle A$: the opposite side is $BC = 8$ and the adjacent side is $AC = 15$.

$$\sin A = \frac{8}{17} \qquad \cos A = \frac{15}{17} \qquad \tan A = \frac{8}{15}$$

For $\angle B$: the opposite side is $AC = 15$ and the adjacent side is $BC = 8$.

$$\sin B = \frac{15}{17} \qquad \cos B = \frac{8}{17} \qquad \tan B = \frac{15}{8}$$

Notice that $\sin A = \cos B$: the side opposite $A$ is the side adjacent to $B$.

Example 2: Finding a side

• (a) In right $\triangle ABC$, $\angle C = 90^\circ$, $\angle A = 35^\circ$ and the hypotenuse $AB = 12$ cm. Find $BC$.
• (b) In a right triangle, an angle of $52^\circ$ has an opposite side of $9$ m. Find the adjacent side $x$.

Solution.

(a) Relative to $\angle A$, $BC$ is opposite and $AB$ is the hypotenuse. Opposite and hypotenuse means SOH:

$$\begin{aligned} \sin 35^\circ &= \frac{BC}{12} \\ BC &= 12 \sin 35^\circ && \text{multiply both sides by } 12 \\ BC &\approx 6.9 \end{aligned}$$

So $BC \approx 6.9$ cm.

(b) Opposite and adjacent means TOA. This time the unknown ends up in the denominator:

$$\begin{aligned} \tan 52^\circ &= \frac{9}{x} \\ x \tan 52^\circ &= 9 && \text{multiply both sides by } x \\ x &= \frac{9}{\tan 52^\circ} && \text{divide by } \tan 52^\circ \\ x &\approx 7.0 \end{aligned}$$

So $x \approx 7.0$ m. Check: an angle bigger than $45^\circ$ has an opposite side longer than its adjacent side, and $9 \gt 7.0$. ✓

Example 3: Finding an angle

A right triangle has legs $7$ cm and $10$ cm. Find the angle $\theta$ that is opposite the $7$ cm side.

Solution. Relative to $\theta$, $7$ is opposite and $10$ is adjacent, so use TOA:

$$\tan \theta = \frac{7}{10} = 0.7$$ $$\theta = \tan^{-1}(0.7) \approx 35^\circ$$

The other acute angle is $90^\circ - 35^\circ = 55^\circ$. Check: the smaller angle is opposite the shorter leg. ✓

Example 4: Solving a right triangle

Solve right $\triangle PQR$, where $\angle Q = 90^\circ$, $\angle P = 28^\circ$ and $PQ = 15$ m.

Solution. Sketch it first: the right angle is at $Q$, so $PR$ is the hypotenuse. Relative to $\angle P$, $PQ$ is adjacent and $QR$ is opposite.

Missing angle:

$$\angle R = 90^\circ - 28^\circ = 62^\circ$$

Side $QR$ (opposite and adjacent, TOA):

$$\tan 28^\circ = \frac{QR}{15} \quad\Rightarrow\quad QR = 15 \tan 28^\circ \approx 8.0 \text{ m}$$

Side $PR$ (adjacent and hypotenuse, CAH):

$$\cos 28^\circ = \frac{15}{PR} \quad\Rightarrow\quad PR = \frac{15}{\cos 28^\circ} \approx 17.0 \text{ m}$$

Check with the Pythagorean theorem, using the unrounded value of $QR$:

$$\sqrt{15^2 + (15 \tan 28^\circ)^2} \approx 17.0 \checkmark$$

So $\angle R = 62^\circ$, $QR \approx 8.0$ m and $PR \approx 17.0$ m.

Common mistakes

Calling the wrong side “opposite”. Opposite and adjacent depend on which angle you’re using. Mark the angle first, then label the sides from that angle’s point of view. The hypotenuse is always across from the right angle.

Calculator in radian mode. If $\sin 30^\circ$ doesn’t give $0.5$, switch to degree mode. Radian mode gives answers that look reasonable but are completely wrong.

Multiplying when you should divide. When the unknown is in the denominator, as in $\tan 52^\circ = \dfrac{9}{x}$, the answer is $x = \dfrac{9}{\tan 52^\circ}$, not $9 \tan 52^\circ$. Multiply both sides by $x$ first, then divide.

Using sin instead of sin⁻¹ to find an angle. To get an angle from a ratio, use the inverse key. $\sin(0.6)$ is a meaningless number here; $\sin^{-1}(0.6) \approx 37^\circ$ is the angle.

Rounding too early. If you round $QR$ to $8.0$ and then use it to find $PR$, small errors can creep in. Keep full values in your calculator and round only at the end.

Using SOH CAH TOA in a triangle with no right angle. These ratios only work in right triangles. For other triangles you’ll use the sine law and the cosine law.

Practice

1. (Warm-up) In right $\triangle DEF$, $\angle E = 90^\circ$.

• (a) Which side is the hypotenuse?
• (b) Relative to $\angle D$, which side is opposite and which is adjacent?
• (c) Relative to $\angle F$, which side is opposite and which is adjacent?

Solution

(a) $DF$, the side across from the right angle at $E$.

(b) Opposite: $EF$. Adjacent: $DE$.

(c) Opposite: $DE$. Adjacent: $EF$.

2. (Warm-up) Use a calculator in degree mode. Round ratios to four decimal places and angles to the nearest degree.

• (a) $\sin 40^\circ$, $\cos 72^\circ$, $\tan 55^\circ$
• (b) Find $\theta$ if $\sin \theta = 0.8$, and if $\tan \theta = 2.5$.

Solution

(a) $\sin 40^\circ \approx 0.6428$, $\cos 72^\circ \approx 0.3090$, $\tan 55^\circ \approx 1.4281$.

(b) $\theta = \sin^{-1}(0.8) \approx 53^\circ$, and $\theta = \tan^{-1}(2.5) \approx 68^\circ$.

3. (Warm-up) In right $\triangle ABC$, $\angle C = 90^\circ$, $BC = 7$, $AC = 24$ and $AB = 25$. Write $\sin A$, $\cos A$ and $\tan A$.

Solution

Relative to $\angle A$: opposite $BC = 7$, adjacent $AC = 24$, hypotenuse $AB = 25$.

$$\sin A = \frac{7}{25} \qquad \cos A = \frac{24}{25} \qquad \tan A = \frac{7}{24}$$

4. (Core) A right triangle has a hypotenuse of $20$ cm and an angle of $41^\circ$. Find the side $x$ opposite the $41^\circ$ angle.

Solution

Opposite and hypotenuse: SOH.

$$\sin 41^\circ = \frac{x}{20} \quad\Rightarrow\quad x = 20 \sin 41^\circ \approx 13.1 \text{ cm}$$

5. (Core) In a right triangle, the side adjacent to a $63^\circ$ angle is $8.5$ m. Find the hypotenuse $x$.

Solution

Adjacent and hypotenuse: CAH. The unknown is in the denominator.

$$\begin{aligned} \cos 63^\circ &= \frac{8.5}{x} \\ x \cos 63^\circ &= 8.5 \\ x &= \frac{8.5}{\cos 63^\circ} \approx 18.7 \text{ m} \end{aligned}$$

Check: the hypotenuse is the longest side, and $18.7 \gt 8.5$. ✓

6. (Core) A right triangle has a hypotenuse of $13$ cm, and one leg is $9$ cm. Find the angle $\theta$ between the $9$ cm leg and the hypotenuse.

Solution

The $9$ cm leg is next to $\theta$, so it’s adjacent. Adjacent and hypotenuse: CAH.

$$\cos \theta = \frac{9}{13} \quad\Rightarrow\quad \theta = \cos^{-1}\left(\frac{9}{13}\right) \approx 46^\circ$$

7. (Core) Solve right $\triangle XYZ$, where $\angle Y = 90^\circ$, $XY = 6.0$ cm and $YZ = 11.0$ cm.

Solution

Hypotenuse $XZ$, by the Pythagorean theorem:

$$XZ = \sqrt{6.0^2 + 11.0^2} = \sqrt{36 + 121} = \sqrt{157} \approx 12.5 \text{ cm}$$

Angle $X$. Relative to $\angle X$, $YZ$ is opposite and $XY$ is adjacent:

$$\tan X = \frac{11.0}{6.0} \quad\Rightarrow\quad \angle X = \tan^{-1}\left(\frac{11}{6}\right) \approx 61^\circ$$

Angle $Z$: $\angle Z \approx 90^\circ - 61.4^\circ \approx 29^\circ$. (Using the unrounded $61.39^\circ$ gives $28.6^\circ$, which still rounds to $29^\circ$.)

8. (Challenge) Let $A$ be an acute angle in a right triangle.

• (a) Explain why $\sin A$ and $\cos A$ are always less than $1$, but $\tan A$ can be greater than $1$.
• (b) For which angle is $\tan A = 1$? Explain using the triangle.
• (c) Explain why $\sin A = \cos(90^\circ - A)$.

Solution

(a) $\sin A$ and $\cos A$ both divide a leg by the hypotenuse. The hypotenuse is the longest side, so each fraction is less than $1$. $\tan A$ divides one leg by the other, and the opposite leg can be longer than the adjacent leg, so $\tan A$ can be greater than $1$ (for example, $\tan 55^\circ \approx 1.43$).

(b) $\tan A = 1$ when the opposite and adjacent legs are equal. Then the triangle is isosceles, so its two acute angles are equal, and each is $\dfrac{90^\circ}{2} = 45^\circ$. So $A = 45^\circ$.

(c) The two acute angles in a right triangle add to $90^\circ$, so the other acute angle is $90^\circ - A$. The side opposite $A$ is the side adjacent to the other angle, and the hypotenuse is the same. So $\sin A = \dfrac{\text{opposite } A}{\text{hypotenuse}} = \dfrac{\text{adjacent to } (90^\circ - A)}{\text{hypotenuse}} = \cos(90^\circ - A)$.

9. (Challenge) An isosceles triangle has two equal sides of $13$ cm and a base of $10$ cm. Find its three angles and its height. (Hint: the height from the top vertex splits the triangle into two congruent right triangles.)

Solution

The height meets the base at its midpoint, so each right triangle has a hypotenuse of $13$ cm and a base of $10 \div 2 = 5$ cm.

Base angle $\theta$: the $5$ cm side is adjacent, so

$$\cos \theta = \frac{5}{13} \quad\Rightarrow\quad \theta = \cos^{-1}\left(\frac{5}{13}\right) \approx 67^\circ$$

Top angle: $180^\circ - 2(67.38^\circ) \approx 45^\circ$.

Height, by the Pythagorean theorem:

$$h = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12 \text{ cm}$$

The angles are about $67^\circ$, $67^\circ$ and $45^\circ$, and the height is $12$ cm. (The rounded angles add to $179^\circ$; the unrounded ones, $67.38^\circ + 67.38^\circ + 45.24^\circ$, add to $180^\circ$.)