Trigonometric Ratios Explained with Examples, Class 10

Sides rel­a­tive to angle A#

• The side oppo­site to $\angle A$ is $BC$ (it does not touch ver­tex $A$).
• The side adja­cent to $\angle A$ is $AB$ (it touches $A$ and is not the hypotenuse).
• The hypotenuse is $AC$.

Sides rel­a­tive to angle C#

The names change when you switch angles. For $\angle C$, the oppo­site side is $AB$ and the adja­cent side is $BC$. The hypotenuse stays $AC$. Always ask "which angle am I work­ing with?" before you write a ratio.

The six trigono­met­ric ratios#

Def­i­n­i­tions#

For an acute angle $A$ in a tri­an­gle right-angled at $B$:

$$\displaystyle \begin{aligned} \sin A &= \frac{\text{side opposite to } \angle A}{\text{hypotenuse}} = \frac{BC}{AC}, & \operatorname{cosec} A &= \frac{AC}{BC}, \\ \cos A &= \frac{\text{side adjacent to } \angle A}{\text{hypotenuse}} = \frac{AB}{AC}, & \sec A &= \frac{AC}{AB}, \\ \tan A &= \frac{\text{side opposite to } \angle A}{\text{side adjacent to } \angle A} = \frac{BC}{AB}, & \cot A &= \frac{AB}{BC}. \end{aligned}$$

The names are short for sine, cosine, tan­gent, cose­cant, secant and cotan­gent. In the tri­an­gle above, $\displaystyle \sin A = \frac{3}{5}$, $\displaystyle \cos A = \frac{4}{5}$ and $\displaystyle \tan A = \frac{3}{4}$.

Rec­i­p­ro­cal rela­tions#

The last three ratios are the first three turned upside down:

$$\displaystyle \operatorname{cosec} A = \frac{1}{\sin A}, \qquad \sec A = \frac{1}{\cos A}, \qquad \cot A = \frac{1}{\tan A}.$$

So if you know $\sin A$, $\cos A$ and $\tan A$, you know all six.

Tan as a quo­tient#

Divide $\sin A$ by $\cos A$:

$$\displaystyle \frac{\sin A}{\cos A} = \frac{BC/AC}{AB/AC} = \frac{BC}{AB} = \tan A.$$

Hence $\displaystyle \tan A = \frac{\sin A}{\cos A}$ and $\displaystyle \cot A = \frac{\cos A}{\sin A}$. These two facts are used con­stantly when sim­pli­fy­ing expres­sions.

Why the ratios depend only on the angle#

A nat­ural worry is that a big­ger tri­an­gle with the same angle $A$ might give a dif­fer­ent value of $\sin A$. It does not, and the rea­son is sim­i­lar­ity.

The sim­i­lar-tri­an­gles argu­ment#

Draw a right tri­an­gle $ABC$ (right angle at $B$) and extend $AB$ to a point $Q$. Draw $QR$ per­pen­dic­u­lar to $AQ$, meet­ing $AC$ extended at $R$. Tri­an­gles $ABC$ and $AQR$ have $\angle A$ in com­mon and a right angle each, so by the AA cri­te­rion $\triangle ABC \sim \triangle AQR$. Cor­re­spond­ing sides of sim­i­lar tri­an­gles are pro­por­tional:

$$\displaystyle \frac{BC}{QR} = \frac{AC}{AR} \quad \Longrightarrow \quad \frac{BC}{AC} = \frac{QR}{AR}.$$

The left side is $\sin A$ worked out from the small tri­an­gle; the right side is $\sin A$ worked out from the large one. They are equal. The same argu­ment works for every other ratio.

This is why we can write $\sin A$ as a prop­erty of the angle alone. Note that $\sin A$ is a sin­gle sym­bol: it is not "sin" mul­ti­plied by $A$.

Why sin A is less than 1 and sec A is at least 1#

For an acute angle $A$, the hypotenuse is the longest side of the right tri­an­gle, so the oppo­site side is shorter than it. There­fore

$$\displaystyle \sin A = \frac{\text{opposite}}{\text{hypotenuse}} \lt 1 \quad \text{and} \quad \cos A = \frac{\text{adjacent}}{\text{hypotenuse}} \lt 1.$$

Tak­ing rec­i­p­ro­cals of pos­i­tive num­bers less than 1 gives num­bers greater than 1, so $\sec A \gt 1$ and $\operatorname{cosec} A \gt 1$ for every acute angle. In the next les­son you will meet $A = 0^\circ$, where $\cos 0^\circ = 1$ and $\sec 0^\circ = 1$. Putting these together, $\sec A \ge 1$ when­ever it is defined, and a value such as $\displaystyle \cos A = \frac{5}{4}$ or $\displaystyle \sec A = \frac{2}{3}$ is impos­si­ble. The tan­gent has no such limit: $\tan A$ can be any pos­i­tive num­ber, because the oppo­site side can be shorter or longer than the adja­cent side.

Method: given one ratio, find the oth­ers#

1. Write the given ratio as a frac­tion of two sides, for exam­ple $\displaystyle \tan A = \frac{12}{35}$ means oppo­site : adja­cent $= 12 : 35$.
2. Let the two sides be $12k$ and $35k$ for some pos­i­tive num­ber $k$.
3. Find the third side with Pythago­ras' the­o­rem.
4. Read off the remain­ing ratios from the def­i­n­i­tions; the $k$ can­cels in every ratio.
5. Use the rec­i­p­ro­cal rela­tions for cosec, sec and cot.
6. Check that $\sin A$ and $\cos A$ are both less than 1.

Worked exam­ples#

Exam­ple 1: all six ratios from the sides#

In $\triangle ABC$, right-angled at $B$, $AB = 12$ cm and $BC = 5$ cm. Find all six trigono­met­ric ratios of $\angle A$.

Solu­tion. By Pythago­ras' the­o­rem,

$$AC^2 = AB^2 + BC^2 = 144 + 25 = 169, \quad AC = 13 \text{ cm}.$$

For $\angle A$: oppo­site $= BC = 5$, adja­cent $= AB = 12$, hypotenuse $= 13$.

$$\displaystyle \sin A = \frac{5}{13},\quad \cos A = \frac{12}{13},\quad \tan A = \frac{5}{12},\quad \operatorname{cosec} A = \frac{13}{5},\quad \sec A = \frac{13}{12},\quad \cot A = \frac{12}{5}.$$

As a check, $\displaystyle \frac{\sin A}{\cos A} = \frac{5/13}{12/13} = \frac{5}{12} = \tan A$.

Exam­ple 2: from tan to the rest#

If $\displaystyle \tan A = \frac{12}{35}$, find the other five ratios.

Solu­tion. Let the oppo­site side be $12k$ and the adja­cent side $35k$. Then

$$\text{hypotenuse} = \sqrt{(12k)^2 + (35k)^2} = \sqrt{1369k^2} = 37k.$$

$$\displaystyle \sin A = \frac{12k}{37k} = \frac{12}{37},\quad \cos A = \frac{35}{37},\quad \operatorname{cosec} A = \frac{37}{12},\quad \sec A = \frac{37}{35},\quad \cot A = \frac{35}{12}.$$

Exam­ple 3: start­ing from sec#

Given $\displaystyle \sec \theta = \frac{25}{7}$, find $\sin \theta$ and $\cot \theta$.

Solu­tion. $\displaystyle \sec \theta = \frac{\text{hypotenuse}}{\text{adjacent}}$, so take hypotenuse $= 25k$ and adja­cent side $= 7k$. The oppo­site side is

$$\sqrt{(25k)^2 - (7k)^2} = \sqrt{625k^2 - 49k^2} = \sqrt{576k^2} = 24k.$$

There­fore $\displaystyle \sin \theta = \frac{24}{25}$ and $\displaystyle \cot \theta = \frac{\text{adjacent}}{\text{opposite}} = \frac{7}{24}$.

Exam­ple 4: eval­u­at­ing an expres­sion#

If $5 \cot A = 12$, find the value of $\displaystyle \frac{1 - \tan^2 A}{1 + \tan^2 A}$.

Solu­tion. $\displaystyle \cot A = \frac{12}{5}$, so $\displaystyle \tan A = \frac{5}{12}$ and $\displaystyle \tan^2 A = \frac{25}{144}$.

$$\displaystyle \frac{1 - \tan^2 A}{1 + \tan^2 A} = \frac{1 - \frac{25}{144}}{1 + \frac{25}{144}} = \frac{\frac{119}{144}}{\frac{169}{144}} = \frac{119}{169}.$$

You can cross-check with the sides $5, 12, 13$: $\displaystyle \cos^2 A - \sin^2 A = \frac{144}{169} - \frac{25}{169} = \frac{119}{169}$, which agrees.

Exam­ple 5: when the sides are hid­den in a con­di­tion#

In $\triangle PQR$, right-angled at $Q$, $PQ = 12$ cm and $PR + QR = 18$ cm. Find $\sin P$, $\cos P$ and $\tan P$.

Solu­tion. By Pythago­ras' the­o­rem, $PR^2 = PQ^2 + QR^2$, so

$$\displaystyle PR^2 - QR^2 = 144 \quad \Longrightarrow \quad (PR - QR)(PR + QR) = 144 \quad \Longrightarrow \quad PR - QR = \frac{144}{18} = 8.$$

Adding $PR + QR = 18$ and $PR - QR = 8$ gives $2PR = 26$, so $PR = 13$ cm and $QR = 5$ cm.

For $\angle P$: oppo­site $= QR = 5$, adja­cent $= PQ = 12$, hypotenuse $= PR = 13$.

$$\displaystyle \sin P = \frac{5}{13}, \qquad \cos P = \frac{12}{13}, \qquad \tan P = \frac{5}{12}.$$

Exam­ple 6: a surd in the answer#

If $A$ is acute and $\displaystyle \sin A = \frac{2}{3}$, find the value of $\sec A + \tan A$.

Solu­tion. Take oppo­site $= 2k$, hypotenuse $= 3k$. The adja­cent side is $\sqrt{9k^2 - 4k^2} = \sqrt{5}\,k$. So

$$\displaystyle \sec A = \frac{3}{\sqrt{5}}, \qquad \tan A = \frac{2}{\sqrt{5}}, \qquad \sec A + \tan A = \frac{5}{\sqrt{5}} = \sqrt{5}.$$

Com­mon mis­takes#

• Nam­ing the oppo­site and adja­cent sides for the wrong angle. Mark the angle on your sketch first.
• Using the hypotenuse as the "adja­cent" side. The adja­cent side is never the hypotenuse.
• Treat­ing $\sin A$ as $\sin \times A$ and "can­celling" the $A$, as in $\displaystyle \frac{\sin A}{A} = \sin$.
• For­get­ting the square root after Pythago­ras' the­o­rem, and writ­ing a side as $169$ instead of $13$.
• Accept­ing an answer such as $\displaystyle \cos A = \frac{13}{12}$. Sine and cosine of an acute angle are always less than 1.
• Leav­ing $\sqrt{5}$ in a denom­i­na­tor when the ques­tion asks for a sim­pli­fied value. Ratio­nalise where needed.

Try these#

1. If $\displaystyle \sin A = \frac{9}{41}$, find $\cos A$ and $\tan A$. Answer: $\displaystyle \cos A = \frac{40}{41}$, $\displaystyle \tan A = \frac{9}{40}$.
2. The two shorter sides of a right tri­an­gle are 20 cm and 21 cm. For the angle oppo­site the 20 cm side, find sin, cos and tan. Answer: $\displaystyle \frac{20}{29}$, $\displaystyle \frac{21}{29}$, $\displaystyle \frac{20}{21}$.
3. If $\displaystyle \cos \theta = \frac{3}{5}$, eval­u­ate $\displaystyle \frac{\sin \theta + \cos \theta}{\sin \theta - \cos \theta}$. Answer: $7$.
4. If $2 \tan A = 1$, find $\sin A \cos A$. Answer: $\displaystyle \frac{2}{5}$.
5. In $\triangle PQR$, right-angled at $Q$, $PQ = 6$ cm and $PR - QR = 2$ cm. Find $\sin P$. Answer: $\displaystyle \frac{4}{5}$ (with $PR = 10$ cm, $QR = 8$ cm).

Key terms#

Hypotenuse

The side oppo­site the right angle; the longest side of a right tri­an­gle.

Oppo­site side

The side of a right tri­an­gle fac­ing the acute angle under dis­cus­sion.

Adja­cent side

The side that forms the acute angle together with the hypotenuse.

Sine

The ratio of the oppo­site side to the hypotenuse.

Cosine

The ratio of the adja­cent side to the hypotenuse.

Tan­gent

The ratio of the oppo­site side to the adja­cent side; also equal to sine divided by cosine.

Rec­i­p­ro­cal ratios

Cose­cant, secant and cotan­gent, which are the rec­i­p­ro­cals of sine, cosine and tan­gent.

Sim­i­lar tri­an­gles

Tri­an­gles with equal cor­re­spond­ing angles, whose cor­re­spond­ing sides are there­fore in the same ratio.

Com­mon ques­tions#

Do I need to know the actual lengths to find a ratio?#

No. Only the ratio of two sides mat­ters. That is why you may take the sides as $12k$ and $35k$; the $k$ can­cels.

Can tan A be greater than 1?#

Yes. If the oppo­site side is longer than the adja­cent side, $\tan A \gt 1$. Only sine and cosine are lim­ited to val­ues below 1 for acute angles.

Why is cosec writ­ten instead of csc?#

Both are abbre­vi­a­tions of cose­cant. The Indian school syl­labus uses cosec, so write cosec in board exam­i­na­tions.

Does the ratio change if the tri­an­gle is drawn in a dif­fer­ent posi­tion?#

No. Turn­ing, flip­ping or enlarg­ing the tri­an­gle keeps the angle the same, so every ratio stays the same.

What does sin² A mean?#

It means $(\sin A)^2$, the square of the value of $\sin A$. It does not mean $\sin (A^2)$.

Ref­er­ences#

1. National Coun­cil of Edu­ca­tional Research and Train­ing. Math­e­mat­ics: Text­book for Class X. NCERT, New Delhi.
2. Loney, S. L. Plane Trigonom­e­try. Cam­bridge Uni­ver­sity Press.
3. Aggar­wal, R. S. Sec­ondary School Math­e­mat­ics for Class 10. Bharati Bhawan.