Six Trigonometric Ratios With Examples And Exercises

Trigonometric ratios are written as the ratios of two sides with respect to an acute angle of a right angled triangle. "T-ratios" is another acronym used for the word trigonometric ratios.

Let's start learning about a right angled triangle and then move on to its trigonometric ratios.

What is a right angled triangle?

A right angled triangle is a triangle which has one angle of measure 90°, out of its three angles. The other two angles of the triangle are always acute angles.

The name of a right angled triangle is written by placing the vertex with 90° at the middle of the other two vertices.

Right angled triangle ABC

ΔABC is a right angled triangle with vertex B with an angle of 90°.
Here, ∠ABC or ∠B = 90°
Also, ∠A and ∠C are acute angles.

Sides of right angled triangle

The three sides of a right angled triangle are named as base, perpendicular and hypotenuse.

Base: Base is the bottom side of the triangle.
Perpendicular: Perpendicular is the side of the right angled triangle which lies at 90° to the base.
Hypotenuse: The side of the right angled triangle which lies directly opposite to its angle of 90°.

Again, in the above diagram, the right angled triangle ΔABC has three sides AB, BC and AC.
The side AB is the base of the triangle, which lies at the bottom of ΔABC, which is called the base of ΔABC.
The side BC is at 90° to the base AB, which is called the perpendicular of ΔABC.
The side AC is opposite to ∠B, which is called as hypotenuse of ΔABC.

The angle formed between base and the perpendicular is always 90°.

Hypotenuse is the longest side than base and perpendicular in a right angled triangle.

What are trigonometric ratios?

Trigonometric ratios are the ratios of two sides of a right angled triangle. These ratios are calculated with respect to an acute angle in the triangle. There are six types of trigonometric ratios which are listed in the following table by their full and short names.

S.N.	Trigonometric ratio name	Abbreviation
1	sine	sin
2	cosine	cos
3	tangent	tan
4	cosecant	cosec
5	secant	sec
6	cotangent	cot

In the right angled ΔABC in the above diagram, the trigonometric ratios w.r.t. acute angle ∠A are written as:

Sine of ∠A as sin A
Cosine of ∠A as cos A
Tangent of ∠A as tan A
Cosecant of ∠A as cosec A
Secant of ∠A as sec A
Cotangent of ∠A as cot A

Cosecant, secant and cotangent are also known as reciprocals of sine, cosine and tangent respectively.

How to calculate trigonometric ratios?

In the right angled ΔABC, we can calculate the trigonometric ratios w.r.t. acute angle ∠A. To calculate them, first find out which are the opposite and adjacent sides to ∠A.
BC is opposite to ∠A
AB is adjacent to ∠A
AC is hypotenuse

Sine ratio

Sine of ∠A = $\frac{Side opposite to ∠A}{Hypotenuse} = \frac{BC}{AC}$

Here, BC is perpendicular and AC is hypotenuse. The ratio can also be written as:
sin A = $\frac{Perpendicular}{Hypotenuse}$

Cosine ratio

Cosine of ∠A = $\frac{Side adjacent to ∠A}{Hypotenuse} = \frac{AB}{AC}$

AB is base and AC is hypotenuse.
∴ cos A = $\frac{Base}{Hypotenuse}$

Tangent ratio

Tangent of ∠A = $\frac{Side opposite to ∠A}{Side adjacent to ∠A} = \frac{BC}{AB}$

Or ratio of tangent can be written as tan A = $\frac{Perpendicular}{Base}$

Cosecant ratio

$cosec A = \frac{1}{sin A}$

Cosecant of ∠A = $\frac{Hypotenuse}{Side opposite to ∠A} = \frac{AC}{BC}$

Secant ratio

Secant of ∠A = $\frac{Hypotenuse}{Side adjacent to ∠A} = \frac{AC}{AB}$

$sec A = \frac{1}{cos A}$

Cotangent ratio

$cot A = \frac{1}{tan A}$

Cotangent of ∠A = $\frac{Side adjacent to ∠A}{Side opposite to ∠A} = \frac{AB}{BC}$

cot A can also be written as: $cot A = \frac{cos A}{sin A}$

The relationship between sin, cos and tan is $tan A = \frac{sin A}{cos A}$

Example of trigonometric ratios calculations

In ΔABC, right angled at B, AB = 8cm, BC = 6cm. Find all six trigonometric ratios of angle ∠A.

In ΔABC,
BC = Perpendicular = 6cm
AB = Base = 8cm
AC = Hypotenuse = ?
Using Pythagoras theorem,
${(AC)}^{2} = {(AB)}^{2} + {(BC)}^{2}$
$= {(8)}^{2} + {(6)}^{2}$
= 64 + 36
= 100
∴ AC = 10cm or Hypotenuse = 10cm
$sin A = \frac{P}{H} = \frac{BC}{AC} = \frac{6}{10}$
$cos A = \frac{B}{H} = \frac{AB}{AC} = \frac{8}{10}$
$tan A = \frac{P}{B} = \frac{BC}{AB} = \frac{6}{8}$
$cosec A = \frac{H}{P} = \frac{AC}{BC} = \frac{10}{6}$
$sec A = \frac{H}{B} = \frac{AC}{AB} = \frac{10}{8}$
$cot A = \frac{B}{P} = \frac{AB}{BC} = \frac{8}{6}$

The values of trigonometric ratios of an angle do not vary with length of sides of a triangle, if the angle involved in the ratios remains same.

Multiplication of trigonometric ratios

When sin, cos and tan are multiplied by cosec, sec and cot respectively, the result obtained is always 1.

Sin multiplied by cosec:
sin A . cosec A = 1

Cos multiplied by sec:
cos A . sec A = 1

Tan multiplied by cot:
tan A . cot A = 1

How to memorize trigonometric ratios?

Trigonometric ratios can be easily remembered using this visual.

Trigonometric ratios table

Trigonometric ratios mnemonics table

To make it more easy to remember, this shorter form can also be used. It can be recalled as "SPH - CBH - TPB".

Solved Examples

1) In ΔABC, right angled at B, AB = 12cm and BC = 5cm. Find the value of sin A and cos A.

Triangle ABC right angled at B with base=12cm, perpendicular=5cm

As we know, sin A = $\frac{BC}{AC}$
and cos A = $\frac{AB}{AC}$
In ΔABC,
AB = 12cm
BC = 5cm
but is AC is not given.
So, by Pythagoras theorem
${(AC)}^{2} = {(AB)}^{2} + {(BC)}^{2}$
${(AC)}^{2} = {(12)}^{2} + {(5)}^{2}$
${(AC)}^{2} = 144 + 225$
${(AC)}^{2} = 169$
AC = 13cm
∴ sin A = $\frac{P}{H} = \frac{BC}{AC} = \frac{5}{13}$
cos A = $\frac{B}{H} = \frac{AB}{AC} = \frac{12}{13}$

2) In ΔPQR, right angled at Q, PQ = 15cm and QR = 8cm. What will be the value of tan θ and cot θ?

Triangle PQR right angled at Q with base=15cm, perpendicular=8cm

Here,
Base = PQ = 15cm
Perpendicular QR = 8cm
By Pythagoras theorem
${(PR)}^{2} = {(PQ)}^{2} + {(QR)}^{2}$
${(PR)}^{2} = {(15)}^{2} + {(8)}^{2}$
${(PR)}^{2} = 225 + 64$
${(PR)}^{2} = 289$
PR = 17cm
∴ tan θ = $\frac{P}{B} = \frac{QR}{PQ} = \frac{8}{15}$
cot θ = $\frac{B}{P} = \frac{QR}{PQ} = \frac{8}{15}$

3) In ΔABC with right angle at B, if sin θ = $\frac{5}{13}$ . Find the value of cos θ and tan θ.

Triangle ABC right angled at B with angle A theta

sin θ = $\frac{5}{13} = \frac{Perpendicular}{Hypotenuse} = \frac{BC}{AB}$
∴ BC = 5
AC = 13
By Pythagoras theorem
${(AC)}^{2} = {(AB)}^{2} + {(BC)}^{2}$
${(13)}^{2} = {(AB)}^{2} + {(5)}^{2}$
$169 = {(AB)}^{2} + 25$
${(13)}^{2} = {(AB)}^{2} + {(5)}^{2}$
${(AB)}^{2} = 169 - 25$
AB = 12
cos θ = $\frac{Base}{Hypotenuse} = \frac{AB}{AC} = \frac{12}{13}$
∴ tan θ = $\frac{Perpendicular}{Base} = \frac{BC}{AB} = \frac{5}{12}$

4) In ΔABC, with ∠B = 90°, if AB = 4cm, AC = 5cm. Find the value of sin A and tan C.

Triangle ABC right angled at B with base=4cm, hypotenuse=5cm

In ΔABC,
AB = 4cm
AC = 5cm
By Pythagoras theorem
${(AC)}^{2} = {(AB)}^{2} + {(BC)}^{2}$
${(5)}^{2} = {(4)}^{2} + {(BC)}^{2}$
$25 = 16 + {(BC)}^{2}$
$25 - 16 = {(BC)}^{2}$
$9 = {(BC)}^{2}$
BC = 3cm
sin A = $\frac{Perpendicular}{Hypotenuse} = \frac{BC}{AC} = \frac{3}{5}$

To find tan C, BC is the adjacent side to ∠C and AB is the opposite side to ∠C.
BC becomes the base and AB becomes the perpendicular.

∴ tan C = $\frac{Perpendicular}{Base} = \frac{AB}{BC} = \frac{4}{3}$

5) In ΔPQR, with ∠Q = 90°, PQ = 4cm, QR = 3cm and PR = 5cm. Find the value of sin P cos R + cos P sin R.

Triangle PQR right angled at Q with base=4cm, perpendicular=3cm

To find sin P and cos P ratios, the adjacent side (base) to ∠P is PQ and opposite side (perpendicular) is QR.
The hypotenuse of ΔPQR is PR.
sin P = $\frac{Perpendicular}{Hypotenuse} = \frac{QR}{PR} = \frac{3}{5}$
cos P = $\frac{Base}{Hypotenuse} = \frac{PQ}{PR} = \frac{4}{5}$

To find sin R and cos R, the adjacent side(base) to ∠R is QR and opposite side(perpendicular) is PQ.

sin R = $\frac{Perpendicular}{Hypotenuse} = \frac{PQ}{PR} = \frac{4}{5}$
cos R = $\frac{Base}{Hypotenuse} = \frac{QR}{PR} = \frac{3}{5}$

Now, fill in the values of sin P, cos P, sin R and cos R in sin P cos R + cos P sin R
sin P cos R + cos P sin R =

(\frac{3}{5}) (\frac{3}{5}) + (\frac{4}{5}) (\frac{4}{5})

\frac{9}{25} + \frac{16}{25}

\frac{25}{25}

= 1

Write True or False Worksheet

Type:	True False
Count:	1

Choose the correct answers from ΔPQR.

Triangle PQR right angled at Q, PQ=5, QR=12 and PR=13 cm

S.N.	Statement	✓ or ✕
1)	The value of sin P is $\frac{12}{13}$
2)	cos P = $\frac{13}{12}$
3)	sec R = $\frac{13}{12}$
4)	tan P = $\frac{12}{5}$
5)	tan P + cot R = $\frac{24}{5}$
6)	sin R cosec R = 1
7)	tan² P = $\frac{12}{5}$
8)	sec² P = $\frac{13}{12}$
9)	cot² P = $\frac{25}{144}$
10)	cos R sin R = 0

True False PDF Worksheet

Match Columns Worksheet

Type:	Matching
Count:	1

Choose the correct answers from ΔLNM.

Trigonometric ratio		Value
1)	sin θ	a)	$\frac{13}{12}$
2)	cos θ	b)	$\frac{12}{5}$
3)	cosec θ	c)	$\frac{5}{12}$
4)	tan θ	d)	$\frac{5}{13}$
5)	cot θ	e)	$\frac{13}{5}$
6)	sec θ	f)	$\frac{12}{13}$

Matching PDF Worksheet

Solve Questions Worksheet

Type:	Solve Questions
Count:	1

If ΔABC is a right angled triangle at B and its base is 7cm and perpendicular is 4cm. Determine the value of cos B and tan B.
If ΔPQR right angled at Q with base 3cm and hypotenuse 7cm. Calculate the value of cot Q and sec Q.
In ΔLMN has ∠M = 90° with hypotenuse 13cm and base 12cm. Find all trigonometric ratios.
In ΔABC with ∠B = 90°. Find hypotenuse if base is 7cm and perpendicular is 24cm. Find sin A and cos A.
In ΔPQR with ∠Q = 90°, base = 2cm and perpendicular = 4cm. Calculate sin P and tan P.

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Learn Trigonometric Ratios, Formulas And Calculations

What is a right angled triangle?

Sides of right angled triangle

What are trigonometric ratios?