Trigonometry is a very important branch of mathematics, which basically deals with triangles and their sides and angles. Trigonometry illustrates six trigonometric functions which are sine, cosine, tangent, cosecant, secant and cotangent. These trigonometric functions are defined in context with right-angled triangle and are expressed as the ratio of two sides of right triangle. These ratios are called trigonometric ratios.

Trigonometric ratios are perhaps the most important part of mathematics. Apart from maths, these ratios are used in physics, chemistry and other branches of science. So, it is essential to have sufficient understanding of trigonometric ratios.

All six trigonometric ratios are defined in right triangle. Let us assume a right triangle $\bigtriangleup$ABC, in which $\angle $ABC is right angle and $\angle $ACB is angle $\theta$(say). Side opposite to right angle is called hypotenuse, side opposite to $\theta$ is known as perpendicular and third side is assumed as the base.
Trigonometric Ratios
Trigonometric ratios sine, cosine, tangent, cosecant, secant and cotangent are abbreviated as sin, cos, tan, cosec (or csc), sec and cot respectively. We are going to define all six trigonometric ratios for angle $\theta$ as shown below:
Trigonometric Ratios In Right Triangles
We can also conclude that:
$\csc \theta$ = $\frac{1}{\sin \theta }$

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

$\cot \theta$ = $\frac{1}{\tan \theta }$
Few values of trigonometric ratios with respect to few important angles are listed in the following table:
Angle $\theta$  0$^o$ 30$^o$ 45$^o$ 60$^o$ 90$^o$
sin $\theta$ 0
cos $\theta$ $\frac{\sqrt{3}}{2}$
tan $\theta$ 0
cosec $\theta$ $\infty$
sec $\theta$ 1
cot $\theta$ $\infty$ 

Inverses of all six trigonometric ratios do exist which are denoted by sin-1  $\theta$, cos-1 $\theta$, tan-1 $\theta$, cosec-1 $\theta$, sec-1 $\theta$ and cot-1 $\theta$ or arc sin $\theta$, arc cos $\theta$, arc tan $\theta$, arc cosec $\theta$, arc sec $\theta$ and arc cot $\theta$.
If $y = \sin \theta $, then $\theta = \sin^{-1}y $. It is similar for all other trigonometric ratios.
Inverse of a function is not equal to reciprocal of that function. It implies that:
$\sin^{-1}x \neq $$\frac{1}{\sin\ x}$ and this is similar for all other trigonometric ratios.
Some examples based on trigonometric ratios are discussed below:

Solved Examples

Question 1: Consider a right triangle $\bigtriangleup$ABC with AB and BC measuring 3 cm and 4 cm. Find all the trigonometric ratios with respect to the angle A.
Trigonometric Ratios Examples
By Pythagorean theorem:



$AC= 5\ cm$

$\sin A$ = $\frac{Perpendicular}{Hypotenuse}$

$\sin A$ = $\frac{4}{5}$

$\cos A$ = $\frac{Base}{Hypotenuse}$

$\cos A$ = $\frac{3}{5}$

$\tan A$ = $\frac{Perpendicular}{Base}$

$\tan A$ = $\frac{4}{3}$

$\csc A$ = $\frac{Hypotenuse}{Perpendicular}$

$\csc A$ = $\frac{5}{4}$

$\sec A$ = $\frac{Hypotenuse}{Base}$

$\sec A$ = $\frac{5}{3}$

$\cot \theta$ = $\frac{Base}{Perpendicular}$

$\cot A$ = $\frac{3}{4}$

Question 2: In the given figure, find the measure of angle C.
Trigonometric Ratios Example
In $\bigtriangleup$ABC:
$\sin C$ = $\frac{Perpendicular}{Hypotenuse}$

$\sin C$ = $\frac{\sqrt{3}}{2}$

$\sin C$ = $\sin60^{\circ}$

$\angle C$ = $60^{\circ}$

Given below are some of the word problems on trigonometric ratios.

Solved Examples

Question 1: A ladder of length 10 m is placed against the wall. At what distance from wall it should be kept to make it inclined at an angle of $60^{\circ}$ from the ground?
Following figure is obtained according to the information mentioned in question.
Trigonometric Ratios Word Problems
In $\bigtriangleup$ABC:
$\cos C$ = $\frac{Base}{Hypotenuse}$

$\cos 60^{\circ}$ = $\frac{BC}{10}$


BC = 5

The ladder is 5 m away from the wall.

Question 2: A helicopter is flying at a constant height from the ground. It makes an angle of $45^{\circ}$, when seen from a fixed point on the ground. After some time, when helicopter moves 2000 feet ahead, it is noted that it makes an angle of $60^{\circ}$ from that fixed point. Calculate the height of the helicopter.
Following figure is obtained according to the given information. Here, A and B are two positions of helicopter and O be the fixed point. Let us suppose that height of helicopter is "h" feet and OC be x feet. AC and BD are perpendicular drawn from point A and B to the ground.
Trigonometric Ratios Word Problem
In $\bigtriangleup$ACO:

$\tan 45^{\circ}$ = $\frac{h}{x}$

1 = $\frac{h}{x}$

h = x    ........1

In $\bigtriangleup$BDO:

$\tan 60^{\circ}$ = $\frac{h}{x - 2000}$

Using equation 1, we get:

$\sqrt{3}$ = $\frac{h}{h - 2000}$

$\sqrt{3}(h - 2000) = h$

$h(\sqrt{3} - 1) = 2000$

$h$ = $\frac{2000}{(\sqrt{3} - 1)}$

$h = 1000(\sqrt{3} + 1)$

h = 2732

Height of the helicopter from the ground is 2732 feet (approx).