Trigonometric Ratios - Definition, Formulas and Examples

Trigonometric ratios are the ratios of the lengths of sides of a triangle. These ratios in trigonometry relate the ratio of sides of a right triangle to the respective angle. In real life, we use trigonometric ratios in navigation, in oceanography. It is also used in the creation of maps.

In this article, we will cover the concept of Trigonometric Ratios. This category falls under the broader category of Trigonometry, which is a crucial Chapter in class 11 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination(JEE Main) and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. Questions based on this topic have been asked frequently in JEE Mains.

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This Story also Contains

1. Trigonometric ratios
2. Trigonometric Functions of Acute Angles
3. Trigonometric Ratios of some Special Angles
4. Summary
5. Solved Examples Based on Trigonometric Ratios

Trigonometric ratios

The trigonometric ratios of a given angle are the ratios of a right-angled triangle's sides with regard to any one of its acute angles. Trigonometric ratios are the ratios of the lengths of sides of a triangle.

Trigonometry ratios connect the side of a triangle with the angle.

The six trigonometric ratios are sine (sin) , cosine (cos) , tangent (tan), cotangent (cot) , secant (sec), cosecant (cosec).

Sine (sin) - sin is defined as a ratio of the side opposite to that angle (perpendicular) and hypotenuse.

Cosine (cos) - cos is defined as the ratio of the side adjacent to that angle (base) and hypotenuse.

Tangent (tan) - tan is defined as the ratio of the side opposite to that angle (perpendicular) and the side adjacent to that angle (base).

cotangent (cot) - cot is defined as the ratio of the side adjacent to that angle (base) and the side opposite to that angle (perpendicular). It is the reciprocal of tan.

secant (sec) - secant is defined as the ratio of the hypotenuse and the side adjacent to that angle (base). It is reciprocal of cos.

Cosecant (cosec) - cosec is defined as the ratio of the hypotenuse and the side opposite to that angle (perpendicular). It is reciprocal of the sin.

Trigonometric Functions of Acute Angles

We can define the trigonometric functions in terms of an angle t and the lengths of the sides of the triangle. The adjacent side (=x) is the side closest to the angle (Adjacent means “next to.”). The opposite side (=y) is the side across from the angle. The hypotenuse (=1) is the side of the triangle opposite the right angle.

$\begin{array}{rl}& \text{Sine}\sin t=\frac{\text{opposite}}{\text{hypotenuse}}\\ & \text{Cosine}\cos t=\frac{\text{adjacent}}{\text{hypotenuse}}\\ & \text{Tangent}\tan t=\frac{\text{opposite}}{\text{adjacent}}\end{array}$

Reciprocal Functions

In addition to sine, cosine, and tangent, there are three more functions. These too are defined in terms of the sides of the triangle. These are the reciprocals of primary trigonometric functions sine, cosine, and tangent. Cosecant is the reciprocal of sine. Secant is the reciprocal of Cosine and Cotangent is the reciprocal of tangent. So, Cosecant, Secant, and Cotangent are referred to as reciprocal functions.

$\begin{array}{rl}& \text{Cosecant}\csc t=\frac{\text{hypotenuse}}{\text{opposite}}=\frac{1}{\sin t}\\ & \text{Secent}\sec t=\frac{\text{hypotenuse}}{\text{adjacent}}=\frac{1}{\cos t}\\ & \text{Cotangent}\cot t=\frac{\text{adjacent}}{\text{opposite}}=\frac{1}{\tan t}\end{array}$

Since the hypotenuse is the greatest side in a right-angle triangle, sin t, and cos t can never be greater than unity, and cosec t and sec t can never be less than unity.

Trigonometric Ratios of some Special Angles

In the trigonometric ratios table, we use the values of trigonometric ratios for standard angles 0°, 30°, 45°, 60°, and 90º. It is easy to predict the values of the table and to use the table as a reference to calculate values of trigonometric ratios for various other angles, using the trigonometric ratio formulas for existing patterns within trigonometric ratios and even between angles.

Below are the trigonometric ratios of some special angles (0, 30, 45, 60, 90)

Summary

Trigonometric ratios enable us to understand and analyze relationships between angles and sides of triangles. Understanding trigonometric ratios provides a powerful framework for solving a wide range of problems.

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Solved Examples Based on Trigonometric Ratios

Example 1: The angle of elevation of the top of a vertical tower standing on a horizontal plane is observed to be ${45}^{\circ }$ from point A on the plane. Let B be the point 30 m vertically above point A. If the angle of elevation of distance (in m) of the foot of the tower from point A is: [JEE MAINS 2019]

$\begin{array}{rl}& \tan {30}^{\circ }=\frac{K}{d}\Rightarrow d=\sqrt{3}K\cdots \cdots \\ & \tan {45}^{\circ }=\frac{K+30}{d}\Rightarrow d=K+30\end{array}$

from (1) and (2)

$\begin{array}{rl}& d=\frac{d}{\sqrt{3}}+30=(1-\frac{1}{\sqrt{3}})d=30\\ & \Rightarrow d=\frac{30\sqrt{3}}{\sqrt{3}-1}\Rightarrow d=\frac{30\sqrt{3}(\sqrt{3}+1)}{2}=15\sqrt{3}(\sqrt{3}+1)\end{array}$

or $d=15(3+\sqrt{3})$

Hence, the required answer is $15(3+\sqrt{3})$.

Example 2: If the angle of elevation of a cloud from a point P which is 25 m above a lake be ${30}^0$ and the angle of depression of reflection of the cloud in the lake from P be ${60}^0$ then the height of the cloud (in meters) from the surface of the lake is : [JEE MAINS 2019]

Solution:

$5\cos A+3=0\Rightarrow \cos A=-3/5$

On solving $9x^2+27x+20=0$

$\begin{array}{rl}& x=-5/3\text{and}x=-4/3\\ & \text{so}x=\sec A;x=\tan A\end{array}$

Example 4: ABC is a triangular park with AB = AC = 100 meters. A vertical tower is situated at the midpoint of BC. if the angles of elevation of the top of the tower at A and B are ${\cot }^{-1}(2\sqrt{2})$ respectively, then the height of the tower (in meters) is : [JEE MAINS 2019]

Solution:

$\begin{array}{rl}& \text{AM}=\mathrm{y};\mathrm{MD}=\mathrm{h}(\text{tower height})\\ & y^2=100-x^2\\ & \cot \theta =3\sqrt{2}\Rightarrow \tan \theta =\frac{1}{3\sqrt{2}}\\ & \frac{h}{y}=\frac{1}{3\sqrt{2}}\\ & \Rightarrow >y=3\sqrt{2}h\\ & \text{from (1)}\\ & 18h^2=100-x^2\\ & \tan \varphi =\frac{1}{\sqrt{7}}=\frac{h}{x}\\ & h\sqrt{7}=x\\ & {100}^2-7h^2=18h^2\\ & {100}^2=25h^2\\ & \frac{100}{5}=h\\ & 20=h\end{array}$

Example 5: If $\cos \theta =p$; then evaluate the value of $\tan \theta +\sec \theta$ (Given $\theta$ is an acute angle)
1) $\frac{\sqrt{p^2-1}+1}{p}$
2) $\frac{\sqrt{1-p^2}+1}{p}$
3) $\frac{p+1}{p}$
4) None of these

Solution:

$\begin{array}{rl}& \cos \theta =p;\sec \theta =\frac{1}{p}\mathrm{\&}\tan \theta =\frac{\sqrt{1-p^2}}{p}\\ & \Rightarrow \tan \theta +\sec \theta =\frac{\sqrt{1-p^2}+1}{n}\end{array}$

Hence, the required answer is option 2.