Trigonometric Ratios of Some Specific Angles: The term “trigonometry” comes from the Greek words “trigonon” (meaning “triangle”) and “metron” (meaning “to measure”). It is an area of mathematics that studies the relationship between a triangle’s angles and sides.

To find the ratios of the sides of a right-angled triangle, we can take two sides at a time from a total of three sides. These are known as trigonometric ratios. Trigonometry is one of the oldest sciences studied by experts all over the world. Sine, cosine, tangent, secant, cosecant, and cotangent are the six basic trigonometric ratios.

What are Trigonometric Ratios?

The values of all trigonometric functions based on the ratio of sides of a right-angled triangle are known as trigonometric ratios. The trigonometric ratios of a right-angled triangle’s sides for any acute angles are known as that angle’s trigonometric ratios.

Six different trigonometric ratios are used: sine, cosine, tangent, secant, cosecant, and cotangent are the six basic trigonometric ratios.

Trigonometric Ratios of Angle

Take a look at the right-angled triangle in the diagram below.

In the right triangle \(ABC\), the trigonometric ratios of the angle \(C\) are:

sine of \(\angle C = \frac{{{\rm{the}}\,{\rm{side}}\,{\rm{opposite}}\,{\rm{to}}\,{\rm{angle}}\,C}}{{{\rm{hypotenuse}}}} = \frac{{BA}}{{CA}}\)

cosine of \(\angle C = \frac{{{\rm{the}}\,{\rm{side}}\,{\rm{adjacent}}\,{\rm{to}}\,{\rm{angle}}\,C}}{{{\rm{hypotenuse}}}} = \frac{{CB}}{{CA}}\)

tangent of \(\angle C = \frac{{{\rm{the}}\,{\rm{side}}\,{\rm{opposite}}\,{\rm{to}}\,{\rm{angle}}\,C}}{{{\rm{the}}\,{\rm{side}}\,{\rm{adjacent}}\,{\rm{to}}\,{\rm{angle}}\,C}} = \frac{{BA}}{{CB}}\)

cosecant of \(\angle C = \frac{1}{{\sin \,\angle C}} = \frac{{{\rm{hypotenuse}}}}{{{\rm{the}}\,{\rm{side}}\,{\rm{opposite}}\,{\rm{to}}\,{\rm{angle}}\,C}} = \frac{{CA}}{{BA}}\)

secant of \(\angle C = \frac{1}{{\cos \,\angle C}} = \frac{{{\rm{hypotenuse}}}}{{{\rm{the}}\,{\rm{side}}\,{\rm{adjacent}}\,{\rm{to}}\,{\rm{angle}}\,C}} = \frac{{CA}}{{CB}}\)

cotangent of \(\angle C = \frac{{{\rm{the}}\,{\rm{side}}\,{\rm{adjacent}}\,{\rm{to}}\,{\rm{angle}}\,C}}{{{\rm{the}}\,{\rm{side}}\,{\rm{opposite}}\,{\rm{to}}\,{\rm{angle}}\,C}} = \frac{{CB}}{{BA}}\)

The ratios defined above are abbreviated as \(\sin C, \cos C, \tan C, \operatorname{cosec} C, \sec C\) and \(\cot C\), respectively. The reciprocals of the ratios \(\sin C, \cos C\), and \(\tan C\) are \(\operatorname{cosec} C, \sec C\), and \(\cot C\).

Trigonometric Ratios of Certain Angles

Trigonometric Ratios of \(45^{\circ}\)

In an \(\triangle A B C\), right-angled triangle at \(B\), if one of the angles is \(45^{\circ}\), then the other angle is also \(45^{\circ}\), i.e., \(\angle A=\angle C=45^{\circ}\).

So, \(B C=A B\)

Now, Suppose \(B C=A B=p\)

By using the Pythagoras theorem, we get,

\(C A^{2}=A B^{2}+B C^{2}\)

\(\Rightarrow C A^{2}=p^{2}+p^{2}\)

\(\Rightarrow C A^{2}=2 p^{2}\)

Then, \(C A=p \sqrt{2}\)

Here we are considering \(\angle C=45^{\circ}\). By using the definition of the trigonometric ratios,

sine of \({45^{\rm{o}}} = \frac{{{\rm{the}}\,{\rm{side}}\,{\rm{opposite}}\,{\rm{to}}\,{\rm{angle}}\,{{45}^{\rm{o}}}}}{{{\rm{hypotenuse}}}} = \frac{{BA}}{{CA}} = \frac{p}{{p\sqrt 2 }} = \frac{1}{{\sqrt 2 }}\)

cosine of \({45^{\rm{o}}} = \frac{{{\rm{the}}\,{\rm{side}}\,{\rm{adjacent}}\,{\rm{to}}\,{\rm{angle}}\,{{45}^{\rm{o}}}}}{{{\rm{hypotenuse}}}} = \frac{{CB}}{{CA}} = \frac{p}{{p\sqrt 2 }} = \frac{1}{{\sqrt 2 }}\)

tangent of \({45^{\rm{o}}} = \frac{{{\rm{the}}\,{\rm{side}}\,{\rm{opposite}}\,{\rm{to}}\,{\rm{angle}}\,{{45}^{\rm{o}}}}}{{{\rm{the}}\,{\rm{side}}\,{\rm{adjacent}}\,{\rm{to}}\,{\rm{angle}}\,{{45}^{\rm{o}}}}} = \frac{{BA}}{{CB}} = \frac{1}{1} = 1\)

cosecant of \({45^{\rm{o}}} = \frac{1}{{\sin \,{{45}^{\rm{o}}}}} = \frac{{{\rm{hypotenuse}}}}{{{\rm{the}}\,{\rm{side}}\,{\rm{opposite}}\,{\rm{to}}\,{\rm{angle}}\,{{45}^{\rm{o}}}}} = \frac{{CA}}{{BA}} = \frac{{p\sqrt 2 }}{p} = \sqrt 2 \)

secant of \({45^{\rm{o}}} = \frac{1}{{\cos \,{{45}^{\rm{o}}}}} = \frac{{{\rm{hypotenuse}}}}{{{\rm{the}}\,{\rm{side}}\,{\rm{adjacent}}\,{\rm{to}}\,{\rm{angle}}\,{{45}^{\rm{o}}}}} = \frac{{CA}}{{CB}} = \sqrt 2 \)

cotangent of \({45^{\rm{o}}} = \frac{{{\rm{the}}\,{\rm{side}}\,{\rm{adjacent}}\,{\rm{to}}\,{\rm{angle}}\,{{45}^{\rm{o}}}}}{{{\rm{the}}\,{\rm{side}}\,{\rm{opposite}}\,{\rm{to}}\,{\rm{angle}}\,{{45}^{\rm{o}}}}} = \frac{{CB}}{{BA}} = 1\)

Trigonometric Ratios of \(30^{\circ}\) and \(60^{\circ}\)

Consider the equilateral triangle \(A B C\).

Since each angle in an equilateral triangle is \(60^{\circ}\), therefore, \(\angle A=\angle B=\angle C=60^{\circ}\).

Then, draw a perpendicular line \(AD\) from \(A\) to the side \(BC\).

Now, \(A D\) is the bisector of the triangle \(A B C\). It forms the two congruent triangles \(\triangle A B D\) and \(\triangle A C D\).

Therefore, \(\triangle A B D \cong \triangle A C D\)

\(B D=D C\)

And \(\angle B A D=\angle C A D\) (CPCT)

Then, by observing the figure, we can say that \(\triangle A B D\) is a right-angle triangle, the right-angled at \(D\) with \(\angle B A D=30^{\circ}\) and \(\angle A B D=60^{\circ}\).

We need to know the lengths of the sides of the triangle’s sides for finding the trigonometric ratios.

If \(A B=a\), then \(B D=\frac{a}{2}\)

Then, by using the Pythagoras theorem, we can find the length of \(A D\).

\(A B^{2}=A D^{2}+B D^{2}\)

\(\Rightarrow A D^{2}=A B^{2}-B D^{2}\)

\(\Rightarrow A D^{2}=a^{2}-\left(\frac{a}{2}\right)^{2}\)

\(\Rightarrow A D^{2}=\frac{4 a^{2}-a^{2}}{4}\)

\(\Rightarrow A D^{2}=\frac{3 a^{2}}{4}\)

\(\Rightarrow A D=\frac{a \sqrt{3}}{2}\)

Therefore, trigonometric ratios of \(30^{\circ}\) are,

\( \sin 30^{\circ}=\frac{B D}{B A}=\frac{\frac{a}{2}}{a}=\frac{a}{2 a}=\frac{1}{2}\)

\( \cos 30^{\circ}=\frac{A D}{B A}=\frac{\frac{a \sqrt{3}}{2}}{a}=\frac{\sqrt{3}}{2}\)

\( \tan 30^{\circ}=\frac{B D}{A D}=\frac{\frac{a}{2}}{\frac{a \sqrt{3}}{2}}=\frac{1}{\sqrt{3}}\)

Also, \({\rm{cosec}}\,{30^{\rm{o}}}=\frac{1}{ \sin 30^{\circ}}=2\)

\( \sec 30^{\circ}=\frac{1}{ \cos 30^{\circ}}=\frac{2}{\sqrt{3}}\)

\( \cot 30^{\circ}=\frac{1}{ \tan 30^{\circ}}=\sqrt{3}\)

Similarly, for trigonometric ratios of \(60^{\circ}\) are,

\( \sin 60^{\circ}=\frac{A D}{B A}=\frac{\frac{a \sqrt{3}}{2}}{a}=\frac{\sqrt{3}}{2}\)

\( \cos 60^{\circ}=\frac{B D}{B A}=\frac{a}{a}=\frac{1}{2}\)

\( \tan 60^{\circ}=\frac{A D}{B D}=\sqrt{3}\)

Also, \({\rm{cosec}}\,{60^{\rm{o}}}=\frac{1}{ \sin 60^{\circ}}=\frac{2}{\sqrt{3}}\)

\( \sec 60^{\circ}=\frac{1}{ \cos 60^{\circ}}=2\)

\( \cot 60^{\circ}=\frac{1}{ \tan 60^{\circ}}=\frac{1}{\sqrt{3}}\)

Trigonometric Ratios of \(0^{\circ}\) and \(90^{\circ}\)

Let us consider a triangle \(A B C\), which is right-angled at \(B\), such that \(\angle A\) is a very small angle.

If \(A\) in the right-angle triangle \(A B C\) is made smaller and smaller until it reaches zero, the length of the side \(B C\) decreases. In this case, point \(C\) gets closer to point \(B\), and finally, when \(\angle A\) becomes very close to \(0^{\circ}, \angle C\) becomes close to \(90^{\circ}, A C\) becomes almost the same as \(A B\), and \(B C\) becomes close to \(0\).

And, when \(\angle A\) becomes actually equal to \(0^{\circ}\), then \(A B\) becomes equal to \(A C\), and \(B C\) becomes equal to \(0\).

Hence, the different trigonometric ratios of \(0^{\circ}\) are,

\(\sin A=\sin 0^{\circ}=\frac{B C}{A C}=\frac{0}{A C}=0\)

\(\cos A=\cos 0^{\circ}=\frac{A B}{A C}=\frac{A C}{A C}=1\)

\(\tan A=\tan 0^{\circ}=\frac{B C}{A B}=\frac{0}{A B}=0\)

And,

\(\cot 0^{\circ}=\frac{ \cos 0^{\circ}}{ \sin 0^{\circ}}=\frac{1}{0}=\) not defined

\( \sec 0^{\circ}=\frac{1}{ \cos 0^{\circ}}=\frac{1}{1}=1\)

\({\rm{cosec}}\,{0^{\rm{o}}}=\frac{1}{ \sin 0^{\circ}}=\frac{1}{0}=\) not defined

Now, in the above figure, as point \(C\) approaches point \(B\), the length of \(A B\) becomes very close to \(A C\), and \(B C\) becomes very close to \(0\)

Thus, when point \(C\) actually coincides with point \(B, \angle C\) is equal to \(90^{\circ}\), then \(A B\) becomes equal to \(A C\), and \(B C\) becomes equal to \(0\).

Hence, the different trigonometric ratios of \(90^{\circ}\) are,

\(\sin 90^{\circ}=1\)

\(\cos 90^{\circ}=0\)

\(\tan \,{90^{\rm{o}}} = \frac{{\sin \,{{90}^{\rm{o}}}}}{{\cos \,{{90}^{\rm{o}}}}} = \frac{1}{0} = \) Not defined

And,

\( \cot 90^{\circ}=\frac{ \cos 90^{\circ}}{\sin 90^{\circ}}=\frac{0}{1}=0\)

\( \sec 0^{\circ}=\frac{1}{ \cos 90^{\circ}}=\frac{1}{0}=\) Not defined

\({\rm{cosec}}\,{0^{\rm{o}}}=\frac{1}{ \sin 90^{\circ}}=\frac{1}{1}=1\)

Trigonometric Ratios Table

The table of trigonometric ratios will help us in determining the values of trigonometric standard angles.

\(0^{\circ}, 30^{\circ}, 45^{\circ}, 60^{\circ}\), and \(90^{\circ}\) are the standard angles for trigonometrical ratios.

To answer trigonometrical problems, the values of trigonometrical ratios of standard angles are crucial. As a result, the value of the trigonometrical ratios of these standard angles must be remembered.

\(\angle A\)	\(0^\circ \)	\(30^\circ \)	\(45^\circ \)	\(60^\circ \)	\(90^\circ \)
\({\mathop{\rm Sin}\nolimits} \,A\)	\(0\)	\(\frac{1}{2}\)	\(\frac{1}{{\sqrt 2 }}\)	\(\frac{{\sqrt 3 }}{2}\)	\(1\)
\({\mathop{\rm Cos}\nolimits} \,A\)	\(1\)	\(\frac{{\sqrt 3 }}{2}\)	\(\frac{1}{{\sqrt 2 }}\)	\(\frac{1}{2}\)	\(0\)
\({\mathop{\rm Tan}\nolimits} \,A\)	\(0\)	\(\frac{1}{{\sqrt 3 }}\)	\(1\)	\({\sqrt 3 }\)	Not defined
\({\mathop{\rm Cosec}\nolimits} \,A\)	Not defined	\(2\)	\(\sqrt 2 \)	\(\frac{2}{{\sqrt 3 }}\)	\(1\)
\({\mathop{\rm Sec}\nolimits} \,A\)	\(1\)	\(\frac{2}{{\sqrt 3 }}\)	\(\sqrt 2 \)	\(2\)	Not defined
\({\mathop{\rm Cot}\nolimits} \,A\)	Not defined	\(\sqrt 3 \)	\(1\)	\(\frac{1}{{\sqrt 3 }}\)	\(0\)

Formulas of Trigonometry

Here are some of the important trigonometric formulas listed below:

1. \(\sin \theta=\frac{\text { Opposite side }}{\text { Hypotenuse }}\)
2. \(\cos \theta=\frac{\text { Adjacent side }}{\text { Hypotenuse }}\)
3. \(\tan \theta=\frac{\text { Opposite side }}{\text { Adjacent side }}\)
4. \(\operatorname{cosec} \theta=\frac{\text { Hypotenuse }}{\text { Opposite side }}\)
5. \(\sec \theta=\frac{\text { Hypotemuse }}{\text { Adjacent side }}\)
6. \(\cot \theta=\frac{\text { Adjacent side }}{\text { Opposite side }}\)

Reciprocal Identities

1. \(\sin \theta=\frac{1}{\operatorname{cosec} \theta}\)
2. \(\cos \theta=\frac{1}{\sec \theta}\)
3. \(\tan \theta=\frac{1}{ \cot \theta}\)
4. \(\operatorname{cosec} \theta=\frac{1}{ \sin \theta}\)
5. \(\sec \theta=\frac{1}{ \cos \theta}\)
6. \(\cot \theta=\frac{1}{ \tan \theta}\)

Trigonometric Ratios of Complementary Angles

1. \(\sin \left(90^{\circ}-A\right)=\cos A\)
2. \(\cos \left(90^{\circ}-A\right)=\sin A\)
3. \(\tan \left(90^{\circ}-A\right)=\cot A\)
4. \(\cot \left(90^{\circ}-A\right)=\tan A\)
5. \(\sec \left(90^{\circ}-A\right)=\operatorname{cosec} A\)
6. \(\operatorname{cosec}\left(90^{\circ}-A\right)=\sec A\)

Trigonometric Ratios of Supplementary Angles

1. \(\sin \left(180^{\circ}-A\right)=\sin A\)
2. \(\cos \left(180^{\circ}-A\right)=-\cos A\)
3. \(\tan \left(180^{\circ}-A\right)=-\tan A\)
4. \(\cot \left(180^{\circ}-A\right)=-\cot A\)
5. \(\sec \left(180^{\circ}-A\right)=-\sec A\)
6. \(\operatorname{cosec}\left(180^{\circ}-A\right)=\operatorname{cosec} A\)

Trigonometric Identities

1. \(\sin ^{2} A+\cos ^{2} A=1\)
2. \(1 + {\tan ^2}A = {\sec ^2}A\)
3. \(1 + {\cot ^2}A = {\rm{cose}}{{\rm{c}}^2}A\)
4. \(\sec ^{2} A-\tan ^{2} A=1\) for \(0^{\circ} \leq A<90^{\circ}\)
5. \({\rm{cose}}{{\rm{c}}^2}A – {\cot ^2}A = 1\)

Sum and Difference Identities

1. \( \sin (x+y)=\sin x \times \cos y+\cos x \times \sin y\)
2. \( \cos (x+y)=\cos x \times \cos y-\sin x \times \sin y\)
3. \(\left. \tan (x+y)=\frac{( \tan x+ \tan y)}{(1-\tan x \times \tan y}\right)\)
4. \( \sin (x-y)=\sin x \times \cos y-\cos x \times \sin y\)
5. \( \cos (x-y)=\cos x \times \cos y+\sin x \times \sin y\)
6. \( \tan (x-y)=\frac{(\tan x- \tan y)}{(1+ \tan x \times \tan y)}\)

Solved Examples – Trigonometric Ratios of Some Specific Angles

Q.1. Find the value of \(\tan 60^{\circ}+\cos 30^{\circ}\).
Ans: We know that \(\tan 60^{\circ}=\sqrt{3}, \cos 30^{\circ}=\frac{\sqrt{3}}{2}\).
So, we have \(\tan 60^{\circ}+\cos 30^{\circ}\).
\(=\sqrt{3}+\frac{\sqrt{3}}{2}\)
\(=\frac{2 \sqrt{3}+\sqrt{3}}{2}\)
\(=\frac{3 \sqrt{3}}{2}\)
Therefore, \(\tan 60^{\circ}+\cos 30^{\circ}=\frac{3 \sqrt{3}}{2}.\)

Q.2. Evaluate the following: \(\sin 60^{\circ} \cos 30^{\circ}+\sin 30^{\circ} \cos 60^{\circ}\)
Ans: We know \(\sin 60^{\circ}=\frac{\sqrt{3}}{2}, \cos 60^{\circ}=\frac{1}{2}, \sin 30^{\circ}=\frac{1}{2}, \cos 30^{\circ}=\frac{\sqrt{3}}{2}\)
So, we have, \(\sin 60^{\circ} \cos 30^{\circ}+\sin 30^{\circ} \cos 60^{\circ}\)
\(=\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}+\frac{1}{2} \times \frac{1}{2}\)
\(=\frac{3}{4}+\frac{1}{4}=\frac{4}{4}=1\)
Therefore, \(\sin 60^{\circ} \cos 30^{\circ}+\sin 30^{\circ} \cos 60^{\circ}=1.\)

Q.3. Find the value of \(\frac{2 \tan 30^{\circ}}{1+\tan ^{2} 30^{\circ}}\).
Ans: We know \(\tan 30^{\circ}=\frac{1}{\sqrt{3}}\)
So, \(\frac{2 \tan 30^{\circ}}{1+\tan ^{2} 30^{\circ}}=\frac{2 \times \frac{1}{\sqrt{3}}}{1+\left(\frac{1}{\sqrt{3}}\right)^{2}}=\frac{2^{2}}{1+\frac{1}{3}}=\frac{2^{\frac{2}{3}}}{\frac{4}{3}}=\frac{2}{\sqrt{3}} \times \frac{3}{4}=\frac{\sqrt{3}}{2}\)
Therefore, \(\frac{2 \tan 30^{\circ}}{1+\tan ^{2} 30^{\circ}}=\frac{\sqrt{3}}{2}.\)

Q.4. In triangle \(A B C\), right-angled at \(B, A B=24 \mathrm{~cm}, B C=7 \mathrm{~cm}\). Determine \(\sin A.\)
Ans: From the given, \(A B=24 \mathrm{~cm}, B C=7 \mathrm{~cm}\)
By using the Pythagoras theorem,
\(A C^{2}=A B^{2}+B C^{2}\)
\(\Rightarrow A C^{2}=24^{2}+7^{2}\)
\(\Rightarrow A C^{2}=576+49\)
\(\Rightarrow A C^{2}=625\)
\(\Rightarrow A C=\sqrt{625}\)
\(\Longrightarrow A C=25\)
Now, \(\sin A=\frac{\text { Opposite of } \angle A}{\text { Hypotenuse }}\)
\(\Rightarrow \sin A=\frac{B C}{A C}=\frac{7}{25}\)
Therefore, \(\sin A=\frac{7}{25}.\)

Q.5. In triangle \(A B C\), right-angled at \(\mathrm{B}, A B=24 \mathrm{~cm}, B C=7 \mathrm{~cm}\). Determine \(\sin \mathrm{C}\).
Ans: From the given, \(A B=24 \mathrm{~cm}, B C=7 \mathrm{~cm}\)
By using the Pythagoras theorem,
\(A C^{2}=A B^{2}+B C^{2}\)
\(\Rightarrow A C^{2}=24^{2}+7^{2}\)
\(\Rightarrow A C^{2}=576+49\)
\(\Longrightarrow A C^{2}=625\)
\(\Rightarrow A C=\sqrt{625}\)
\(\Rightarrow A C=25\)
Now, \(\sin C=\frac{\text { Opposite of } \angle A}{\text { Hypotenuse }}\)
\(\Longrightarrow \sin C=\frac{A B}{A C}=\frac{24}{25}\)
Therefore, \(\sin C=\frac{24}{25}\).

Summary

In this article, we discussed different trigonometric ratios, how to derive trigonometric ratios of some specific angles of \(0^{\circ}, 30^{\circ}, 45^{\circ}, 60^{\circ}\) and \(90^{\circ}\), the trigonometric ratio value table. We provided a collection of the trigonometric ratios of complementary and supplementary angles, a few basic trigonometric basic identities and formulas for the sum and the difference of angles.

This article helps in better understanding the topic of trigonometric ratios of some specific angles. This article’s outcome helps in learning about trigonometric ratios of some specific angles.

FAQs

Q.1. What are trigonometric ratios?
Ans: The values of all trigonometric functions based on the value of the ratio of sides of a right-angled triangle are known as trigonometric ratios.

Q.2. What are the trigonometric ratios of special angles?
Ans: Consider the right-triangle \(CAB\), with a right angle at \(A\). The trigonometric ratios of the angle \(C\) are:
sine of \(\angle C = \frac{{{\rm{the}}\,{\rm{side}}\,{\rm{opposite}}\,{\rm{to}}\,{\rm{angle}}\,C}}{{{\rm{hypotenuse}}}} = \frac{{BA}}{{CB}}\)
cosine of \(\angle C = \frac{{{\rm{the}}\,{\rm{side}}\,{\rm{adjacent}}\,{\rm{to}}\,{\rm{angle}}\,C}}{{{\rm{hypotenuse}}}} = \frac{{CA}}{{CB}}\)
tangent of \(\angle C = \frac{{{\rm{the}}\,{\rm{side}}\,{\rm{opposite}}\,{\rm{to}}\,{\rm{angle}}\,C}}{{{\rm{the}}\,{\rm{side}}\,{\rm{adjacent}}\,{\rm{to}}\,{\rm{angle}}\,C}} = \frac{{BA}}{{CA}}\)
cosecant of \(\angle C = \frac{1}{{\sin \,\angle C}} = \frac{{{\rm{hypotenuse}}}}{{{\rm{the}}\,{\rm{side}}\,{\rm{opposite}}\,{\rm{to}}\,{\rm{angle}}\,C}} = \frac{{CB}}{{BA}}\)
secant of \(\angle C = \frac{1}{{\cos \,\angle C}} = \frac{{{\rm{hypotenuse}}}}{{{\rm{the}}\,{\rm{side}}\,{\rm{adjacent}}\,{\rm{to}}\,{\rm{angle}}\,C}} = \frac{{CB}}{{CA}}\)
cotangent of \(\angle C = \frac{{{\rm{the}}\,{\rm{side}}\,{\rm{adjacent}}\,{\rm{to}}\,{\rm{angle}}\,C}}{{{\rm{the}}\,{\rm{side}}\,{\rm{opposite}}\,{\rm{to}}\,{\rm{angle}}\,C}} = \frac{{CA}}{{BA}}\)

Q.3. What are the 6 trigonometric ratios?
Ans: Sine, cosine, tangent, secant, cosecant, and cotangent are the six basic trigonometric ratios.

Q.4. Write examples of trigonometric ratios of some specific angles.
Ans: Examples of trigonometric ratios of some specific angles are given in the table.

\(\angle A\)	\(0^\circ \)	\(30^\circ \)	\(45^\circ \)	\(60^\circ \)	\(90^\circ \)
\({\mathop{\rm Sin}\nolimits} \,A\)	\(0\)	\(\frac{1}{2}\)	\(\frac{1}{{\sqrt 2 }}\)	\(\frac{{\sqrt 3 }}{2}\)	\(1\)
\({\mathop{\rm Cos}\nolimits} \,A\)	\(1\)	\(\frac{{\sqrt 3 }}{2}\)	\(\frac{1}{{\sqrt 2 }}\)	\(\frac{1}{2}\)	\(0\)
\({\mathop{\rm Tan}\nolimits} \,A\)	\(0\)	\(\frac{1}{{\sqrt 3 }}\)	\(1\)	\({\sqrt 3 }\)	Not defined
\({\mathop{\rm Cosec}\nolimits} \,A\)	Not defined	\(2\)	\(\sqrt 2 \)	\(\frac{2}{{\sqrt 3 }}\)	\(1\)
\({\mathop{\rm Sec}\nolimits} \,A\)	\(1\)	\(\frac{2}{{\sqrt 3 }}\)	\(\sqrt 2 \)	\(2\)	Not defined
\({\mathop{\rm Cot}\nolimits} \,A\)	Not defined	\(\sqrt 3 \)	\(1\)	\(\frac{1}{{\sqrt 3 }}\)	\(0\)

Q.5. What are the trigonometric ratios of an angle?
Ans: Sine, cosine, tangent, cosecant, secant, and cotangent are the six trigonometric ratios. Sin, cos, tan, cosec, sec, and cot are the abbreviations for these six trigonometric ratios, respectively. They can be represented in terms of the sides of a right-angled triangle for a particular angle.

Q.6. What are the specific angles?
Ans: In trigonometry, by the specific angles, we mean angles \(0^{\circ}, 30^{\circ}, 45^{\circ}, 60^{\circ}\) and \(90^{\circ}\). The numerical values of the trigonometric angles are simple to remember.

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