NCERT Solutions for Class 10 Maths Chapter 8: Introduction to Trigonometry - Exercise 8.3

EXERCISE – 8.3

1. Evaluate 

(i) \[\dfrac{{\sin {{18}^0}}}{{\cos {{72}^0}}}\]

Ans: \[\dfrac{{\sin {{18}^0}}}{{\cos {{72}^0}}}\]= \[\dfrac{{\sin ({{90}^0} - {{72}^0})}}{{\cos {{72}^0}}}\]  (\[\because {\text{sin(9}}{{\text{0}}^{\text{0}}}{{ -\theta) = cos\theta }}\])

\[ = \dfrac{{\cos {{72}^0}}}{{\cos {{72}^0}}} = 1\]

(ii) \[\dfrac{{\tan {{26}^0}}}{{\cot {{64}^0}}}\]

Ans: \[\dfrac{{\tan {{26}^0}}}{{\cot {{64}^0}}}\]= \[\dfrac{{\tan ({{90}^0} - {{64}^0})}}{{\cot {{64}^0}}}\] (\[\because {\text{tan(9}}{{\text{0}}^{\text{0}}}{{ - \theta ) = cot\theta }}\])

\[ = \dfrac{{\cot {{64}^0}}}{{\cot {{64}^0}}} = 1\]

(iii) \[\cos {48^0} - \sin {42^0}\]

Ans: \[\cos {48^0} - \sin {42^0}\]\[ = \cos ({90^0} - {42^0}) - \sin {42^0}\]

\[ = \sin {42^0} - \sin {42^0}\]

\[ = 0\]

(iv) \[\cos ec{31^0} - \sec {59^0}\]

Ans: \[\cos ec{31^0} - \sec {59^0}\]\[{\text{ = sec5}}{{\text{9}}^{\text{0}}}{\text{ - sec5}}{{\text{9}}^{\text{0}}}\]

\[ = 0\]

2. Show that  

(i) \[\tan {48^0}\tan {23^0}\tan {42^0}\tan {67^0} = 1\]

Ans: Considering LHS,

\[\tan {48^0}\tan {23^0}\tan {42^0}\tan {67^0}\]           ......(i)

We know that \[{\text{tan(9}}{{\text{0}}^{\text{0}}}{{ - \theta ) =cot\theta }}\]

Now, using this formula in equation (i)

\[ = \tan ({90^0} - {42^0})\tan ({90^0} - {67^0})\tan {42^0}\tan {67^0}\]

\[ = \cot {42^0}\cot {67^0}\tan {42^0}\tan {67^0}\]

\[ = (\cot {42^0}\tan {42^0})(\cot {67^0}\tan {67^0})\]

\[ = (1)(1)\]

\[ = 1\]

(ii) \[\cos {38^0}\cos {52^0} - \sin {38^0}\sin {52^0} = 0\]

Considering LHS,

\[\cos {38^0}\cos {52^0} - \sin {38^0}\sin {52^0}\]    ......(i) 

\[ = \cos ({90^0} - {52^0})\cos ({90^0} - {38^0}) - \sin {38^0}\sin {52^0}\]

\[ = \sin {52^0}\sin {38^0} - \sin {38^0}\sin {52^0}\]

\[ = 0\]

3. If \[{\text{tan2A = cot(A - 1}}{{\text{8}}^{\text{0}}}{\text{)}}\], where \[{\text{2A}}\] is an acute angle, find the value of A.

Ans: Given, 

\[{\text{tan2A = cot(A - 1}}{{\text{8}}^{\text{0}}}{\text{)}}\]      ......(i)

We know that \[{\text{tan2A = cot(9}}{{\text{0}}^{\text{0}}}{\text{ - 2A)}}\]

\[{\text{tan2A = cot(9}}{{\text{0}}^{\text{0}}}{\text{ - 2A)}}\]     (by using above formula in equation (i))

\[ \Rightarrow {\text{cot(9}}{{\text{0}}^{\text{0}}}{\text{ - 2A) = cot(A - 1}}{{\text{8}}^{\text{0}}}{\text{)}}\]

\[ \Rightarrow {\text{9}}{{\text{0}}^{\text{0}}}{\text{ - 2A = A - 1}}{{\text{8}}^{\text{0}}}\]

\[ \Rightarrow {\text{A + 2A = 9}}{{\text{0}}^{\text{0}}}{\text{ + 1}}{{\text{8}}^{\text{0}}}\]

\[ \Rightarrow {\text{3A = 10}}{{\text{8}}^{\text{0}}}\]

\[\therefore {\text{A = 3}}{{\text{6}}^{\text{0}}}\]

4. If \[{\text{tanA = cotB}}\], prove that \[{\text{A + B = 9}}{{\text{0}}^{\text{0}}}\]

Ans: Given, 

\[{\text{tanA = cotB}}\]     ......(i)

We know that \[{\text{tanA = cot(9}}{{\text{0}}^{\text{0}}}{\text{ - A)}}\]

\[ \Rightarrow {\text{tanA = tan(9}}{{\text{0}}^{\text{0}}}{\text{ - B)}}\]    (by using above formula in equation (i))

\[{\text{A = 9}}{{\text{0}}^{\text{0}}}{\text{ - B}}\]

\[\therefore {\text{A + B = 9}}{{\text{0}}^{\text{0}}}\]

5. If \[{\text{sec4A = cosec(A - 2}}{{\text{0}}^{\text{0}}}{\text{)}}\], where \[{\text{4A}}\]is an acute angle, find the value of A.

Ans: Given, 

\[{\text{sec4A = cosec(A - 2}}{{\text{0}}^{\text{0}}}{\text{)}}\]      .....(i)

We know that \[{\text{secA = cosec(9}}{{\text{0}}^{\text{0}}}{\text{ - A)}}\],

  (by using above formula in equation (i))

\[ \Rightarrow {\text{cosec(9}}{{\text{0}}^{\text{0}}}{\text{ - 4A) = cosec(A -2}}{{\text{0}}^{\text{0}}}{\text{)}}\]

\[ \Rightarrow {\text{9}}{{\text{0}}^{\text{0}}}{\text{  -  4A  =  A  -  2}}{{\text{0}}^{\text{0}}}\]

\[ \Rightarrow {\text{5A = 11}}{{\text{0}}^{\text{0}}}\]

\[\therefore {\text{A}} = {22^0}\]

6. if A, B and C are interior angles of a triangle ABC then show that

\[{\text{sin}}\left( {\dfrac{{{\text{B + C}}}}{{\text{2}}}} \right){\text{ = cos}}\dfrac{{\text{A}}}{{\text{2}}}\]

Ans: In \[\vartriangle {\text{ABC}}\] we know,

\[\angle {\text{A + }}\angle {\text{B + }}\angle {\text{C = 18}}{{\text{0}}^{\text{0}}}\]

\[\angle {\text{B}} + \angle {\text{C}} = {180^0} - \angle {\text{A}}\]     

Dividing \[2\]both the sides,

\[ \Rightarrow \dfrac{{\angle {\text{B}} + \angle {\text{C}}}}{2} = \dfrac{{{{180}^0}}}{2} - \dfrac{{\angle {\text{A}}}}{2}\]

Applying sine angle on both the sides,

\[ \Rightarrow \sin \left( {\dfrac{{\angle {\text{B}} + \angle {\text{C}}}}{2}} \right) = \sin \left( {{{90}^0} - \dfrac{{\angle {\text{A}}}}{2}} \right)\]

\[ \Rightarrow \cos \left( {\dfrac{{\text{A}}}{2}} \right)\]

7. Express \[\sin {67^0} + \cos {75^0}\] in terms of trigonometric ratios of angles between \[{0^0}\] and \[{45^0}\].

Ans: \[\sin {67^0} + \cos {75^0}\]

We know that \[{\text{cos(9}}{{\text{0}}^{\text{0}}}{{ - \theta ) = sin\theta }}\] and \[{\text{sin(9}}{{\text{0}}^{\text{0}}}{{ - \theta ) = cos\theta }}\]

\[ \Rightarrow \sin ({90^0} - {23^0}) + \cos ({90^0} - {15^0})\]

\[ \Rightarrow \cos {23^0} + \sin {15^0}\]

NCERT Solution Of Maths Class 10 Chapter 8 Exercise 8.3 - Free PDF Download

NCERT Solutions for Class 10 Maths Chapter 8 Exercise 8.3: History of Trigonometry

The word trigonometry, as you will be informed in exercise 8.3 class 10 Maths, finds its root in ancient Greek words where “tri” means three, “gon” means sides, and “metron” means measure. Thus the sides and angles of a triangle are the main base of calculation in trigonometry. Egypt and Babylon, ex 8.3 class 10 Maths solutions suggest, were among the first countries where trigonometry was used for calculating the distance of Earth from other planets and stars. The advancement of technology is dependent on trigonometry to a great extent and be it Engineering or Physical Science- it is used by and large.

Important Topics under NCERT Solutions for Class 10 Maths Chapter 8 Exercise 8.3

NCERT Solutions Class 10 Maths Chapter 8 cover the concepts of Introduction to Trigonometry. This is one of the most important chapters covered in Class 10 and is divided into 6 major parts. The following table comprises the important topics covered in the chapter Introduction to Trigonometry. 

We recommend that students go through each of these topics carefully to be able to master this chapter and solve problems in their exams based on Trigonometry.

Importance of Trigonometry in Mathematics

Trigonometry is the branch of mathematics dealing with relations between the sides and angles of triangles, along with the relevant functions that are related to the angles.

Trigonometry is a very important area that is covered in Class 10 Mathematics. It integrates the understanding of some of the most significant concepts and problem-solving abilities among students. Trigonometry has numerous applications in our day-to-day lives. It is thus recommended that students gather as much knowledge as they can on the topics that are covered in the chapter Introduction to Trigonometry to be able to score well in their exams.

Class 10 Maths Exercise 8.3: What is it All About

While in the first portions of the chapter you get to know about the trigonometric ratio of angles, class 10 Maths ch 8 ex 8.3 focuses on trigonometric ratios of complementary angles. Here you will again be reminded of what makes two angles complementary. When the sum of two angles equals 90o, they are referred to as complementary angles. In class 10 Maths chapter 8 exercise 8.3 you will come across an example of a triangle that has a 90o angle. Now if the other two angles sum up to be 90o, then each of them will be 45o and hence, complementary.

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In the triangle given as an example, you will see that the ∠ B is the right angle and ∠ A and ∠ C sums up to 90o. 

Now, according to this, we can say that angles A + C = 90o and thus form a pair.

So, sin A= \[\frac{BC}{AC}\]

cos A= \[\frac{AB}{AC}\]

tan A= \[\frac{BC}{AB}\]

cosec A= \[\frac{AC}{BC}\]

sec A= \[\frac{AC}{AB}\]

cot A= \[\frac{AB}{BC}\]

CBSE Class 10 Maths Chapter 8 Exercise 8.3 Solutions: Important Considerations

As you carry on with the ratios of contemporary angles you will finally conclude that:

sin (90° – A) = cos A

cos (90° – A) = sin A

tan (90° – A) = cot A

cot (90° – A) = tan A 

sec (90° – A) = cosec A 

cosec (90° – A) = sec A

The class 10 maths chapter 8 exercise 8.3 solutions will also inform you of another important point that you must keep in mind while solving solutions. Whatever calculations you may come across, the following always holds true. 

tan 0° = cot 90° = 0

sec 0° = cosec 90° = 1

Values of sec 90°, tan 90°, cosec 0°, and cot 0° cannot be determined. These are undefined entities.

NCERT Solution of Maths Class 10 Chapter 8 Exercise 8.3: Solved Examples

While going through the chapter in the textbook and the solutions you will come across the following examples of problems based on what you learned about the ratio of angles in a contemporary triangle. In the following, we have some solutions ready for you for such problems.

Problem 1: Evaluate \[\frac{tan tan 65^{0}}{cott cot 25^{0}}\]

The solution, as mentioned in the maths class 10 chapter 8 exercise 8.3 solved examples is as follows:

We know that:

cot A = tan (90° – A)

Therefore, the value of cot 25° is tan (90° – 25°) which is equal to tan 65°

This can also be rewritten as \[\frac{tan tan 65^{0}}{cot cot 25^{0}}\] = \[\frac{tan tan 65^{0}}{cott cot 65^{0}}\] = 1

Problem 2: If Sin 3A = Cos (A – 26°), Where 3A is an Acute Angle, Find the Value of A

You will find the solution to this problem also in exercise 8.3 class 10 Maths NCERT solutions. The solution will be:

According to the problem, sin 3A = cos (A – 26°)

Let us term this as equation (i)

As, sin 3A = cos (90° – 3A)

Equation (i) can be rewritten as, cos (90° – 3A) = cos (A – 26°)

Now, again, both (90o – 3A) and (A – 26o) are acute angles. 

So, we can say that (90o – 3A) = (A – 26o)

This gives A = 29°

Again, another fun question from the textbook exercise that you can expect in your examination:

If tan A = cot B, then prove A + B = 90°. 

tan A = cot B

tan A = tan (90o – B) [as tan (90o – θ) = cot θ]

A = 90o – B

A + B = 90o (Hence proved)

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NCERT Solutions for Class 10 Maths

• Chapter 1 - Real Numbers

• Chapter 2 - Polynomials

• Chapter 3 - Pair of Linear Equations in Two Variables

• Chapter 4 - Quadratic Equations

• Chapter 5 - Arithmetic Progressions

• Chapter 6 - Triangles

• Chapter 7 - Coordinate Geometry

• Chapter 8 - Introduction to Trigonometry

• Chapter 9 - Some Applications of Trigonometry

• Chapter 10 - Circles

• Chapter 11 - Constructions

• Chapter 12 - Areas Related to Circles

• Chapter 13 - Surface Areas and Volumes

• Chapter 14 - Statistics

• Chapter 15 - Probability

Access Other Exercises of Class 10 Maths Chapter 8