Class 10 Maths Ncert Solutions Chapter Trigonometry

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry - Chapter 8 of NCERT explains the relation between the angles and sides of a right angle triangle. Students appearing in the Class 10 Board exams must check the NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry. These Class 10 Maths chapter 8 NCERT solutions are strictly based on the NCERT books for Class 10 Maths. NCERT Class 10 maths solutions chapter 8 gives a detailed explanation of each and every question given in the textbook. Moreover, NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry offers many tips and tricks to solve the questions in an easy way. You can also check the NCERT solutions for Class 10 for other subjects as well.

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NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry

NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry Excercise: 8.1

Q1 In $\Delta \: ABC$ , right-angled at $B, AB = 24 \: cm$ , $BC = 7 \: cm$ . Determine : $(i)\; \sin A, \cos A$ $(ii)\; \sin C, \cos C$

Answer:

1635935005859 We have,
In $\Delta \: ABC$ , $\angle$ B = 90, and the length of the base (AB) = 24 cm and length of perpendicular (BC) = 7 cm
So, by using Pythagoras theorem,
$\\AC^2 = AB^2 + BC^2\\ AC = \sqrt{AB^2+BC^2}$
Therefore, $AC = \sqrt{576+49}$
$AC = \sqrt{625}$
AC = 25 cm

Now,
(i) $\sin A = P/H = BC/AB = 7/25$
$\cos A = B/H = BA/AC = 24/25$

(ii) For angle C, AB is perpendicular to the base (BC). Here B indicates to Base and P means perpendicular wrt angle $\angle$ C
So, $\sin C = P/H = BA/AC = 24/25$
and $\cos C = B/H = BC/AC = 7/25$

Q2 In Fig. 8.13, find $\tan P - \cot R$ .

1635935041656

Answer:

We have, $\Delta$ PQR is a right-angled triangle, length of PQ and PR are 12 cm and 13 cm respectively.
So, by using Pythagoras theorem,
$QR = \sqrt{13^2-12^2}$
$QR = \sqrt{169-144}$
$QR = \sqrt{25} = 5\ cm$

Now, According to question,
$\tan P -\cot R$ = $\frac{RQ}{QP}-\frac{QR}{PQ}$
= 5/12 - 5/12 = 0

Q5 Given $\sec \theta =\frac{13}{12},$ calculate all other trigonometric ratios.

Answer:

We have,
$\sec \theta =\frac{13}{12},$

It means the Hypotenuse of the triangle is 13 units and the base is 12 units.
Let ABC is a right-angled triangle in which $\angle$ B is 90 and AB is the base, BC is perpendicular height and AC is the hypotenuse.

1635935082533 By using Pythagoras theorem,
$BC = \sqrt{169-144}=\sqrt{25}$
BC = 5 unit

Therefore,
$\sin \theta = \frac{BC}{AC}=\frac{5}{13}$
$\cos \theta = \frac{BA}{AC}=\frac{12}{13}$

$\tan \theta = \frac{BC}{AB}=\frac{5}{12}$

$\cot \theta = \frac{BA}{BC}=\frac{12}{5}$

$\sec \theta = \frac{AC}{AB}=\frac{13}{12}$

$\csc \theta = \frac{AC}{BC}=\frac{13}{5}$

Q11 State whether the following are true or false. Justify your answer.

(i) The value of $\tan A$ is always less than 1.
(ii) $\sec A=\frac{12}{5}$ for some value of angle A.
(iii) $\cos A$ is the abbreviation used for the cosecant of angle A.
(iv) $\cot A$ is the product of cot and A.
(v) $\sin \Theta =\frac{4}{3}$ for some angle $\Theta .$

Answer:

(i) False,
because $\tan 60 = \sqrt{3}$ , which is greater than 1

(ii) TRue,
because $\sec A \geq 1$

(iii) False,
Because $\cos A$ abbreviation is used for cosine A.

(iv) False,
because the term $\cot A$ is a single term, not a product.

(v) False,
because $\sin \theta$ lies between (-1 to +1) [ $-1\leq \sin \theta\leq 1$ ]

Q1 Evaluate the following :

$(iv)\: \frac{\sin 30^{o}+\tan 45^{o}-cosec 60^{o}}{\sec 30^{o}+\cos 60^{o}+\cot 45^{o}}$

Answer:

$\frac{\sin 30^{o}+\tan 45^{o}-cosec 60^{o}}{\sec 30^{o}+\cos 60^{o}+\cot 45^{o}}$ ..................(i)
It is known that the values of the given trigonometric functions,
$\\\sin 30^0 = 1/2=cos 60^0\\ \tan 45^0 = 1=\cot 45^0\\ \sec 30^0 = 2/\sqrt{3}=cosec 60^0\\$
Put all these values in equation (i), we get;
$\\\Rightarrow \frac{1/2+1-2/\sqrt{3}}{2/\sqrt{3}+1/2+1}\\ \Rightarrow\frac{3/2-2/\sqrt{3}}{3/2+2/\sqrt{3}}\\ \Rightarrow \frac{3\sqrt{3}-4}{3\sqrt{3}+4}\times\frac{4-3\sqrt{3}}{4-3\sqrt{3}}\\ \Rightarrow \frac{12\sqrt{3}-27-16+12\sqrt{3}}{-11}\\ \Rightarrow \frac{43-24\sqrt{3}}{11}$

Q1 Evaluate the following :

$(v)\frac{5\cos^{2}60^{o}+ 4\sec^{2}30^{o}-\tan^{2}45^{o}}{\sin^{2}30^{o}+\cos^{2}30^{o}}$

Answer:

$\frac{5\cos^{2}60^{o}+ 4\sec^{2}30^{o}-\tan^{2}45^{o}}{\sin^{2}30^{o}+\cos^{2}30^{o}}$ .....................(i)
We know the values of-
$\\\cos 60^0 = 1/2= \sin 30^0\\ \sec 30^0 = 2/\sqrt{3}\\ \tan 45^0 = 1\\ \cos 30^0 = \sqrt{3}/2$
By substituting all these values in equation(i), we get;

$\\\Rightarrow \frac{5.(\frac{1}{2})^2+4.(\frac{2}{\sqrt{3}}) ^2-1}{(\frac{1}{2})^2+(\frac{\sqrt{3}}{2})^2}\\ \Rightarrow \frac{5/4-1+16/3}{1}\\ \Rightarrow \frac{1/4+16/3}{1}\\ \Rightarrow\frac{67}{12}$

NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry Excercise: 8.3

Q1 Evaluate :

$(i)$ $\frac{\sin 18^{o}}{\cos 72^{o}}$

Answer:

$\frac{\sin 18^{o}}{\cos 72^{o}}$
We can write the above equation as;
$=\frac{\sin (90^0-72^0)}{\cos 72^0}$
By using the identity of $\sin (90^o-\theta) = \cos \theta$
Therefore, $\frac{\cos 72^0}{\cos 72^0} = 1$

So, the answer is 1.

Q1 Evaluate :

$(ii) \frac{\tan 26^{o}}{\cot 64^{o}}$

Answer:

$\frac{\tan 26^{o}}{\cot 64^{o}}$
The above equation can be written as ;

$\tan (90^o-64^o)/\cot 64^o$ .........(i)
It is known that, $\tan (90^o-\theta) = \cot \theta$
Therefore, equation (i) becomes,
$\cot64^o/\cot 64^o = 1$

So, the answer is 1.

Q1 Evaluate :

$(iii) \cos 48^{o}-\sin 42^{o}$

Answer:

$\cos 48^{o}-\sin 42^{o}$
The above equation can be written as ;
$\cos (90^o-42^{o})-\sin 42^{o}$ ....................(i)
It is known that $\cos (90^o-\theta) = \sin \theta$
Therefore, equation (i) becomes,
$\sin42^{o}-\sin 42^{o} = 0$

So, the answer is 0.

Q1 Evaluate :

$(iv) cosec \: 31^{o}-\sec 59^{o}$

Answer:

$cosec \: 31^{o}-\sec 59^{o}$

This equation can be written as;
$cosec 31^o - \sec(90^o-31^o)$ .................(i)
We know that $\sec(90^o-\theta) = cosec \theta$

Therefore, equation (i) becomes;
$cosec 31^o - cosec\ 31^o$ = 0

So, the answer is 0.

Q2 Show that :

$(i) \tan 48^{o}\tan 23^{o}\tan 42^{o}\tan 67^{o}= 1$

Answer:

$\tan 48^{o}\tan 23^{o}\tan 42^{o}\tan 67^{o}= 1$
Taking Left Hand Side (LHS)
= $\tan 48^{o}\tan 23^{o}\tan 42^{o}\tan 67^{o}$
$\Rightarrow \tan 48^{o}\tan 23^{o}\tan (90^o-48^{o})\tan (90^o-23^{o})$
$\Rightarrow \tan 48^{o}\tan 23^{o}\cot 48^{o}\cot23^{o}$ [it is known that $\tan (90^0-\theta = \cot\theta)$ and $\cot\theta\times \tan \theta =1$
$=1$

Hence proved.

Q2 Show that :

$(ii) \cos 38^{o}\cos 52^{o}-\sin 38^{o}\sin 52^{o}= 0$

Answer:

$\cos 38^{o}\cos 52^{o}-\sin 38^{o}\sin 52^{o}= 0$

Taking Left Hand Side (LHS)
= $\cos 38^{o}\cos 52^{o}-\sin 38^{o}\sin 52^{o}$
= $\cos 38^{o}\cos (90^o-38^{o})-\sin 38^{o}\sin (90^o-38^{o})$
= $\cos 38^{o}\sin38^{o}-\sin 38^{o}\cos 38^{o}$ [it is known that $\sin(90^0-\theta) =\cos \theta$ and $\cos(90^0-\theta) =\sin \theta$ ]
= 0

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Excercise: 8.4

Q3 Evaluate :

$(i)\frac{\sin ^{2}63^{o}+\sin ^{2}27^{o}}{\cos ^{2}17^{o}+\cos ^{2}73^{o}}$

Answer:

$\frac{\sin ^{2}63^{o}+\sin ^{2}27^{o}}{\cos ^{2}17^{o}+\cos ^{2}73^{o}}$ ....................(i)

The above equation can be written as;

$\\=\frac{\sin ^{2}63^{o}+\sin ^{2}(90^0-63^{o})}{\cos ^{2}(90^0-73^{o})+\cos ^{2}73^{o}}\\\\ =\frac{\sin ^{2}63^{o}+\cos ^{2}63^{o}}{\sin ^{2}73^{o}+\cos ^{2}73^{o}}\\\\ = 1$
(Since $\sin^2\theta +\cos^2\theta = 1$ )

About NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry

Unit 5 "Trigonometry" holds 12 marks out of 80 marks in the maths paper of CBSE board examination and we can expect 2-3 questions from this chapter of total around 8 marks. There is a total of 4 exercises with 27 questions in the NCERT solutions for class 10 maths chapter 8. These NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry are designed to provide assistance for homework and for preparing the board examinations.

The trigonometric ratios of the angle A in right triangle ABC are defined as follows-

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$\\sine \:of\: \angle A=\frac{side \:opposite \:to \:angle \:A}{hypotenuse}=\frac{BC}{AC}\\\\cosine \:of\: \angle A=\frac{side \:adjacent\: to \:angle \:A}{hypotenuse}=\frac{AB}{AC}\\\\tangent \:of\: \angle A=\frac{side \:opposite \:to \:angle \:A}{side \:adjacent\: to \:angle \:A}=\frac{BC}{AB}\\\\cosecant \:of\:\angle A=\frac{hypotenuse}{side \:opposite \:to \:angle \:A}=\frac{AC}{BC}\\\\secant \:of \:\angle A= \frac{hypotenuse}{side \:adjacent\: to \:angle \:A}=\frac{AC}{AB}\\\\cotangent \:of\: \angle A=\frac{side \:adjacent\: to \:angle \:A}{side \:opposite \:to \:angle \:A}=\frac{AB}{BC}$

The values of all the trigonometric ratios of 0°, 30°, 45°, 60°, and 90° are-

$\angle A$
Sin A	0	$\frac{1}{2}$	$\frac{1}{\sqrt{2}}$	$\frac{\sqrt{3}}{2}$	1
Cos A	1	$\frac{\sqrt{3}}{2}$		$\frac{1}{2}$	0
Tan A	0	$\frac{1}{\sqrt{3}}$	1	$\sqrt{3}$	Not defined
Cosec A	Not defined	2	$\sqrt{2}$	$\frac{2}{\sqrt{3}}$	1
Sec A	1	$\frac{2}{\sqrt{3}}$	$\sqrt{2}$	2	Not defined
Cot A	Not defined	$\sqrt{3}$	1	$\frac{1}{\sqrt{3}}$	0

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Benefits of NCERT Solutions for Class 10 Maths Chapter 8

These Class 10 Maths Chapter 8 NCERT solutions are prepared by the experts. Hence these solutions are 100 per cent reliable.
The NCERT Class 10 maths solutions chapter 8 will be beneficial for Class 10 board exams and for higher studies as well.
These solutions will help in building the basic concepts of trigonometry and bring forth some easy ways to solve the questions.

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How to use NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry?

Firstly, learn all the concepts given in the NCERT book. Memorise all the trigonometric ratios, angle values, and trigonometric identities.
Now practice exercises by referring to the NCERT Class 10 maths solutions chapter 8.
As the NCERT Solutions for Class 10 Maths Chapter 8 PDF Download is not available. So you can save the webpage to practice the solutions offline.
After doing all these you can practice the last 5 years question papers of board examinations.

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Frequently Asked Question (FAQs) - NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry

Question: Whether this unit introduction to trigonometry is helpful for higher studies?

Answer:

Trigonometry is a most important field in mathematics which is useful in almost every field including architecture, electronics, seismology, meteorology, oceanography etc.

Question: How many chapters are there in the class 10 maths?

Answer:

There are a total of 15 chapters in the class 10 maths NCERT.

Question: What book is best for CBSE class 10 maths?

Answer:

NCERT textbook is only book which you should know very well, there is no need of supplementary book as CBSE board class 10 paper is entirely based on the NCERT.

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Class 10 Maths Ncert Solutions Chapter Trigonometry

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Class 10 Maths Ncert Solutions Chapter Trigonometry

NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry Excercise: 8.1

Q1 In $\Delta \: ABC$ , right-angled at $B, AB = 24 \: cm$ , $BC = 7 \: cm$ . Determine : $(i)\; \sin A, \cos A$ $(ii)\; \sin C, \cos C$

Answer:

Answer:

Answer:

Answer:

Answer:

NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry Excercise: 8.3

Answer:

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NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Excercise: 8.4

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About NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry

NCERT Solutions for Class 10 Maths All Chapters

Benefits of NCERT Solutions for Class 10 Maths Chapter 8

How to use NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry?

Frequently Asked Question (FAQs) - NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry

Question: Whether this unit introduction to trigonometry is helpful for higher studies?

Question: How many chapters are there in the class 10 maths?

Question: What book is best for CBSE class 10 maths?

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Class 10 Maths Ncert Solutions Chapter Trigonometry

NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry Excercise: 8.1

Q1 In , right-angled at , . Determine :

Answer:

Answer:

Answer:

Answer:

Answer:

NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry Excercise: 8.3

Answer:

Answer:

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Answer:

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NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Excercise: 8.4

Answer:

About NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry

NCERT Solutions for Class 10 Maths All Chapters

Benefits of NCERT Solutions for Class 10 Maths Chapter 8

How to use NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry?

Frequently Asked Question (FAQs) - NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry

Question: Whether this unit introduction to trigonometry is helpful for higher studies?

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Q1 In $\Delta \: ABC$ , right-angled at $B, AB = 24 \: cm$ , $BC = 7 \: cm$ . Determine : $(i)\; \sin A, \cos A$ $(ii)\; \sin C, \cos C$