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NCERT Solutions for Exercise 8.1 Class 10 Maths Chapter 8 - Introduction to Trigonometry

NCERT Solutions for Exercise 8.1 Class 10 Maths Chapter 8 - Introduction to Trigonometry

Edited By Ramraj Saini | Updated on Nov 27, 2023 08:44 AM IST | #CBSE Class 10th

NCERT Solutions For Class 10 Maths Chapter 8 Exercise 8.1

NCERT Solutions for Exercise 8.1 Class 10 Maths Chapter 8 Introduction to Trigonometry are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. Class 10 maths ex 8.1 deal with the concept of trigonometry and trigonometric ratios. The Branch of Mathematics that deals with the measurement of sides and angles of a triangle is known as trigonometry. Trigonometric Ratios are nothing but the ratios of the sides of a right triangle with respect to its acute angles. Trigonometrically, there are six ratios.

This Story also Contains
  1. NCERT Solutions For Class 10 Maths Chapter 8 Exercise 8.1
  2. Download Free Pdf of NCERT Solutions for Class 10 Maths Chapter 8 Exercise 8.1
  3. Assess NCERT Solutions for Class 10 Maths Chapter 8 Exercise 8.1
  4. More About NCERT Solutions for Class 10 Maths Exercise 8.1: Trigonometric Ratios
  5. Benefits of NCERT Solutions for Class 10 Maths Exercise 8.1:
  6. NCERT Solutions Subject Wise
  7. Subject Wise NCERT Exemplar Solutions
NCERT Solutions for Exercise 8.1 Class 10 Maths Chapter 8 - Introduction to Trigonometry
NCERT Solutions for Exercise 8.1 Class 10 Maths Chapter 8 - Introduction to Trigonometry

NCERT solutions for exercise 8.1 Class 10 Maths chapter 8 Introduction to Trigonometry focuses on trigonometry and trigonometric ratios including sine, cosine, tangent, cosecant, secant and cotangent. 10th class Maths exercise 8.1 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

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Assess NCERT Solutions for Class 10 Maths Chapter 8 Exercise 8.1

Introduction to Trigonometry class 10 chapter 8 Exercise: 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

Q3 If \sin A=\frac{3}{4}, calculate \cos A and \tan A .

Answer:

Suppose \Delta ABC is a right-angled triangle in which \angle B = 90^0 and we have \sin A=\frac{3}{4},
So,
1635935058122

Let the length of AB be 4 unit and the length of BC = 3 unit So, by using Pythagoras theorem,
AB = \sqrt{16-9} = \sqrt{7} units
Therefore,
\cos A = \frac{AB}{AC} = \frac{\sqrt{7}}{4} and \tan A = \frac{BC}{AB} = \frac{3}{\sqrt{7}}

Q4 Given 15 \: \cot A=8, find \sin A and \sec A .

Answer:

We have,
15 \: \cot A=8, \Rightarrow \cot A =8/15
It implies that In the triangle ABC in which \angle B =90^0 . The length of AB be 8 units and the length of BC = 15 units

Now, by using Pythagoras theorem,
AC = \sqrt{64 +225} = \sqrt{289}
\Rightarrow AC =17 units

So, \sin A = \frac{BC}{AC} = \frac{15}{17}
and \sec A = \frac{AC}{AB} = \frac{17}{8}

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}

Q6 If \angle A and \angle B are acute angles such that \cos A = \cos B , then show that \angle A =\angle B .

Answer:

We have, A and B are two acute angles of triangle ABC and \cos A =\cos B

sdfgjhkjhdcgv According to question, In triangle ABC,
\cos A =\cos B


\frac{AC}{AB}=\frac{BC}{AB}
\Rightarrow AC = AB
Therefore, \angle A = \angle B [angle opposite to equal sides are equal]

Q7 If \cot \theta =\frac{7}{8}, evaluate: (i)\; \frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)} (ii)\; \cot ^{2}\theta

Answer:

Given that,
\cot \theta =\frac{7}{8}
\therefore perpendicular (AB) = 8 units and Base (AB) = 7 units
Draw a right-angled triangle ABC in which \angle B =90^0
Now, By using Pythagoras theorem,
AC^2 = AB^2+BC^2
AC = \sqrt{64 +49} =\sqrt{113}

So, \sin \theta = \frac{BC}{AC} = \frac{8}{\sqrt{113}}
and \cos \theta = \frac{AB}{AC} = \frac{7}{\sqrt{113}}

\Rightarrow \cot \theta =\frac{\cos \theta}{\sin \theta} = \frac{7}{8}
(i)\; \frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}
\Rightarrow \frac{(1-\sin^2\theta)}{(1-\cos^2\theta)} = \frac{\cos^2\theta}{\sin^2\theta} = \cot ^2\theta
=(\frac{7}{8})^2 = \frac{49}{64}

(ii)\; \cot ^{2}\theta
=(\frac{7}{8})^2 = \frac{49}{64}

Q8 If 3\cot A=4, check wether \frac{1-\tan ^{2}A}{1+\tan ^{2}A}=\cos ^{2}A-\sin ^{2}A or not.

Answer:

Given that,
3\cot A=4,
\Rightarrow \cot = \frac{4}{3} = \frac{base}{perp.}
ABC is a right-angled triangle in which \angle B =90^0 and the length of the base AB is 4 units and length of perpendicular is 3 units
1635935097081

By using Pythagoras theorem, In triangle ABC,
\\AC^2=AB^2+BC^2\\ AC = \sqrt{16+9}\\ AC = \sqrt{25}
AC = 5 units

So,
\tan A = \frac{BC}{AB} = \frac{3}{4}
\cos A = \frac{AB}{AC} = \frac{4}{5}
\sin A = \frac{BC}{AC} = \frac{3}{5}
\frac{1-\tan ^{2}A}{1+\tan ^{2}A}=\cos ^{2}A-\sin ^{2}A
Put the values of above trigonometric ratios, we get;
\Rightarrow \frac{1-9/4}{1+9/4} = \frac{16}{25}-\frac{9}{25}
\Rightarrow -\frac{5}{13} \neq \frac{7}{25}
LHS \neq RHS

Q9 In triangle ABC , right-angled at B , if \tan A =\frac{1}{\sqrt{3}}, find the value of:

(i) \sin A\: \cos C + \cos A\: \sin C
(ii) \cos A\: \cos C + \sin A\: \sin C

Answer:

Given a triangle ABC, right-angled at B and \tan A =\frac{1}{\sqrt{3}} \Rightarrow A=30^0

FireShot%20Capture%20151%20-%20Careers360%20CMS%20-%20cms-articles%20-%20learn According to question, \tan A =\frac{1}{\sqrt{3}} = \frac{BC}{AB}
By using Pythagoras theorem,
\\AC^2 = AB^2+BC^2\\ AC = \sqrt{1+3} =\sqrt{4}
AC = 2
Now,
\\\sin A = \frac{BC}{AC} = \frac{1}{2}\\ \sin C =\frac{AB}{AC} = \frac{\sqrt{3}}{2}\\ \cos A = \frac{AB}{AC} = \frac{\sqrt{3}}{2}\\ \cos C = \frac{BC}{AC} = \frac{1}{2}

Therefore,

(i) \sin A\: \cos C + \cos A\: \sin C
\\\Rightarrow \frac{1}{2}\times\frac{1}{2}+\frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2}\\ \Rightarrow1/4 +3/4\\ \Rightarrow4/4 = 1

(ii) \cos A\: \cos C + \sin A\: \sin C
\\\Rightarrow \frac{\sqrt{3}}{2}\times \frac{1}{2}+\frac{1}{2}\times\frac{\sqrt{3}}{2}\\ \Rightarrow \frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{4}\\ \Rightarrow \frac{\sqrt{3}}{2}


Q10 In \Delta \: PQR , right-angled at Q , PR + QR = 25\: cm and PQ = 5 \: cm . Determine the values of \sin P, \cos P \: and\: \tan P.

Answer:

sdfghjkdfghj We have, PR + QR = 25 cm.............(i)
PQ = 5 cm
and \angle Q =90^0
According to question,
In triangle \Delta PQR,
By using Pythagoras theorem,
\\PR^2 = PQ^2+QR^2\\ PQ^2 =PR^2-QR^2 \\ 5^2= (PR-QR)(PR+QR)\\ 25 = 25(PR-QR) \\
PR - QR = 1........(ii)

From equation(i) and equation(ii), we get;
PR = 13 cm and QR = 12 cm.

therefore,
\\\sin P= \frac{QR}{PR}= 12/13\\ \cos P = \frac{PQ}{RP} = 5/13\\ \therefore \tan P = \frac{\sin P}{\cos P} = 12/5

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 ]

More About NCERT Solutions for Class 10 Maths Exercise 8.1: Trigonometric Ratios

The NCERT solutions for Class 10 Maths exercise 8.1 also focused on the relationship between the trigonometric ratios. Exercise 8.1 Class 10 Maths consists of 11 questions based on trigonometric ratios and their relations. The sine is the ratio of the opposing side to the hypotenuse of a right-angle triangle.

The cosine is the ratio of the neighboring side to the hypotenuse of a right-angle triangle. The tangent is the ratio of the opposite side to the adjacent side of a right-angle triangle. The multiplicative inverse of sine is known as cosecant. Similarly, the multiplicative inverse of cosine is known as secant. The multiplicative inverse of the tangent is known as cotangent. Usage of Pythagoras theorem is also covered in exercise 8.1 Class 10 Maths. Students can use these Introduction to Trigonometry Class 10 notes to quick revision of the important concepts discussed in this chapter.

Benefits of NCERT Solutions for Class 10 Maths Exercise 8.1:

  • NCERT solutions for Class 10 Maths exercise 8.1 are important for first and second term exams as well as for higher grades. We should practice all the questions present in NCERT solutions for Class 10 Maths exercise 8.1 twice or thrice to attain good and high marks.
  • Exercise 8.1 Class 10 Maths, Trigonometric ratios are used in finding the missing sides or angles of the right-angle triangle.
  • By solving the NCERT solution for Class 10 Maths chapter 8 exercise 8.1 exercises, students can do any type of questions related to trigonometry easily by exploring the derivation of these formulas.

Also, see-

NCERT Solutions Subject Wise

Subject Wise NCERT Exemplar Solutions

Frequently Asked Questions (FAQs)

1. What are the three basic trigonometric ratios according to NCERT solutions for Class 10 Maths chapter 8 exercise 8.1 ?

The three basic trigonometric ratios are sine, cosine and tangent. 

2. What is the sine function and write it’s formula?

The ratio of the opposite side to the hypotenuse of the right angle triangle is known as the sine.

Sin θ=opposite side/hypotenuse 

3. What is Cos ?

Cos means Cosine is the ratio of Adjacent Side and Hypotenuse

4. What is the relationship between sine, cosine and tangent functions ?

tan θ=sin θ/cos θ

5. What is the relationship between sine and cosecant functions ?

The multiplicative inverse of sine is known as the cosecant

6. What are the six trigonometric ratios according to NCERT solutions for Class 10 Maths chapter 8 exercise 8.1 ?

The six trigonometric ratios are 

  • Sine 

  • Cosine 

  • Tangent 

  • Cotangent 

  • Cosecant  

  • Secant.

7. How many questions and what type of questions are there in the NCERT solutions for Class 10 Maths chapter 8 exercise 8.1 ?

NCERT solutions for Class 10 Maths chapter 8 exercise 8. 1 consists of 11 Questions in which 7 are short answers, 3 of them are long answers and the remaining one is a short answer with reasoning and all the questions are based on trigonometric ratios. 

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sai raja 01 September,2024
show the previous year exams ?

Hello Aspirant,  Hope your doing great,  your question was incomplete and regarding  what exam your asking.

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SWETCHCHA MISKA 10 July,2024
as an icse student in class 10 is above 80 percent good enough to get into commerce in 11th cbse

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Manasvini Gupta 23 July,2024
i had cleared 10th from cbse in 2022 2 years ago now i wish to apply for 12th as private candidate in 2025. can i do so?

hello Zaid,

Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.

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Ashvadha 03 July,2024
i had cleared 10th in 2022 from cbse can i apply for 12th as private candidate this year?

According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.

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