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

Komal MiglaniUpdated on 30 Apr 2025, 03:20 PM IST

The exercise establishes basic trigonometric foundations by teaching about the six vital ratios, which include sine, cosine, tangent, cosecant, secant and cotangent. The ratios help us to understand angle-per-length relationships through right-angled triangles. We have to master the exercise because it creates essential foundations for higher-level trigonometric subjects and serves as a key part of the Class 10 mathematics syllabus.

This Story also Contains

  1. NCERT Solutions Class 10 Maths Chapter 8: Exercise 8.1
  2. Access Solution of Introduction to Trigonometry Class 10 Chapter 8 Exercise: 8.1
  3. Topics covered in Chapter 8, Introduction to Trigonometry: Exercise 8.1
  4. NCERT Solutions of Class 10 Subject Wise
  5. NCERT Exemplar Solutions of Class 10 Subject Wise
NCERT Solutions for Exercise 8.1 Class 10 Maths Chapter 8 - Introduction to Trigonometry
exercise 8.1

The NCERT Solutions for Class 10 Maths present practical methods for using trigonometric ratios according to the content found in NCERT textbooks. The exercises guide students to evaluate trigonometric ratios when given dimensional triangle measurements and help students determine missing lengths through ratio analysis while confirming trigonometric identity relations. Exercise 8.1 in the NCERT Books helps students strengthen their trigonometry knowledge as it develops their ability to solve complex mathematical problems required for higher educational levels.

NCERT Solutions Class 10 Maths Chapter 8: Exercise 8.1

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Access Solution of 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:

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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 $\tan P - \cot R$ .

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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.

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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 $\cos A =\cos B$

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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 $\tan A =\frac{1}{\sqrt{3}}$ $\Rightarrow A=30^0$

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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:

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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) [ $-1\leq \sin \theta\leq 1$ ]

Topics covered in Chapter 8, Introduction to Trigonometry: Exercise 8.1

1. Calculating Trigonometric Ratios: Individuals can determine the values of trigonometric ratios when given the lengths of right-angled triangles.

2. Determining Side Lengths: Calculating the unknown side lengths of right-angled triangles becomes possible when applying known trigonometric ratios.

3. Verifying Trigonometric Identities: By using both calculation and reasoning, students verify fundamental trigonometric identities.

4. Real-Life Applications: Applying trigonometric principles helps solve real-life situations where trigonometry measures sides and angles in practical problems.

5. Definition of Trigonometric Ratios: Understanding sine, cosine, tangent, cosecant, secant, and cotangent in the context of right-angled triangles.

Trigonometric Ratios

Formula

sin θ

$\frac{\text{Perpendicular}}{\text{Hypotenuse}}$

cos θ

$\frac{\text{Base}}{\text{Hypotenuse}}$

tan θ

$\frac{\text{Perpendicular}}{\text{Base}}$

cosec θ

$\frac{\text{Hypotenuse}}{\text{Perpendicular}}$

sec θ

$\frac{\text{Hypotenuse}}{\text{Base}}$

cot θ

$\frac{\text{Base}}{\text{Perpendicular}}$

NCERT Solutions of Class 10 Subject Wise

Students must check the NCERT solutions for class 10 of Mathematics and Science Subjects.

NCERT Exemplar Solutions of Class 10 Subject Wise

Students must check the NCERT Exemplar solutions for class 10 of Mathematics and Science Subjects.

Frequently Asked Questions (FAQs)

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

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

Q: What is the sine function and write it’s formula?
A:

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

Sin θ=opposite side/hypotenuse 

Q: What is Cos ?
A:

Cos means Cosine is the ratio of Adjacent Side and Hypotenuse

Q: What is the relationship between sine, cosine and tangent functions ?
A:

tan θ=sin θ/cos θ

Q: What is the relationship between sine and cosecant functions ?
A:

The multiplicative inverse of sine is known as the cosecant

Q: What are the six trigonometric ratios according to NCERT solutions for Class 10 Maths chapter 8 exercise 8.1 ?
A:

The six trigonometric ratios are 

  • Sine 

  • Cosine 

  • Tangent 

  • Cotangent 

  • Cosecant  

  • Secant.

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

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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