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.

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

$(i)\; \sin A, \cos A$ $(ii)\; \sin C, \cos C$

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 $\cos C = B/H = BC/AC = 7/25$

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

Suppose $\Delta$ ABC is a right-angled triangle in which $\angle B = 90^0$ and we have $\dpi{100} \sin A=\frac{3}{4},$
So,

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 $\tan A = \frac{BC}{AB} = \frac{3}{\sqrt{7}}$

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 $\sec A = \frac{AC}{AB} = \frac{17}{8}$

We have,
$\dpi{100} \sec \theta =\frac{13}{12},$

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.

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

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

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

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

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

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

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

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]

$(i)\; \frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}$ $(ii)\; \cot ^{2}\theta$

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 $\cos \theta = \frac{AB}{AC} = \frac{7}{\sqrt{113}}$

$\Rightarrow \cot \theta =\frac{\cos \theta}{\sin \theta} = \frac{7}{8}$
$\dpi{100} (i)\; \frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}$
$\Rightarrow \frac{(1-\sin^2\theta)}{(1-\cos^2\theta)} = \frac{\cos^2\theta}{\sin^2\theta} = \cot ^2\theta$
$=(\frac{7}{8})^2 = \frac{49}{64}$

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

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

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

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

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

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,

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

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

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

(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 $\Theta .$

(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-

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## Subject Wise NCERT Exemplar Solutions

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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