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

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## NCERT Solutions for Class 10 Maths Chapter 8 Exercise 8.1 - FREE PDF Download

NCERT Solutions for Class 10 Maths Chapter 8, Introduction to Trigonometry, Exercise 8.1, is available here in PDF format. The solutions can be downloaded easily and can be referred by the students from anywhere. The NCERT solutions created by our subject experts are explained in a step-by-step procedure along with the related geometric figure. The solutions are prepared per the latest NCERT syllabus and guidelines of the CBSE board and it aims to help the students to excel in their board and competitive examination.

Table of Content
1. NCERT Solutions for Class 10 Maths Chapter 8 Exercise 8.1 - FREE PDF Download
2. Glance on NCERT Solutions Maths Chapter 8 Exercise 8.1 Class 10 | Vedantu
3. Access Class 10 Maths NCERT Solutions Chapter 8 Introduction to Trigonometry Exercise 8.1
4. Class 10 Maths Chapter 8: Exercises Breakdown
5. CBSE Class 10 Maths Chapter 8 Other Study Materials
6. Chapter-Specific NCERT Solutions for Class 10 Maths
FAQs

## Glance on NCERT Solutions Maths Chapter 8 Exercise 8.1 Class 10 | Vedantu

• This article deals with Trigonometry which is the branch of mathematics that is related to the specific functions of angles and their application to calculations.

• There are six functions of an angle that are widely used in trigonometry to solve unknown sides. Their six functions of trigonometry and their abbreviations are sine (sin), cosine (cos), tangent (tan), cotangent (cot), secant (sec), and cosecant (cosec).

• Trigonometry is a concept that represents the relation between the angles and sides of a right-angled triangle.

• Understanding the concept of right-angled triangles and their components (perpendicular, base, hypotenuse).

• Determining missing angles using trigonometric ratios and inverse trigonometric functions and missing side lengths in a right triangle when one side length and a trigonometric ratio are given.

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## Access Class 10 Maths NCERT Solutions Chapter 8 Introduction to Trigonometry Exercise 8.1

1. In $\Delta ABC$ right angled at $B$, $AB=24\text{ cm}$, $BC=7\text{ cm}$. Determine

i) $\sin A,\cos A$

Ans: Given that in the right angle triangle $\Delta ABC$, $AB=24\text{ cm}$, $BC=7\text{ cm}$.

Let us draw a right triangle $\Delta ABC$, also $AB=24\text{ cm}$, $BC=7\text{ cm}$. We get

We have to find $\sin A,\cos A$.

We know that for right triangle

$\sin \theta =\dfrac{\text{opposite side}}{\text{hypotenuse}}$ and

$\cos \theta =\dfrac{\text{adjacent side}}{\text{hypotenuse}}$

Now, we need to apply the Pythagoras theorem to find the measure of hypotenuse.

In $\Delta ABC$, by Pythagoras theorem ,

${{\left( \text{hypotenuse} \right)}^{2}}={{\left( base \right)}^{2}}+{{\left( perpendicular \right)}^{2}}$

Here, $AB=24\text{ cm}$, $BC=7\text{ cm}$

We get

$\Rightarrow {{\left( \text{AC} \right)}^{2}}={{\left( AB \right)}^{2}}+{{\left( BC \right)}^{2}}$

$\Rightarrow {{\left( \text{AC} \right)}^{2}}={{\left( 24 \right)}^{2}}+{{\left( 7 \right)}^{2}}$

$\Rightarrow {{\left( \text{AC} \right)}^{2}}=576+49$

$\Rightarrow {{\left( \text{AC} \right)}^{2}}=625\text{ }c{{m}^{2}}$

$\Rightarrow AC=25\text{ cm}$

Now,

$\sin \theta =\dfrac{\text{opposite side}}{\text{hypotenuse}}$

$\Rightarrow \sin A=\dfrac{BC}{AC}$

$\therefore \sin A=\dfrac{7}{25}$

$\cos \theta =\dfrac{\text{adjacent side}}{\text{hypotenuse}}$

$\Rightarrow \cos A=\dfrac{AB}{AC}$

$\therefore \cos A=\dfrac{24}{25}$

ii) $\sin C,\cos C$

Ans: Given that in the right angle triangle $\Delta ABC$, $AB=24\text{ cm}$, $BC=7\text{ cm}$.

Let us draw a right triangle $\Delta ABC$, also $AB=24\text{ cm}$, $BC=7\text{ cm}$. We get

We have to find $\sin C,\cos C$.

We know that for right triangle

$\sin \theta =\dfrac{\text{opposite side}}{\text{hypotenuse}}$ and

$\cos \theta =\dfrac{\text{adjacent side}}{\text{hypotenuse}}$

Now, we need to apply the Pythagoras theorem to find the measure of hypotenuse.

In $\Delta ABC$, by Pythagoras theorem ,

${{\left( \text{hypotenuse} \right)}^{2}}={{\left( base \right)}^{2}}+{{\left( perpendicular \right)}^{2}}$

Here, $AB=24\text{ cm}$, $BC=7\text{ cm}$

We get

$\Rightarrow {{\left( \text{AC} \right)}^{2}}={{\left( AB \right)}^{2}}+{{\left( BC \right)}^{2}}$

$\Rightarrow {{\left( \text{AC} \right)}^{2}}={{\left( 24 \right)}^{2}}+{{\left( 7 \right)}^{2}}$

$\Rightarrow {{\left( \text{AC} \right)}^{2}}=576+49$

$\Rightarrow {{\left( \text{AC} \right)}^{2}}=625\text{ }c{{m}^{2}}$

$\Rightarrow AC=25\text{ cm}$

Now,

$\sin \theta =\dfrac{\text{opposite side}}{\text{hypotenuse}}$

$\Rightarrow \sin C=\dfrac{AB}{AC}$

$\therefore \sin C=\dfrac{24}{25}$

$\cos \theta =\dfrac{\text{adjacent side}}{\text{hypotenuse}}$

$\Rightarrow \cos C=\dfrac{BC}{AC}$

$\therefore \cos A=\dfrac{7}{25}$

2. In the given figure find $\tan P-\cot R$.

Ans: Given in the figure,

$PQ=12\text{ cm}$

$PQ=13\text{ cm}$

We know that for right triangle

$\tan \theta =\dfrac{\text{opposite side}}{\text{adjacent side}}$ and

$\cot \theta =\dfrac{\text{adjacent side}}{\text{opposite side}}$

Now, we need to apply the Pythagoras theorem to find the measure of adjacent side/base.

In $\Delta PQR$, by Pythagoras theorem ,

${{\left( \text{hypotenuse} \right)}^{2}}={{\left( base \right)}^{2}}+{{\left( perpendicular \right)}^{2}}$

We get

$\Rightarrow {{\left( PR \right)}^{2}}={{\left( PQ \right)}^{2}}+{{\left( QR \right)}^{2}}$

$\Rightarrow {{\left( 13 \right)}^{2}}={{\left( 12 \right)}^{2}}+{{\left( QR \right)}^{2}}$

$\Rightarrow 169=144+{{\left( QR \right)}^{2}}$

$\Rightarrow {{\left( QR \right)}^{2}}=169-144$

$\Rightarrow {{\left( QR \right)}^{2}}=25\text{ }c{{m}^{2}}$

$\Rightarrow QR=5\text{ cm}$

Now,

$\tan P=\dfrac{\text{opposite side}}{\text{adjacent side}}$

$\Rightarrow \tan P=\dfrac{QR}{PQ}$

$\therefore \tan P=\dfrac{5}{12}$

$\cot R=\dfrac{\text{adjacent side}}{\text{opposite side}}$

$\Rightarrow \cot R=\dfrac{QR}{PQ}$

$\therefore \cot R=\dfrac{5}{12}$

$\Rightarrow \tan P-\cot R=\dfrac{5}{12}-\dfrac{5}{12}$

$\therefore \tan P-\cot R=0$

3. If $\sin A=\dfrac{3}{4}$, calculate $\cos A$ and $\tan A$.

Ans: Let us consider a right angled triangle $\Delta ABC$. We get

Given that $\sin A=\dfrac{3}{4}$.

We know that  $\sin \theta =\dfrac{\text{opposite side}}{\text{hypotenuse}}$.

From the above figure, we get

$\sin A=\dfrac{BC}{AC}$

Therefore, we get

$\Rightarrow BC=3$ and

$\Rightarrow AC=4$

Now, we have to find the values of $\cos A$ and $\tan A$.

We know that $\cos \theta =\dfrac{\text{adjacent side}}{\text{hypotenuse}}$ and $\tan \theta =\dfrac{\text{opposite side}}{\text{adjacent side}}$.

Now, we need to apply the Pythagoras theorem to find the measure of adjacent side/base.

In $\Delta ABC$, by Pythagoras theorem ,

${{\left( \text{hypotenuse} \right)}^{2}}={{\left( base \right)}^{2}}+{{\left( perpendicular \right)}^{2}}$

Here, $AC=4\text{ cm}$, $BC=3\text{ cm}$

We get

$\Rightarrow {{\left( \text{AC} \right)}^{2}}={{\left( AB \right)}^{2}}+{{\left( BC \right)}^{2}}$

$\Rightarrow {{4}^{2}}=A{{B}^{2}}+{{3}^{2}}$

$\Rightarrow 16=A{{B}^{2}}+9$

$\Rightarrow A{{B}^{2}}=16-9$

$\Rightarrow A{{B}^{2}}=7$

$\Rightarrow AB=\sqrt{7}\text{ cm}$

Now, we get

$\cos A=\dfrac{AB}{AC}$

$\therefore \cos A=\dfrac{\sqrt{7}}{4}$

And $\tan A=\dfrac{BC}{AB}$

$\therefore \tan A=\dfrac{3}{\sqrt{7}}$

4. Given $15\cot A=8$. Find $\sin A$ and $\sec A$.

Ans: Let us consider a right angled triangle $\Delta ABC$. We get

Given that $15\cot A=8$.

We get $\cot A=\dfrac{8}{15}$.

We know that $\cot \theta =\dfrac{\text{adjacent side}}{\text{opposite side}}$.

From the above figure, we get

$\cot A=\dfrac{AB}{BC}$

Therefore, we get

$\Rightarrow BC=15$ and

$\Rightarrow AB=8$

Now, we have to find the values of $\sin A$ and $\sec A$.

We know that $\sin \theta =\dfrac{\text{opposite side}}{\text{hypotenuse}}$ and $\sec \theta =\dfrac{\text{hypotenuse}}{\text{adjacent side}}$.

Now, we need to apply the Pythagoras theorem to find the measure of hypotenuse.

In $\Delta ABC$, by Pythagoras theorem ,

${{\left( \text{hypotenuse} \right)}^{2}}={{\left( base \right)}^{2}}+{{\left( perpendicular \right)}^{2}}$

We get

$\Rightarrow {{\left( \text{AC} \right)}^{2}}={{\left( AB \right)}^{2}}+{{\left( BC \right)}^{2}}$

$\Rightarrow A{{C}^{2}}={{8}^{2}}+{{15}^{2}}$

$\Rightarrow A{{C}^{2}}=64+225$

$\Rightarrow A{{C}^{2}}=289$

$\Rightarrow AC=17\text{ cm}$

Now, we get

$\sin A=\dfrac{BC}{AC}$

$\therefore \sin A=\dfrac{15}{17}$

And $\sec A=\dfrac{AC}{AB}$

$\therefore \sec A=\dfrac{17}{8}$

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

Ans: Let us consider a right angled triangle $\Delta ABC$. We get

Given that $\sec \theta =\dfrac{13}{12}$.

We know that $\sec \theta =\dfrac{\text{hypotenuse}}{\text{adjacent side}}$.

From the above figure, we get

$\sec \theta =\dfrac{AC}{AB}$

Therefore, we get

$\Rightarrow AC=13$ and

$\Rightarrow AB=12$

Now, we need to apply the Pythagoras theorem to find the measure of the perpendicular/opposite side.

In $\Delta ABC$, by Pythagoras theorem ,

${{\left( \text{hypotenuse} \right)}^{2}}={{\left( base \right)}^{2}}+{{\left( perpendicular \right)}^{2}}$

We get

$\Rightarrow {{\left( \text{AC} \right)}^{2}}={{\left( AB \right)}^{2}}+{{\left( BC \right)}^{2}}$

$\Rightarrow {{13}^{2}}={{12}^{2}}+B{{C}^{2}}$

$\Rightarrow 169=144+B{{C}^{2}}$

$\Rightarrow B{{C}^{2}}=25$

$\Rightarrow BC=5\text{ cm}$

Now, we know that

$\sin \theta =\dfrac{\text{opposite side}}{\text{hypotenuse}}$

Here, $\sin \theta =\dfrac{BC}{AC}$

$\therefore \sin \theta =\dfrac{5}{13}$

We know that $\cos \theta =\dfrac{\text{adjacent side}}{\text{hypotenuse}}$

Here, $\cos \theta =\dfrac{AB}{AC}$

$\therefore \cos \theta =\dfrac{12}{13}$

We know that $\tan \theta =\dfrac{\text{opposite side}}{\text{adjacent side}}$

Here, $\tan \theta =\dfrac{BC}{AB}$

$\therefore \tan \theta =\dfrac{5}{12}$

We know that $\operatorname{cosec}\theta =\dfrac{\text{hypotenuse}}{\text{opposite side}}$

Here, $\operatorname{cosec}\theta =\dfrac{AC}{BC}$

$\therefore \operatorname{cosec}\theta =\dfrac{13}{5}$

We know that $\cot \theta =\dfrac{\text{adjacent side}}{\text{opposite side}}$

Here, $\cot \theta =\dfrac{\text{AB}}{BC}$

$\therefore \cot \theta =\dfrac{12}{5}$.

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

Ans: Let us consider a right angled triangle $\Delta ABC$. We get

Given that $\cos A=\cos B$.

In a right triangle $\Delta ABC$, we know that

$\cos \theta =\dfrac{\text{adjacent side}}{\text{hypotenuse}}$

Here,

$\cos A=\dfrac{AC}{AB}$

And $\cos B=\dfrac{BC}{AB}$

As given $\cos A=\cos B$, we get

$\Rightarrow \dfrac{AC}{AB}=\dfrac{BC}{AB}$

$\Rightarrow AC=AB$

Now, we know that angles opposite to the equal sides are also equal in measure.

Then, we get

$\angle A=\angle B$

Hence proved.

7. Evaluate the following if $\cot \theta =\dfrac{7}{8}$

i) $\dfrac{\left( 1+\sin \theta \right)\left( 1-\sin \theta \right)}{\left( 1+\cos \theta \right)\left( 1-\cos \theta \right)}$

Ans: Let us consider a right angled triangle $\Delta ABC$. We get

Now, in a right triangle we know that $\cot \theta =\dfrac{\text{adjacent side}}{\text{opposite side}}$.

Here, from the figure $\cot \theta =\dfrac{BC}{AB}$ .

We get

$AB=8$ and

$BC=7$

Now, we need to apply the Pythagoras theorem to find the measure of hypotenuse.

In $\Delta ABC$, by Pythagoras theorem ,

${{\left( \text{hypotenuse} \right)}^{2}}={{\left( base \right)}^{2}}+{{\left( perpendicular \right)}^{2}}$

We get

$\Rightarrow {{\left( \text{AC} \right)}^{2}}={{\left( AB \right)}^{2}}+{{\left( BC \right)}^{2}}$

$\Rightarrow {{\left( \text{AC} \right)}^{2}}={{8}^{2}}+{{7}^{2}}$

$\Rightarrow {{\left( \text{AC} \right)}^{2}}=64+49$

$\Rightarrow {{\left( \text{AC} \right)}^{2}}=113$

$\Rightarrow AC=\sqrt{113}$

Now, we know that

$\sin \theta =\dfrac{\text{opposite side}}{\text{hypotenuse}}$

Here, we get

$\sin \theta =\dfrac{AB}{AC}=\dfrac{8}{\sqrt{113}}$ and

$\cos \theta =\dfrac{\text{adjacent side}}{\text{hypotenuse}}$

Here, we get

$\cos \theta =\dfrac{BC}{AC}=\dfrac{7}{\sqrt{113}}$

Now, we have to evaluate

$\dfrac{\left( 1+\sin \theta \right)\left( 1-\sin \theta \right)}{\left( 1+\cos \theta \right)\left( 1-\cos \theta \right)}$

Applying the identity $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$, we get

$\Rightarrow \dfrac{\left( 1+\sin \theta \right)\left( 1-\sin \theta \right)}{\left( 1+\cos \theta \right)\left( 1-\cos \theta \right)}=\dfrac{1-{{\sin }^{2}}\theta }{1-{{\cos }^{2}}\theta }$

Substituting the values, we get

$\Rightarrow \dfrac{\left( 1+\sin \theta \right)\left( 1-\sin \theta \right)}{\left( 1+\cos \theta \right)\left( 1-\cos \theta \right)}=\dfrac{1-{{\left( \dfrac{8}{\sqrt{113}} \right)}^{2}}}{1-{{\left( \dfrac{7}{\sqrt{113}} \right)}^{2}}}$

$\Rightarrow \dfrac{\left( 1+\sin \theta \right)\left( 1-\sin \theta \right)}{\left( 1+\cos \theta \right)\left( 1-\cos \theta \right)}=\dfrac{1-\dfrac{64}{113}}{1-\dfrac{49}{113}}$

$\Rightarrow \dfrac{\left( 1+\sin \theta \right)\left( 1-\sin \theta \right)}{\left( 1+\cos \theta \right)\left( 1-\cos \theta \right)}=\dfrac{\dfrac{113-64}{113}}{\dfrac{113-49}{113}}$

$\Rightarrow \dfrac{\left( 1+\sin \theta \right)\left( 1-\sin \theta \right)}{\left( 1+\cos \theta \right)\left( 1-\cos \theta \right)}=\dfrac{\dfrac{49}{113}}{\dfrac{64}{113}}$

$\therefore \dfrac{\left( 1+\sin \theta \right)\left( 1-\sin \theta \right)}{\left( 1+\cos \theta \right)\left( 1-\cos \theta \right)}=\dfrac{49}{64}$

ii) ${{\cot }^{2}}\theta$

Ans: Given that $\cot \theta =\dfrac{7}{8}$.

Now, ${{\cot }^{2}}\theta ={{\left( \dfrac{7}{8} \right)}^{2}}$

$\therefore {{\cot }^{2}}\theta =\dfrac{49}{64}$

8. If $3\cot A=4$, check whether $\dfrac{1-{{\tan }^{2}}A}{1+{{\tan }^{2}}A}={{\cos }^{2}}A-{{\sin }^{2}}A$ or not.

Ans: Let us consider a right angled triangle $\Delta ABC$. We get

Given that $3\cot A=4$.

We get $\cot A=\dfrac{4}{3}$.

We know that $\cot \theta =\dfrac{\text{adjacent side}}{\text{opposite side}}$.

From the above figure, we get

$\cot A=\dfrac{AB}{BC}$

Therefore, we get

$\Rightarrow BC=3$ and

$\Rightarrow AB=4$

Now, we need to apply the Pythagoras theorem to find the measure of hypotenuse.

In $\Delta ABC$, by Pythagoras theorem ,

${{\left( \text{hypotenuse} \right)}^{2}}={{\left( base \right)}^{2}}+{{\left( perpendicular \right)}^{2}}$

We get

$\Rightarrow {{\left( \text{AC} \right)}^{2}}={{\left( AB \right)}^{2}}+{{\left( BC \right)}^{2}}$

$\Rightarrow A{{C}^{2}}={{4}^{2}}+{{3}^{2}}$

$\Rightarrow A{{C}^{2}}=16+9$

$\Rightarrow A{{C}^{2}}=25$

$\Rightarrow AC=5$

Now, let us consider LHS of the expression $\dfrac{1-{{\tan }^{2}}A}{1+{{\tan }^{2}}A}={{\cos }^{2}}A-{{\sin }^{2}}A$, we get

$LHS=\dfrac{1-{{\tan }^{2}}A}{1+{{\tan }^{2}}A}$

Now, we know that $\tan \theta =\dfrac{\text{opposite side}}{\text{adjacent side}}$

Here, we get

$\tan A=\dfrac{BC}{AB}=\dfrac{3}{4}$

Substitute the value, we get

$\Rightarrow \dfrac{1-{{\tan }^{2}}A}{1+{{\tan }^{2}}A}=\dfrac{1-{{\left( \dfrac{3}{4} \right)}^{2}}}{1+{{\left( \dfrac{3}{4} \right)}^{2}}}$

$\Rightarrow \dfrac{1-{{\tan }^{2}}A}{1+{{\tan }^{2}}A}=\dfrac{1-\dfrac{9}{16}}{1+\dfrac{9}{16}}$

$\Rightarrow \dfrac{1-{{\tan }^{2}}A}{1+{{\tan }^{2}}A}=\dfrac{\dfrac{16-9}{16}}{\dfrac{16+9}{16}}$

$\Rightarrow \dfrac{1-{{\tan }^{2}}A}{1+{{\tan }^{2}}A}=\dfrac{\dfrac{7}{16}}{\dfrac{25}{16}}$

$\Rightarrow \dfrac{1-{{\tan }^{2}}A}{1+{{\tan }^{2}}A}=\dfrac{7}{25}$

Now, let us consider RHS of the expression $\dfrac{1-{{\tan }^{2}}A}{1+{{\tan }^{2}}A}={{\cos }^{2}}A-{{\sin }^{2}}A$, we get

$RHS={{\cos }^{2}}A-{{\sin }^{2}}A$

We know that $\sin \theta =\dfrac{\text{opposite side}}{\text{hypotenuse}}$ and $\cos \theta =\dfrac{\text{adjacent side}}{\text{hypotenuse}}$.

Here, we get

$\sin A=\dfrac{BC}{AC}=\dfrac{3}{5}$

And $\cos A=\dfrac{AB}{AC}=\dfrac{4}{5}$

Substitute the values, we get

$\Rightarrow {{\cos }^{2}}A-{{\sin }^{2}}A={{\left( \dfrac{4}{5} \right)}^{2}}-{{\left( \dfrac{3}{5} \right)}^{2}}$

$\Rightarrow {{\cos }^{2}}A-{{\sin }^{2}}A=\dfrac{16}{25}-\dfrac{9}{25}$

$\Rightarrow {{\cos }^{2}}A-{{\sin }^{2}}A=\dfrac{7}{25}$

Hence, we get LHS=RHS

$\therefore \dfrac{1-{{\tan }^{2}}A}{1+{{\tan }^{2}}A}={{\cos }^{2}}A-{{\sin }^{2}}A$.

9. In $ABC$, right angled at $B$. If $\tan A=\dfrac{1}{\sqrt{3}}$, find the value of

i) $\sin A\cos C+\cos A\sin C$

Ans: Let us consider a right angled triangle $\Delta ABC$. We get

Given that $\tan A=\dfrac{1}{\sqrt{3}}$.

In a right triangle, we know that $\tan \theta =\dfrac{\text{opposite side}}{\text{adjacent side}}$

Here, from the figure we get

$\tan A=\dfrac{BC}{AB}=\dfrac{1}{\sqrt{3}}$

We get $BC=1$ and $AB=\sqrt{3}$ .

Now, we need to apply the Pythagoras theorem to find the measure of hypotenuse.

In $\Delta ABC$, by Pythagoras theorem ,

${{\left( \text{hypotenuse} \right)}^{2}}={{\left( base \right)}^{2}}+{{\left( perpendicular \right)}^{2}}$

We get

$\Rightarrow {{\left( \text{AC} \right)}^{2}}={{\left( AB \right)}^{2}}+{{\left( BC \right)}^{2}}$

$\Rightarrow A{{C}^{2}}={{\left( \sqrt{3} \right)}^{2}}+{{1}^{2}}$

$\Rightarrow A{{C}^{2}}=3+1$

$\Rightarrow A{{C}^{2}}=4$

$\Rightarrow AC=2$

We know that $\sin \theta =\dfrac{\text{opposite side}}{\text{hypotenuse}}$ and $\cos \theta =\dfrac{\text{adjacent side}}{\text{hypotenuse}}$.

Here, we get

$\sin A=\dfrac{BC}{AC}=\dfrac{1}{2}$  and $\sin C=\dfrac{AB}{AC}=\dfrac{\sqrt{3}}{2}$

And $\cos A=\dfrac{AB}{AC}=\dfrac{\sqrt{3}}{2}$and $\cos C=\dfrac{BC}{AC}=\dfrac{1}{2}$

Now, we have to find the value of the expression $\sin A\cos C+\cos A\sin C$.

Substituting the values we get

$\Rightarrow \sin A\cos C+\cos A\sin C=\dfrac{1}{2}\times \dfrac{1}{2}+\dfrac{\sqrt{3}}{2}\times \dfrac{\sqrt{3}}{2}$

$\Rightarrow \sin A\cos C+\cos A\sin C=\dfrac{1}{4}+\dfrac{3}{4}$

$\Rightarrow \sin A\cos C+\cos A\sin C=\dfrac{4}{4}$

$\therefore \sin A\cos C+\cos A\sin C=1$

ii) $\cos A\cos C-\sin A\sin C$

Ans: Let us consider a right angled triangle $\Delta ABC$. We get

Given that $\tan A=\dfrac{1}{\sqrt{3}}$.

In a right triangle, we know that $\tan \theta =\dfrac{\text{opposite side}}{\text{adjacent side}}$

Here, from the figure we get

$\tan A=\dfrac{BC}{AB}=\dfrac{1}{\sqrt{3}}$

We get $BC=1$ and $AB=\sqrt{3}$ .

Now, we need to apply the Pythagoras theorem to find the measure of hypotenuse.

In $\Delta ABC$, by Pythagoras theorem ,

${{\left( \text{hypotenuse} \right)}^{2}}={{\left( base \right)}^{2}}+{{\left( perpendicular \right)}^{2}}$

We get

$\Rightarrow {{\left( \text{AC} \right)}^{2}}={{\left( AB \right)}^{2}}+{{\left( BC \right)}^{2}}$

$\Rightarrow A{{C}^{2}}={{\left( \sqrt{3} \right)}^{2}}+{{1}^{2}}$

$\Rightarrow A{{C}^{2}}=3+1$

$\Rightarrow A{{C}^{2}}=4$

$\Rightarrow AC=2$

We know that $\sin \theta =\dfrac{\text{opposite side}}{\text{hypotenuse}}$ and $\cos \theta =\dfrac{\text{adjacent side}}{\text{hypotenuse}}$.

Here, we get

$\sin A=\dfrac{BC}{AC}=\dfrac{1}{2}$  and $\sin C=\dfrac{AB}{AC}=\dfrac{\sqrt{3}}{2}$

And $\cos A=\dfrac{AB}{AC}=\dfrac{\sqrt{3}}{2}$and $\cos C=\dfrac{BC}{AC}=\dfrac{1}{2}$

Now, we have to find the value of the expression $\cos A\cos C-\sin A\sin C$.

Substituting the values we get

$\Rightarrow \cos A\cos C-\sin A\sin C=\dfrac{\sqrt{3}}{2}\times \dfrac{1}{2}-\dfrac{1}{2}\times \dfrac{\sqrt{3}}{2}$

$\Rightarrow \cos A\cos C-\sin A\sin C=\dfrac{\sqrt{3}}{4}-\dfrac{\sqrt{3}}{4}$

$\therefore \Rightarrow \cos A\cos C-\sin A\sin C=0$

10. In $\Delta PQR$, right angled at $Q$, $PR+QR=25\text{ cm}$ and $PQ=5\text{ cm}$. Determine the values of $\sin P,\cos P$ and $\tan P$.

Ans: Let us consider a right angled triangle $\Delta PQR$, we get

Given that $PR+QR=25\text{ cm}$ and $PQ=5\text{ cm}$.

Let $QR=25-PR$

Now, applying the Pythagoras theorem in $\Delta PQR$, we get

${{\left( \text{hypotenuse} \right)}^{2}}={{\left( base \right)}^{2}}+{{\left( perpendicular \right)}^{2}}$

We get

$\Rightarrow {{\left( PR \right)}^{2}}={{\left( PQ \right)}^{2}}+{{\left( QR \right)}^{2}}$

$\Rightarrow P{{R}^{2}}={{5}^{2}}+{{\left( 25-PR \right)}^{2}}$

$\Rightarrow P{{R}^{2}}=25+{{25}^{2}}+P{{R}^{2}}-50PR$

$\Rightarrow P{{R}^{2}}=P{{R}^{2}}+25+625-50PR$

$\Rightarrow 50PR=650$

$\Rightarrow PR=13\text{ cm}$

Therefore,

$QR=25-13$

$\Rightarrow QR=12\text{ cm}$

Now, we know that in right triangle,

$\sin \theta =\dfrac{\text{opposite side}}{\text{hypotenuse}}$, $\cos \theta =\dfrac{\text{adjacent side}}{\text{hypotenuse}}$ and $\tan \theta =\dfrac{\text{opposite side}}{\text{adjacent side}}$.

Here, we get

$\sin P=\dfrac{QR}{PR}$

$\therefore \sin P=\dfrac{12}{13}$

$\cos P=\dfrac{PQ}{PR}$

$\therefore \cos P=\dfrac{5}{13}$

$\tan P=\dfrac{QR}{PQ}$

$\therefore \tan P=\dfrac{12}{5}$

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

i) The value of $\tan A$ is always less than $1$.

Ans: The given statement is false. The value of $\tan A$ depends on the length of sides of a right triangle and sides of a triangle may have any measure.

ii) For some value of angle $A$, $\sec A=\dfrac{12}{5}$.

Ans: We know that in the right triangle $\sec A=\dfrac{\text{hypotenuse}}{\text{adjacent side of }\angle \text{A}}$ .

We know that in the right triangle the hypotenuse is the largest side.

Therefore, the value of $\sec A$ must be greater than $1$.

In the given statement $\sec A=\dfrac{12}{5}$, which is greater than $1$.

Therefore, the given statement is true.

iii) $\cos A$ is the abbreviation used for the cosecant of angle $A$.

Ans: The given statement is false because $\cos A$ is the abbreviation used for the cosine of angle $A$. Abbreviation used for the cosecant of angle $A$ is $\operatorname{cosec}A$.

iv) $\cot A$ is the product of $\cot$ and $A$.

Ans: $\cot A$ is the abbreviation used for the cotangent of angle $A$. Hence the given statement is false.

v) For some angle $\theta$, $\sin \theta =\dfrac{4}{3}$.

Ans: We know that in the right triangle $\sin \theta =\dfrac{\text{opposite side}}{\text{hypotenuse}}$ .

We know that in the right triangle the hypotenuse is the largest side.

Therefore, the value of $\sin \theta$ must be less than $1$.

In the given statement $\sin \theta =\dfrac{4}{3}$, which is greater than $1$.

Therefore, the given statement is false.

## Conclusion

Class 10 Exercise 8.1 of Maths Chapter 8 -Introduction to Trigonometry, is crucial for a solid foundation in math. This exercise introduces the students to the fundamentals of trigonometry. Chapter 8 Maths Class 10 Ex 8.1 NCERT solutions deals with the types of problems like determining the trigonometric ratios given the sides of the triangle and finding the other trigonometric ratios given one of them by constructing a triangle of the respective ratio.

## Class 10 Maths Chapter 8: Exercises Breakdown

 Exercise Number of Questions Exercise 8.2 4 Questions & Solutions Exercise 8.3 4 Questions & Solutions

## CBSE Class 10 Maths Chapter 8 Other Study Materials

 S.No. Important Links for Chapter 8 Introduction to Trigonometry 1 Class 10 Introduction To Trigonometry Revision Notes 2 Class 10 Introduction To Trigonometry Important Questions 3 Class 10 Introduction To Trigonometry Formula 4 Class 10 Introduction To Trigonometry NCERT Exemplar Solution

## Chapter-Specific NCERT Solutions for Class 10 Maths

Given below are the chapter-wise NCERT Solutions for Class 10 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

## FAQs on NCERT Solutions For Class 10 Maths Chapter 8 Exercise 8.1 - Introduction to Trigonometry

1. What are the formulas to calculate the trigonometric ratio of the angles?

Consider a right-angled triangle of an acute angle A

sin A = Side opposite to angle A / Hypotenuse

cos A = Side adjacent to angle A / Hypotenuse

tan A = Side opposite to angle A / Side adjacent to angle A

cosec A = Hypotenuse / Side opposite to angle A

sec A = Hypotenuse / Side adjacent to angle A

cot A = Side adjacent to angle A / Side opposite to angle A

are the formulas for calculating the trigonometric ratio of the angles.

2. Why is the Class 10 Trigonometry Exercise 8.1 important?

Class 10 Trigonometry Exercise 8.1 covers the fundamentals of the vast branch of mathematics, namely trigonometry. This concept is crucial for understanding the concepts of the student's higher education irrespective of the stream they choose. These concepts are explained and are broken down to their simplest form in Exercise 8.1 Class 10 Maths Ncert solutions. Learning these concepts by heart helps the students score significantly high in their examinations. As the CBSE Board question paper gives a pretty high weightage to this chapter, it's easy to score well once the concepts are well-learned.

3. How Vedantu NCERT solutions help in examination times?

NCERT solutions of Vedantu help students to frame mind maps of the type of problems that they may encounter in their examination. The mentors of Vedantu train the students to face the exams with confidence. Easier learning of concepts lessens the preparation time, which in turn reflects in the student's Examination tactics. Knowing their strengths and weaknesses helps the student to plan their methodology to attend to the questions with the time constraint in mind.

4. What are the topics discussed in exercise 8.1 ?

Chapter 8 in Class 10 Maths is trigonometry. Trigonometry studies ratios and angles of triangles. Exercise 8.1 of the textbook focuses on the basics of trigonometry. It introduces them to questions with acute angles. It forms a basis for topics that would be discussed in further exercises and classes. It is a fairly easy exercise. The students can find step by step solutions in the solutions PDF provided by Vedantu online on the website.

5. How many questions does Exercise 8.1 have?

Exercise 8.1 is the first exercise in Chapter 8 of Class 10 Maths. The chapter introduces the basics and fundamentals of trigonometry. It is an important topic that is studied in classes ahead as well. There are 11 questions in the exercise discussing ratios and angels. The solutions for all questions can be found in the solution pdf for the reference of students. They are explained in an easy language for the students to understand better.

6. Why is Chapter 8 of Class 10 Maths important?

Chapter 8 of Class 10 is called trigonometry. It introduces students to the basics of the topic and aims at helping them form a strong base. This topic is important as it is discussed in further classes and in colleges as well depending on the subjects chosen by the student. It is a topic that is discussed in daily life as well. The topic requires the concentration of the students and regular practice. The students can score highly in it if they practice enough.

7. How many exercises are there in chapter 8 Trigonometry?

There are a total of four exercises in chapter 8- Exercise 8.1, 8.2, 8.3 and 8.4. The exercises cover all the topics and formulas discussed throughout the chapter. Each exercise covers a few particular topics and formulas. The exercises aim at making the student comfortable and well-versed with the topic. The students can refer to the solutions PDF for answers and extra questions for Trigonometry. Answers to the NCERT textbook exercises are explained step by step in easy language.

8. How should I prepare chapter 8 for the final exam?

Trigonometry requires the students to be thorough and regular with the topic and all the questions in the NCERT textbook. They should also practice extra questions from the PDF to form a stronger base on the subject. Students should regularly revise what they are being taught in school. They should also write down all the formulas so that it is easy for them to revise and they can retain them better. If they understand the topic they will be able to score good marks in the subject.

9. What is the formula for trigonometry 8.1 Class 10?

Here are the key formulas used in this exercise:

• Trigonometric Ratios:

• $\sin \theta = \dfrac{\text{Opposite}}{\text{Hypotenuse}}$

• $\cos \theta = \dfrac{\text{Adjacent}}{\text{Hypotenuse}}$

• $\tan \theta = \dfrac{\text{Opposite}}{\text{Adjacent}}$

10. Who invented trigonometry?

The father of trigonometry is Hipparchus