## Algebraic solution of linear equations problems that are not easy

Challenging linear equations problems for class 10 solved by linear equations solving methods and problem solving techniques. NCERT math Ex. 3.7 solved.

In this seventh and last part of NCERT Solutions for Chapter 3, you'll** **

Learn how to solve the clallenging set of linear equations problems in exercise 3.7 quickly by standard methods and also by not so well-known new problem solving techniques.

### Solution to challenging linear equations problems in Exercise 3.7 of NCERT book on Class 10 maths

#### Problem1.

The ages of two friends Ani and Biju differ by 3 years. Ani's father Dharam is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam differ by 30 years. Find the ages of Ani and Biju.

#### Problem 2.

One says, "Give me a hundred, friend! I shall then become twice as rich as you". The other replies, "If you give me ten, I shall be six times as rich as you". Tell me what are the amounts of their (respective) capital.

#### Problem 3.

A train covered a certain distance at a uniform speed. If the train would have been 10km/h faster, it would have taken two hours less than the scheduled time. And, if the train were slower by 10km/h, it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.

#### Problem 4.

The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be two rows more. Find the number of students in the class.

#### Problem 5.

In a $\triangle ABC$, $\angle C=3\angle B=2(\angle A+\angle B)$. Find the three angles.

#### Problem 6.

Draw the graphs of the equations, $5x-y=5$ and $3x-y=3$. Determine the coordinates of the vertices of the triangle formed by these lines and $y$ axis.

#### Problem 7.

Solve the following pair of linear equations.

- $px+qy=p-q$ and $qx-py=p+q$
- $ax+by=c$ and $bx+cy=1+c$
- $\displaystyle\frac{x}{a}-\displaystyle\frac{y}{b}=0$ and $ax+by=a^2+b^2$
- $(a-b)x+(a+b)y=a^2-2ab-b^2$ and $(a+b)(x+y)=a^2+b^2$
- $152x-378y=-74$ and $-378x+152y=-604$

#### Problem 8.

$\text{ABCD}$ is a cyclic quadrilateral as in the figure below. Find the angles of the cyclic quadrilateral.

### Solution to the problems

#### Problem 1.

The ages of two friends Ani and Biju differ by 3 years. Ani's father Dharam is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam differ by 30 years. Find the ages of Ani and Biju.

#### Solution to Problem 1.

Let the ages of Ani and Biju be $x$ and $y$ and ages of Dharam and Cathy be $a$ and $b$, all variables in years.

Our first objective is to **form a pair of linear equations in $x$ and $y$ as these are the values that you have to find out.**

At this point, you *can't be sure about who among Ani and Biju is older*, but it is certain that Ani's father Dharam is older than Cathy.

Let's then form the equations we are certain of.

As Dharam is twice as old as Ani,

$a=2x$.

Similarly, as Biju is twice as old as his sister Cathy,

$y=2b$,

Or, $b=\frac{1}{2}y$.

We have obtained $a$ and $b$ in terms of $x$ and $y$ respectively that would be substituted in the age difference equation for Dharam and Cathy in the next step.

As Dharam and Cathy differ by 30 years in their ages,

$a-b=30$.

Substitute values of $a$ and $b$ in terms of $x$ and $y$ respectively,

$2x-\frac{1}{2}y=30$,

Or, $x-\frac{1}{4}y=15$.

This is the first linear equation in $x$ and $y$.

Variables $a$ and $b$ will no longer be required to solve the problem. We may call these as **Intermediary or Helper variables.**

Even at this late stage you are not certain who among Ani and Biju is older.

**First assume** Ani is older than Biju by 3 years. The age difference equation between Ani and Biju is then,

$x-y=3$.

This is the *first possibility of second linear equation.*

Subtract this second equation from the first to eliminate $x$. The result is,

$\frac{3}{4}y=12$,

Or, $y=16$.

Substitute this in second simpler equation to get Ani's age as,

$x=16+3=19$.

**First answer:** Ages of Ani and Biju are 19 years and 16 years.

**Verification:** Dharam being twice as old as Ani, his age is 38 years. Biju being twice as old as Cathy, her age is 8 years. That makes the difference of ages of Dharam and Cathy as 30 years. **All conditions satisfied.**

**Second possibility of Biju older than Ani by 3 years**

In this case the age difference of Ani and Biju is,

$y-x=3$.

Add this now with the first linear equation $x-\frac{1}{4}y=15$ to get,

$\frac{3}{4}y=18$,

Or, $y=24$.

As Biju is older than Ani by 3 years, age of Ani is, $24-3=21$ years.

**Second answer:** Ages of Ani and Biju are respectively 21 years and 24 years.

**Verification:** Being twice as old as Ani, Dharam is 42 years old. Biju being twice as old as his sister Cathy, age of Cathy is half the age of Biju, that is, 12 years.

Difference of ages of Dharam and Cathy is $42-12=30$ years. **All conditions satisfied.**

**Note:** If you are not aware of the uncertainty in forming the second equation, you will miss one of the two possible solutions. This is **exhaustive enumeration of possibilities.**

#### Problem 2.

One says, "Give me a hundred, friend! I shall then become twice as rich as you". The other replies, "If you give me ten, I shall be six times as rich as you". Tell me what are the amounts of their (respective) capital.

#### Solution to Problem 2.

Let us assume the amounts of Capital of the first and second person be $x$ and $y$.

By the first statement then,

$x+100=2(y-100)$,

Or, $2y-x=300$.

This is the first linear equation.

By the second statement,

$y+10=6(x-10)$,

Or, $6x-y=70$.

This is the second linear equation.

Care should be taken to visualize what exactly happens by each of the transfers of money.

Multiply the second equation by 2. Result is,

$12x-2y=140$.

Add this result to first equation to eliminate $y$. Result is,

$11x=440$,

So, $x=40$.

Substitute this value in simpler first equation. Result is,

$2y-40=300$,

Or, $2y=340$,

So, $y=170$.

**Answer:** Capital of first person is Rs.40 and that of second person is Rs.170.

**Verification:**

**First transaction:** Second person gives Rs.100 to the first thus depleting his capital to Rs.70, but capital of first person increases to Rs.140, exactly double as that of the **present capital of the second person.**

This is the **crucial point to remember**—*in the relationships between the two capitals after transaction*, **the two changed values after the transaction must be considered.**

**Second transaction:** First person gives Rs.10 to the second thus depleting his own capital to Rs.30 but increasing the capital of the second person to Rs.180 which is exactly six times of the present capital of the first.

**All conditions having satisfied,** the solution is certainly correct.

**Note:** Awareness of exactly what happens in a problem is called **Event mapping.**

#### Problem 3.

A train covered a certain distance at a uniform speed. If the train would have been 10km/h faster, it would have taken two hours less than the scheduled time. And, if the train were slower by 10km/h, it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.

#### Solution to Problem 3.

Speed time distance relation is given by,

$\text{Distance covered }=\text{ Speed }\times{\text{ Travel time}}$.

Let the distance the train covers be $d$ km, the normal uniform speed as $s$ km/hour, and the scheduled time to cover the distance be $t$ hours.

So in the normal case,

$d=st$.

When the train runs faster at speed $(s+10)$ km/hour, it takes a shorter time of $(t-2)$ hours to cover distance $d$ at this speed. So,

$d=st=(s+10)(t-2)$,

Or, $st=st-2s+10t-20$,

Or, $-2s+10t=20$,

Or, $-s+5t=10$.

This is your first linear equation in $s$ and $t$.

When the train runs slower at speed $(s-10)$ km/hour, it takes a longer time of $(t+3)$ hours to cover distance $d$ at this speed. So,

$d=st=(s-10)(t+3)$,

Or, $st=st+3s-10t-30$,

Or, $3s-10t=30$.

This is the second linear equation in $s$ and $t$.

Multiply the first linear equation by 3 and add the result with the second linear equation to eliminate $s$ and get,

$5t=60$,

Or, $t=12$ hours.

Substituting the value of $t$ in the first linear equation, $-s+5t=10$,

$s=60-10=50$ km/hour.

Finally then,

Distance, $d=st=50\times{12}=600$ km.

**Answer:** The distance covered is 600 km.

#### Problem 4.

The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be two rows more. Find the number of students in the class.

#### Solution to Problem 4.

Let the number of rows and columns in the normal arrangement be $r$ and $c$.

Total number of students is then,

$N=rc$.

With 3 extra students in each row, number of columns increases to $(c+3)$ and number of rows decreases by 1 to $(r-1)$. Total number of students is given by,

$N=rc=(c+3)(r-1)=rc-c+3r-3$,

Or, $3r-c=3$.

This is the first linear equation.

With 3 less students in each row, the number of columns decreases to $(c-3)$ and number of rows increases by 2 to $(r+2)$. Total number of students is given by,

$N=rc=(c-3)(r+2)=rc+2c-3r-6$,

Or, $3r-2c=-6$.

This is the second linear equation.

Subtract second equation from first to cancel out $r$. Result of addition is,

$c=9$.

Substitute this value of $c$ in simpler first equation. Result is,

$3r-9=3$,

Or, $3r=12$,

Or, $r=4$

So, total number of students, $4\times{9}=36$.

**Answer:** Total number of students is 36.

**Verification:**

In normal arrangement 36 students are arranged standing in 4 rows and 9 columns.

When number of students in each row increases by 3, effectively number of columns increases by 3 to 12 and number of rows decreases by 1 to 3. The total number of students still remains same at $3\times{12}=36$.

When number of students in each row decreases by 3, number of columns as an effect reduces by 3 to 6, and number of rows increases by 2 and becomes 6. Total number of students still remains same at $6\times{6}=36$.

**All conditions satisfied.**

#### Visualization in terms of Areas of rectangles—domain mapping

The arrangements in the form of rows and columns can be imagined as similar to rectangles with Lengths as Number of columns and Breadths as Number of Rows.

The following graphic captures the three stages of the problem—normal starting stage (rows:4, columns:9), first changed stage (rows:3, columns:12) and second changed stage (rows:6, columns:6).

This way of representing the essence of a problem in one topic domain onto another topic domain is called **domain mapping**.

By this mapping of the problem onto Planar Geometry domain, **Columns, Rows and number of students are equivalent respectively to Lengths, Breadths and Areas of Rectangles.**

If you keep the corresponding values of variables same, all analysis and results would also be same.

**First advantage of this approach** is, **ability to visualize.**

**Second advantage** is the visual help to *understand the powerful new method of problem solving by change analysis.*

#### The powerful new approach of solution by Change analysis technique

Look at the graphic again.

The original area (or total number of students) has been rectangle ABCD with length (or number of columns $c$) 9 units and breadth (or number of rows $r$) 4 units. We have to find these values by solving the problem. The two unknown variables are $c$ (equivalent to length of rectangle) and $r$ (equivalent to breadth of rectangle).

By the first change, the rectangle ABCD is transformed to another rectangle EBGF of same area (36 units). But length increases by 3 units to $(c+3)$ and breadth reduces by 1 unit to $(r-1)$. *As the total area remains unchanged*, the **reduction in area represented by rectangle AEHD must be equal to the increase in area represented by the rectangle HCGF.**

So **equating the two changes in area** you get,

$c\times{1}=3\times{(r-1)}$,

Or, $3r-c=3$.

You have got the first linear equation in $r$ and $c$ practically in a single step.

Similarly by the second change in rows and columns (or breadth and length), ABCD changes to rectangle JBNK.

Increase in area is represented by area of rectangle JALK. This is,

$2\times(c-3)$.

And the decrease in area is represented by area of rectangle, LNCD. This is,

$3r$.

Equating the two changes you get,

$2\times(c-3)=3r$,

So, $3r-2c=-6$.

Subtract the second from the first as before to get the value of $c=9$, and then substituting it to first equation get the value of $r$ as,

$3r-9=3$,

Or, $r=4$.

**Solution in least number of steps and time.**

**Saving in time and number of steps is due to bypassing dealing with WHOLE AREAS. Instead we have equated only the changes in the area.**

This is what we call **Problem solver's approach.**

Domain mappingandChange analysis techniqueare two of the more powerful generalProblem Solving Resourcesthat can be applied to solve varieties of problems in different topic or subject domains resulting in more innovative solutions at less cost and time.

#### Problem 5.

In a $\triangle ABC$, $\angle C=3\angle B=2(\angle A+\angle B)$. Find the three angles.

#### Solution to Problem 5.

The given chained equation is formed by joining two distinct equations,

$\angle C=3\angle B$, and

$3\angle B=2(\angle A +\angle B)$,

Or, $\angle B=2\angle A$.

These are two distinct equations in three variables. To evaluate the three angles you need one more distinct equation in the three angle variables.

This third relation you can form from the property of a triangle that sum of three angles is $180^0$,

$\angle A+\angle B +\angle C=180^0$.

Substitute in this equation, value of $\angle C=3\angle B$ and $\angle A=\frac{1}{2}\angle B$,

$(\frac{1}{2}+1+3)\angle B=180^0$,

Or, $\displaystyle\frac{9}{2}\angle B=180^0$,

So, $\angle B=40^0$,

$\angle C=3\times{40^0}=120^0$, and,

$\angle A=\frac{1}{2}\angle B=20^0$.

The sum of three angles is, $180^0$.

**Answer:** $\angle A=20^0$, $\angle B=40^0$ and $\angle C=120^0$.

#### Problem 6.

Draw the graphs of the equations, $5x-y=5$ and $3x-y=3$. Determine the coordinates of the vertices of the triangle formed by these lines and $y$ axis.

#### Solution to Problem 6.

To plot a linear equation on $(x-y)$ coordinate system, you need two pairs of coordinate values of two points on the straight line that represents the equation.

Often, the easiest way to find the two points is to put $y=0$ in the equation to get the value of $x$ and so the coordinates of the point of intersection of the line with the $x\text{-axis}$, and similarly put $x=0$ in the equation to get the value of $y$ as the coordinates of the point of intersection of the line with $y\text{-axis}$.

Following this approach, get the coordinates for the two points of intersection of the first line with $y\text{-axis}$ (for $x=0$) and $x\text{-axis}$ (for $y=0$) as, $\text{A (0, -5)}$ and $\text{B (1, 0)}$. Joining the two points and extending in both directions outwards, the line corresponding to the equation, $5x-y=5$ is formed.

In the same way, for the second equation, $3x-y=3$, the points of intersection with the $y\text{-axis}$ (for $x=0$) and $x\text{-axis}$ (for $y=0$) are, $\text{C (0, -3)}$ and the same previous point $\text{B (1, 0)}$.

$\text{B (1, 0)}$ is the point of intersection of the two lines and $\text{A (0, -5)}$ and $\text{C (0, -3)}$ are the other two vertices of the triangle formed by the point of intersection and $y\text{-axis}$.

The corresponding graphic is shown below.

**Answer:** The three vertices of the triangle are, A (0, -5), B (1, 0) and C (0, -3).

#### Problem 7.i

Solve the following pair of linear equations.

- $px+qy=p-q$ and $qx-py=p+q$

#### Solution to Problem 7.i.

The two equations are,

$px+qy=p-q$,

Or, $p(x-1)+q(y+1)=0$, and,

$qx-py=p+q$,

Or, $q(x-1)-p(y+1)=0$.

These are two linear equations, but in two compound variables, $(x-1)$ and $(y+1)$.

Multiply the first equation by $p$ to transform it to,

$p^2(x-1)+pq(y+1)=0$.

And multiply the second equation by $q$ to change it to,

$q^2(x-1)-pq(y+1)=0$.

Add the two to eliminate $(y+1)$. Result of addition is,

$(p^2+q^2)(x-1)=0$.

The first factor being a sum of squares, it cannot be zero.

So,

$(x-1)=0$,

Or, $x=1$.

Substitute this value in the first equation, $p(x-1)+q(y+1)=0$. Result is,

$q(y+1)=0$,

Or, $y=-1$.

**Answer:** $x=1$, $y=-1$.

**Verification:** Substitute in the original two equations to verify correctness of the solution,

$p-q=p-q$, and

$p+q=p+q$.

#### Problem 7.ii.

Solve the following pair of linear equations.

- $ax+by=c$ and $bx+ay=1+c$

#### Solution to Problem 7.ii.

The two equations are,

$ax+by=c$,

Or, $ax+by-c=0$, and,

$bx+ay=1+c$,

Or, $bx+ay-(c+1)=0$.

In the cross-multiplication solution scheme, form the chained equation as,

$\displaystyle\frac{x}{-b(c+1)+ca}=\frac{y}{-bc+a(c+1)}=\frac{1}{a^2-b^2}$.

So,

$x=\displaystyle\frac{c(a-b)-b}{a^2-b^2}$, and

$y=\displaystyle\frac{c(a-b)+a}{a^2-b^2}$.

**Answer:** $x=\displaystyle\frac{c(a-b)-b}{a^2-b^2}$, and $y=\displaystyle\frac{c(a-b)+a}{a^2-b^2}$.

#### Problem 7.iii.

Solve the following pair of linear equations.

- $\displaystyle\frac{x}{a}-\displaystyle\frac{y}{b}=0$ and $ax+by=a^2+b^2$.

#### Solution to Problem 7.iii.

The two equations are,

$\displaystyle\frac{x}{a}-\displaystyle\frac{y}{b}=0$, and,

$ax+by=a^2+b^2$,

Or, $ax+by-(a^2+b^2)=0$.

Form the chained equation by cross-multiplication scheme as,

$\displaystyle\frac{x}{\displaystyle\frac{a^2+b^2}{b}}=\frac{y}{\displaystyle\frac{a^2+b^2}{a}}=\frac{1}{\displaystyle\frac{b}{a}+\displaystyle\frac{a}{b}}=\frac{ab}{a^2+b^2}$.

So the values for $x$ and $y$ are,

$x=\displaystyle\frac{a^2+b^2}{b}\times{\displaystyle\frac{ab}{a^2+b^2}}=a$, and

$y=\displaystyle\frac{a^2+b^2}{a}\times{\displaystyle\frac{ab}{a^2+b^2}}=b$.

**Answer:** $x=a$, and $y=b$.

**Verification:** Substitute the two values of $x$ and $y$ in the two original equations to verify,

$1-1=0$, and,

$a^2+b^2=a^2+b^2$.

**Solution verified.**

#### Problem 7.iv.

Solve the following pair of linear equations.

- $(a-b)x+(a+b)y=a^2-2ab-b^2$ and $(a+b)(x+y)=a^2+b^2$

#### Solution to Problem 7.iv.

The two equations are,

$(a-b)x+(a+b)y=a^2-2ab-b^2$ and,

$(a+b)(x+y)=a^2+b^2$,

Or, $(a+b)x+(a+b)y=a^2+b^2$.

Subtract the second equation from the first to eliminate $y$. Result is,

$-2bx=-2ab-2b^2=-2b(a+b)$,

Or, $x=a+b$.

Substitute value of $x$ in second equation. Result is,

$(a+b)^2+(a+b)y=a^2+b^2$,

Or, $y=-\displaystyle\frac{2ab}{a+b}$.

**Answer:** $x=a+b$, and $y=-\displaystyle\frac{2ab}{a+b}$.

**Verification:** Substitute the values of $x$ and $y$ in the two original equations. The results are,

$a^2-b^2-2ab=a^2-2ab-b^2$, and,

$(a+b)\left(a+b-\displaystyle\frac{2ab}{a+b}\right)=\displaystyle\frac{(a+b)(a^2+b^2)}{a+b}=a^2+b^2$.

**Conditions satisfied.**

#### Problem 7.v.

Solve the following pair of linear equations.

- $152x-378y=-74$ and $-378x+152y=-604$

#### Solution to Problem 7.v.

The two equations are,

$152x-378y=-74$ and,

$-378x+152y=-604$.

**First key pattern discovered:** Absolute values of the coefficients of $x$ and $y$ in the first equation are interchanged in the second equation. This pattern must be utilized for a quick solution.

**Second belief created from experience:** There must be a quick solution. No problem in School tests (or Competitive tests) should ask you to carry out such humongous arithmetic calculations needed for elimination or cross-multiplication.

Based on these two, when the values of coefficients and the constants are examined more deeply **the second key pattern** could immediately be discovered.

**Let's show you the effect of the second key pattern** rather than explaining.

Subtract the second equation from the first. Both the pairs of coefficients of $x$ and $y$ will be added to become 530 in absolute. And most favorably, the result of subtraction of the two constants also becomes 530. Let's show you.

The result of subtraction is,

$530x-530y=530$.

Or, $x-y=1$, no large number in any parameter as common factor 530 is eliminated from all three parameters,

Or, $y=x-1$.

Substitute this simple expression in the first equation $152x-378y=-74$. Result is,

$152x-378(x-1)=-74$,

Or, $-226x=-452$,

Or $x=2$, and $y=x-1=1$.

**Answer:** $x=2$, $y=1$.

**Verification:** Substitute values of $x$ and $y$ to the two original equations. Results are,

$152\times{2}-378=304-378=-74$, and

$-378\times{2}+152=-756+152=-604$.

**Solution verified.**

This is a **Problem solver's solution** based on **pattern discovery** and associated methods.

**Comment:** This is a beautifully engineered problem with hidden patterns (two patterns) for quick and elegant solution.

#### Problem 8.

$\text{ABCD}$ is a cyclic quadrilateral as in the figure below. Find the angles of the cyclic quadrilateral.

#### Solution to Problem 8.

By the property of a cyclic quadrilateral inscribed in a circle,

Sum of opposite angles is $180^0$.

Sum of $\angle A$ and $\angle C$ is then,

$\angle A+\angle C=4y+20-4x=180$,

Or, $4(y-x)=160$,

Or, $y-x=40$.

This is the first linear equation in $x$ and $y$.

Applying the same property on the two remaining angles,

$\angle B+\angle D=3y-5+(-7x+5)=180$,

Or, $3y-7x=180$.

This is the second linear equation in $x$ and $y$.

We'll multiply first equation by 3 and subtract this result from the second equation to eliminate $y$. The result of multiplying first equation by 3 is,

$3y-3x=120$.

Subtract this result from second equation. Result is,

$-4x=60$,

Or, $x=-15$.

Substitute value of $x$ in the simpler first equation. Result is,

$y-x=y+15=40$,

So, $y=25$.

Now we can get the values of the four angles as,

$\angle A=4y+20=4\times{25}+20=120^0$,

$\angle B=3y-5=3\times{25}-5=70^0$,

$\angle C=-4x=(-4)\times{(-15)}=60^0$, and

$\angle D=-7x+5=(-7)\times{(-15)}+5=110^0$.

**Answer:** $\angle A=120^0$, $\angle B=70^0$, $\angle C=60^0$, and $\angle D=110^0$.

**Verify** that the sum of opposite angles in both cases are $180^0$.

### NCERT Solutions for Class 10 Maths

#### Chapter 1: Real Numbers

**NCERT Solutions for Class 10 Maths on Real numbers part 1, Euclid’s division lemma puzzle solutions**

#### Chapter 2: Polynomials

#### Chapter 3: Linear Equations

**NCERT solutions for class 10 maths Chapter 3 Linear equations 7 Problem Collection**

**NCERT solutions for class 10 maths Chapter 3 Linear equations 6 Reducing non-linear to linear form**

**NCERT solutions for class 10 maths Chapter 3 Linear Equations 4 Algebraic solution by Elimination**

**NCERT solutions for class 10 maths Chapter 3 Linear Equations 3 Algebraic solution by Substitution**

**NCERT solutions for class 10 maths Chapter 3 Linear Equations 2 Graphical solutions**

**NCERT solutions for class 10 maths Chapter 3 Linear Equations 1 Graphical representation.**

#### Chapter 4: Quadratic equations

**NCERT solutions for class 10 maths Chapter 4 Quadratic Equations 1 What are quadratic equations**

**NCERT solutions for class 10 maths Chapter 4 Quadratic Equations 2 Solving by factorization**

**NCERT solutions for class 10 maths Chapter 4 Quadratic Equations 3 Solution by Completing the square**

#### Chapter 6: Triangles

**NCERT solutions for class 10 maths chapter 6 Triangles 1 Similarity of Triangles and Polygons**

**Solutions to Exercise 2 Chapter 6 NCERT X Maths, Characteristics of Similar triangles**

#### Chapter 8: Introduction to Trigonometry, Concepts and solutions to exercise problems

*NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry 1 Trigonometric Ratios*

*NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry 4 Trigonometric identities*

#### Chapter 8: Introduction to Trigonometry, only solutions to selected problems

**NCERT Solutions for Class 10 Maths on Trigonometry, solution set 6**

**NCERT Solutions for Class 10 Maths on Trigonometry, solution set 5**

**NCERT Solutions for Class 10 Maths on Trigonometry, solution set 4**

**NCERT Solutions for Class 10 Maths on Trigonometry, solution set 3**

**NCERT Solutions for Class 10 Maths on Trigonometry, solution set 2**

**NCERT Solutions for Class 10 Maths on Trigonometry, solution set 1**