# Exercise E6.2 Lines and Angles NCERT Solutions Class 9

Exercise 6.2

## Question 1

In the given figure, find the values of \(x\) and \(y\) and then show that \(AB ‖ CD\).

### Solution

**Video Solution**

**Reasoning:**

- When two lines intersect, vertically opposite angles formed are equal.
- Also, when a ray intersects a line sum of adjacent angles formed is \(180^ {\circ}\).
- If a transversal intersects two lines such that a pair of alternate angles is equal, then the two lines are parallel to each other.

**Steps:**

Line \(CD\) is intersected with line \(P\), hence the vertically opposite angles so formed are equal. \(y = 130^ {\circ}.\)

Similarly, line \(AB\) is intersected by line \(P\) hence the sum of adjacent angles formed is \(180^ {\circ}.\)

\[\begin{align} x + 50 ^ { \circ } & = 180 ^ { \circ } \\ x & = 180 ^ { \circ } - 50 ^ { \circ } \\ & = 130 ^ { \circ } \end{align}\]

We know that, if a transversal intersects two lines such that a pair of alternate interior angles is equal, then the two lines are parallel. Here we can see that the pair of alternate angles formed when lines \(AB\) and \(CD\) are intersected by transversal \(P\) are equal. Therefore, \(x = y = 130^ {\circ}.\) So we can say the two lines \(AB\) and \(CD\) are parallel. Hence \(AB ‖ CD\) is proved.

## Question 2

In the given figure, if \(AB\, ‖ \,CD\,,\; CD\, ‖ \,EF\) and \(y:z = 3:7\), find \(x\).

### Solution

**Video Solution**

**What is known?**

\(AB ‖ CD\)

\(CD ‖ EF\)

\(y:z = 3:7\) implies \(y = 3a\) and \(z = 7a\)

**What is unknown?**

Value of \(x.\)

**Reasoning:**

- Lines which are parallel to the same line are parallel to each other.
- When two parallel lines are cut by a transversal, co-interior angles formed are supplementary.

**Steps:**

We know that, lines which are parallel to the same line are parallel to each other.

If \(AB ‖ CD\), \(CD ‖ EF\), we can say that \(AB ‖ EF\).

Therefore, the angles \(x\) and \(z\) are alternate interior angles and hence are equal.

\[x = z \qquad\ldots \ldots \ldots (1)\]

\(AB\) and \(CD\) are parallel lines cut by transversal. So the co-interior angles formed are supplementary.

\[x + y =180^ {\circ}\]

\(\text{Since}\ x = z\), we get \(y + z =180^ {\circ}\)

Given \(y = 3a, \;z = 7a\)

\[\begin{align} 3 a + 7 a &= 180 ^ { \circ } \\ 10 a &= 180 ^ { \circ } \\ a &= \frac { 180^ { \circ } } { 10 } \\ a &= 18 ^ { \circ } \end{align}\]

\[\begin{align} \therefore \qquad y & = 3 a \\ y &= 3 \times 18 ^ { \circ } \\ y &= 54 ^ { \circ } \end{align}\]

\[\begin{align} \therefore \qquad x + y &= 180 ^ { \circ } \\ x + 54 ^ { \circ } &= 180 ^ { \circ } \\ x &= (180 ^ { \circ } - 54 ^ { \circ }) \\ x &= 126 ^ { \circ } \end{align}\]

## Question 3

In the given figure, if \(AB \,‖\, CD\), \(EF \,\bot \,CD\) and \(\angle GED = 126^ {\circ},\) find \(\angle AGE, \; \angle GEF\) and \(\angle FGE.\)

### Solution

**Video Solution**

**What is known?**

\(AB ‖ CD\), \(\text{EF} \bot \text{CD}\) and \(\angle GED = 126^ {\circ},\)

**What is unknown?**

\(\angle AGE =?,\;\; \angle GEF=?\) and \(\angle FGE =?\)

**Reasoning:**

- When two lines intersect, adjacent angles formed are supplementary.
- When two parallel lines are cut by a transversal, alternate interior angles formed are equal.

**Steps:**

Let \( \angle AGE = x,\;\; \angle GED= y\) and \( \angle FGE = z\).

From the figure, we can see that,

\[\begin{align} \angle GED& = \angle GEF + \angle FED \\ y & = (126 ^ { \circ } - 90 ^ { \circ }) \\ \angle GEF & = y = 36 ^ { \circ } \end{align}\]

\(AB\) and \(CD\) are parallel lines cut by a transversal, the pair of alternative angles formed are equal.

\[\begin{align} \angle AGE &= \angle GED \\ \angle AGE &= x = 126 ^ { \circ } \end{align}\]

Line \(AB\) is intersected by line \(GE\) hence adjacent angles formed are supplementary.

\[\begin{align} x + z &= 180 ^ { \circ } \\ 126 ^ { \circ } + z &= 180 ^ { \circ } \\ z &= 180 ^ { \circ } - 126 ^ {\circ } \\ \quad &= 54 ^ {\circ} \\ \angle FGE = z &= 54 ^ { \circ } \end{align}\]

## Question 4

In the given figure, if \(PQ || ST\), \(\angle PQR = 110^ \circ\) and \(\angle RST = 130^ {\circ}\), find \(\angle QRS\).

### Solution

**Video Solution**

**What is known?**

\(PQ \| ST, \angle PQR = 110 ^ { \circ }\) and \( \angle RST = 130 ^ { \circ }\)

**What is unknown?**

\(\angle QRS =?\)

**Reasoning:**

- Lines which are parallel to the same line are parallel to each other.
- When two parallel lines are cut a transversal, co-interior angles formed are supplementary.

**Steps:**

Draw a line \(AB\) parallel to \(ST\) through point \(R\). Since \(AB\|ST\) and \(PQ\|ST\). So, \(AB\|PQ\).

Let \(\angle SRQ = x,\;\; \angle SRB = y\) and \(\angle QRA = z\)

Lines \(ST\) and \(AB\) are parallel with transversal \(SR\) intersecting. Therefore, the co-interior angles are supplementary.

\[\begin{align} \angle RST + \angle SRB & = 180 ^ { \circ } \\ 130 ^ { \circ } + y & = 180 ^ { \circ } \\ y & = (180 ^ { \circ } - 130 ^ { \circ }) \\ & = 50 ^ { \circ } \\ \angle SRB = y & = 50 ^ { \circ } \end{align}\]

Similarly, lines \(PQ\) and \(AB\) are parallel with transversal \(QR\) intersecting. Therefore, the co-interior angles are supplementary.

\[\begin{align} \angle PQR + \angle QRA & = 180 ^ { \circ } \\ 110 ^ { \circ } + z & = 180 ^ { \circ } \\ z & = (180 ^ { \circ } - 110 ^ { \circ }) \\ & = 70 ^ { \circ } \\ \angle QRA = z & = 70 ^ { \circ } \end{align}\]

\(AB\) is a line, \(RQ\) and \(RS\) are rays on \(AB\). Hence,

\[\begin{align} \angle QRA + \angle QRS + &\angle SRB = 180 ^ { \circ } \\ 70 ^ { \circ } + x + 50 ^ { \circ } & = 180 ^ { \circ } \\ 120 ^ { \circ } + x & = 180 ^ { \circ } \\ x & = (180 ^ { \circ } - 120 ^ { \circ }) \\ x & = 60 ^ { \circ } \\ \angle QRS = x & = 60 ^ { \circ } \end{align}\]

## Question 5

In the given figure, if \(AB \| CD,\;\; \angle APQ = 50 ^ { \circ }\) and \(\angle PRD = 127 ^ { \circ }\), find \(x\) and \(y\).

### Solution

**Video Solution**

**What is known?**

\(AB \| CD,\;\; \angle APQ = 50 ^ { \circ }\) and \(\angle PRD = 127 ^ { \circ }\)

**What is unknown?**

\(x =?\) and \(y = ?\)

**Reasoning:**

- When a ray intersects a line, sum of adjacent angles formed is \(180^ {\circ}\).
- When two parallel lines are cut by a transversal, alternate interior angles formed are equal.

**Steps:**

\(AB\) and \(CD\) are parallel lines cut by transversal \(PQ\) hence the alternate interior angles formed are equal.

\(\angle APQ = \angle PQR\) and hence \(x = 50^ {\circ}\).

Similarly, \(AB\) and \(CD\) are parallel lines cut by transversal \(PR\) hence the alternate angles formed are equal.

\[\begin{align} \angle APR+ \angle PRD &= 127 ^ { 0 } \\ \angle APQ + \angle QPR &= \angle PRD \\&= 127 ^ { \circ } \\ 50 ^ { \circ } + y &= 127 ^ { 0 } \\ y &= (127 ^ { 0 } - 50 ^ { \circ }) \\ y &= 77 ^ { \circ } \end{align}\]

## Question 6

In the given figure, \(PQ\) and \(RS\) are two mirrors placed parallel to each other. An incident ray \(AB\) strikes the mirror \(PQ\) at \(B\), the reflected ray moves along the path \(BC\) and strikes the mirror \(RS\) at \(C\) and again reflects back along \(CD\).

Prove that \(AB \|CD\).

### Solution

**Video Solution**

**What is known?**

\(PQ \| RS\)

**What is unknown?**

To prove: \(AB \| CD\)

**Reasoning:**

When two parallel lines are cut by a transversal, alternate angles formed are equal.

In optics the angle of incidence (the angle which an incident ray makes with a perpendicular to the surface at the point of incidence) and the angle of reflection (the angle formed by the reflected ray with a perpendicular to the surface at the point of incidence) are equal.

**Steps:**

Draw perpendicular lines \(BL\) and \(CM\) at the point of incident on both mirrors since \(PQ\) and \(RS\) parallel to each other, perpendiculars drawn are parallel \(BL \| CM\). Since \(BC\) is a transversal to lines \(BL\) and \(CM\), alternate angles are equal so we get

\[\angle LBC = \angle BCM = x \;(\text{say})\; \ldots ( 1 )\]

By laws of reflection, at the first point of incidence \(B\), we get:

\[\begin{align} \angle ABL & = \angle LBC = x \\ \therefore \angle ABC& = \angle ABL+ \angle LBC \\ & = x + x \\ \therefore \angle ABC & = 2 x \ldots \ldots \ldots ( 2 ) \end{align}\]

By laws of reflection, at the first point of incidence \(C\), we get:

\[\begin{align} \angle MCD & = \angle BCM = x \\ \therefore \angle BCD& = \angle BCM + \angle MCD \\ & = x + x \\ \angle BCD &= 2 x \ldots \ldots ( 3 ) \end{align}\]

From equations (\(2\)) and (\(3\)), we get \(\angle ABC = \angle BCD\).

We know that, if a transversal intersects two lines such that a pair of alternate interior angles is equal, then the two lines are parallel. As Alternate angles are equal we can say \(AB \| CD\).

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