# Lines And Angles - Class 9 - Mathematics

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## Examples - Lines And Angles | NCERT | Mathematics | Class 9

In Fig. 6.9, lines PQ and RS intersect each other at point $O$. If $\angle \mathrm{POR}: \angle \mathrm{ROQ}=5: 7$, find all the angles.

In the given figure, lines PQ and RS intersect each other at point $O$. Given that $\angle \mathrm{POR} : \angle \mathrm{ROQ} = 5 : 7$, we can find all the angles using the properties of angles formed by intersecting lines.

Firstly, recall that the angles around a point sum up to $360^\circ$. Let $x$ be the common ratio for the angles, so we have $\angle \mathrm{POR} = 5x$ and $\angle \mathrm{ROQ} = 7x$.

Since the lines intersect, we can also say that $\angle \mathrm{POR} + \angle \mathrm{ROQ} + \angle \mathrm{POS} + \angle \mathrm{SOQ} = 360^\circ$. And since opposite angles are equal when two lines intersect, $\angle \mathrm{POR} = \angle \mathrm{SOQ}$ and $\angle \mathrm{ROQ} = \angle \mathrm{POS}$.

Thus, the equation becomes $5x + 7x + 5x + 7x = 360^\circ$, which simplifies to $24x = 360^\circ$.

Let's solve for (x), and then we can calculate each angle using the respective values:

$\angle \mathrm{POR} = 5x$

$\angle \mathrm{ROQ} = 7x$

$\angle \mathrm{POS} = 7x$

$\angle \mathrm{SOQ} = 5x$

We find that $x = 15^\circ$. Now we can calculate each angle:

$\angle \mathrm{POR} = 5x = 5 \times 15^\circ = 75^\circ$

$\angle \mathrm{ROQ} = 7x = 7 \times 15^\circ = 105^\circ$

$\angle \mathrm{POS} = 7x = 7 \times 15^\circ = 105^\circ$

$\angle \mathrm{SOQ} = 5x = 5 \times 15^\circ = 75^\circ$

Therefore, the angles are as follows:

$\angle \mathrm{POR} = 75^\circ$

$\angle \mathrm{ROQ} = 105^\circ$

$\angle \mathrm{POS} = 105^\circ$

$\angle \mathrm{SOQ} = 75^\circ$

In Fig. 6.10, ray OS stands on a line POQ. Ray OR and ray OT are angle bisectors of $\angle \mathrm{POS}$ and $\angle \mathrm{SOQ}$, respectively. If $\angle \mathrm{POS}=x$, find $\angle \mathrm{ROT}$.

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Sign up nowIn Fig. 6.11, OP, OQ, OR and OS are four rays. Prove that $\angle \mathrm{POQ}+\angle \mathrm{QOR}+\angle \mathrm{SOR}+$ $\angle \mathrm{POS}=360^{\circ}$.

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Sign up nowIn Fig. 6.19, if $\mathrm{PQ} \| \mathrm{RS}, \angle \mathrm{MXQ}=135^{\circ}$ and $\angle \mathrm{MYR}=40^{\circ}$, find $\angle \mathrm{XMY}$.

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Sign up nowIf a transversal intersects two lines such that the bisectors of a pair of corresponding angles are parallel, then prove that the two lines are parallel.

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In Fig. 6.22, $\mathrm{AB} \| \mathrm{CD}$ and $\mathrm{CD} \| \mathrm{EF}$. Also $\mathrm{EA} \perp \mathrm{AB}$. If $\angle \mathrm{BEF}=55^{\circ}$, find the values of $x, y$ and $z$.

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## Exercise 6.1 - Lines And Angles | NCERT | Mathematics | Class 9

In Fig. 6.13, lines $\mathrm{AB}$ and $\mathrm{CD}$ intersect at $\mathrm{O}$. If $\angle \mathrm{AOC}+\angle \mathrm{BOE}=70^{\circ}$ and $\angle \mathrm{BOD}=40^{\circ}$, find $\angle \mathrm{BOE}$ and reflex $\angle \mathrm{COE}$.

From the given figure and data, let's deduce the values systematically.

First, note that lines $AB$ and $CD$ intersect at $O$, forming vertically opposite angles at $O$. Vertically opposite angles are equal. Hence, $\angle AOC = \angle BOD$.

Given that $\angle AOC + \angle BOE = 70^\circ$ and knowing that $\angle BOD = 40^\circ$, we can also conclude that $\angle AOC = 40^\circ$, considering point 1.

Substituting $\angle AOC = 40^\circ$ into the equation $\angle AOC + \angle BOE = 70^\circ$, we can find the value of $\angle BOE$.

Let's calculate $\angle BOE$ first.

$\angle BOE = 70^\circ - \angle AOC$

$\angle BOE = 70^\circ - 40^\circ = 30^\circ$

To find the reflex of $\angle COE$,we can add $\angle AOC$, $\angle BOE$ and $\angle AOB$. Now, $\angle AOB$ is $180^\circ$, and $\angle AOC + \angle BOE = 70^\circ$

Therefore, the reflex of $\angle COE$ is $180^\circ + 70^\circ$, which is $250^\circ$

In Fig. 6.14, lines $X Y$ and $M N$ intersect at $O$. If $\angle \mathrm{POY}=90^{\circ}$ and $a: b=2: 3$, find $c$.

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In Fig. 6.15, $\angle \mathrm{PQR}=\angle \mathrm{PRQ}$, then prove that $\angle \mathrm{PQS}=\angle \mathrm{PRT}$.

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In Fig. 6.16, if $x+y=w+z$, then prove that $\mathrm{AOB}$ is a line.

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In Fig. 6.17, $\mathrm{POQ}$ is a line. Ray $\mathrm{OR}$ is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that

$\angle \mathrm{ROS}=\frac{1}{2}(\angle \mathrm{QOS}-\angle \mathrm{POS})$.

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It is given that $\angle \mathrm{XYZ}=64^{\circ}$ and $\mathrm{XY}$ is produced to point $\mathrm{P}$. Draw a figure from the given information. If ray $\mathrm{YQ}$ bisects $\angle \mathrm{ZYP}$, find $\angle \mathrm{XYQ}$ and reflex $\angle \mathrm{QYP}$.

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## Exercise 6.2 - Lines And Angles | NCERT | Mathematics | Class 9

In Fig. 6.23, if $\mathrm{AB}\|\mathrm{CD}, \mathrm{CD}\| \mathrm{EF}$ and $y: z=3: 7$, find $x$.

We know that $ \mathrm{AB} \parallel \mathrm{CD} \parallel \mathrm{EF} $.

This tells us that angles $x$ and $y$ are alternate interior angles, which means that:

$$ x + y = 180^\circ $$

Now, if we consider the supplement of the angle $z$ on the line $\mathrm{EF}$, we can also surmise that angles $y$ and $180^\circ - z$ are equal (corresponding angles axiom)

$$ y = 180^\circ - z $$

Using the fact that $y : z = 3 : 7$, we can write $y$ as $\frac{3}{7} z$

$$ \frac{3}{7} z = 180^\circ - z $$

$$ \frac{10}{7} z = 180^\circ $$

$$ z = \frac{180^\circ \times 7}{10} $$

$$ z = 126^\circ $$

We can now calculate the value of $y$ to be $\frac{3 \times 126^\circ}{7} = 54^\circ$

Now,

$$ x = 180^\circ - y = 180^\circ - 54^\circ = 126^\circ $$

In Fig. 6.24, if $\mathrm{AB} \| \mathrm{CD}, \mathrm{EF} \perp \mathrm{CD}$ and $\angle \mathrm{GED}=126^{\circ}$, find $\angle \mathrm{AGE}, \angle \mathrm{GEF}$ and $\angle \mathrm{FGE}$.

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In Fig. 6.25, if $\mathrm{PQ} \| \mathrm{ST}, \angle \mathrm{PQR}=110^{\circ}$ and $\angle \mathrm{RST}=130^{\circ}$, find $\angle \mathrm{QRS}$.

[Hint : Draw a line parallel to ST through point R.]

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In Fig. 6.26, if $\mathrm{AB} \| \mathrm{CD}, \angle \mathrm{APQ}=50^{\circ}$ and $\angle \mathrm{PRD}=127^{\circ}$, find $x$ and $y$.

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In Fig. 6.27, PQ and RS are two mirrors placed parallel to each other. An incident ray $\mathrm{AB}$ strikes the mirror $\mathrm{PQ}$ at $\mathrm{B}$, the reflected ray moves along the path $\mathrm{BC}$ and strikes the mirror RS at $\mathrm{C}$ and again reflects back along $\mathrm{CD}$. Prove that $\mathrm{AB} \| \mathrm{CD}$.

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## Extra Questions - Lines And Angles | NCERT | Mathematics | Class 9

Which of the following angles can be constructed with the help of just a ruler and a pair of compasses?

Option A) $42.5^\circ$

Option B) $40^\circ$

Option C) $67.5^\circ$

Option D) $55^\circ$

To determine which of these angles can be constructed using only a ruler and a pair of compasses, we need to analyze if each angle can be derived from constructible angles through a series of bisecting or adding/subtracting angles that can already be constructed.

**Option A: $42.5^\circ$**

This angle does not seem to be constructible directly as it is not resulting from halving or combining common constructible angles such as $45^\circ, 30^\circ$, or $60^\circ$.**Option B: $40^\circ$**

Constructing $40^\circ$ is not possible.**Option C: $67.5^\circ$**

This can be constructed by bisecting $135^\circ$ (which itself can be constructed by doubling $67.5^\circ$). Since $135^\circ$ is constructible as a result of subtracting $45^\circ$ from $180^\circ$, where both $45^\circ$ and $180^\circ$ are**constructible**, $67.5^\circ$ is**constructible**.**Option D: $55^\circ$**

Like $42.5^\circ$, $55^\circ$ is not directly constructible from basic constructible angles and does not have a straightforward method of construction using only the basic tools.

Hence, the constructible angles from the given options are:

**Option B: $40^\circ$****Option C: $67.5^\circ$**

In the figure $AB = CB$ and $O$ is the center of the circle. Prove that $BO$ bisects $\angle ABC$.

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(a) If $AB | DE$, $\angle AOD = 50^{\circ}$, and $\angle DPC = 160^{\circ}$, find the value of $\angle DBE$.

(b) A $15$ m long ladder reached a window $12$ m high from the ground when placed against a wall at a distance '$a$'. Find the distance of the foot of the ladder from the wall.

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Angle between 2 planes $\bar{r} \cdot \hat{n}*{1}=d*{1}$, $\bar{r} \cdot \hat{n}*{2}=d*{2}$ can always be given by:

(A) $\theta=\sin^{-1}\left|\hat{n}*{1} \times \hat{n}*{2}\right|$

(B) $\theta=\cos^{-1}\left(\hat{n}*{1} \cdot \hat{n}*{2}\right)$

(C) $\theta=\sin^{-1}\left|\hat{r}*{1} \cdot \hat{r}*{2}\right|$

(D) $\theta=\cos^{-1}\left|\left(\hat{n}*{1} \cdot \hat{n}*{2}\right)\right|$

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Find an angle which is $1/4$ of its complement.

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The measure of the angle which is equal to its complementary angle is

(A) $60^{\circ}$

(B) $70^{\circ}$

(C) $45^{\circ}$

(D) $40^{\circ}$

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When a line is drawn parallel to a given line through a point not lying on the line, using a ruler and compass, two arcs are drawn, one on the line and the other at the point. If one of the arcs is of the radius $3$ cm from the respective center, the measure of the other arc will be:

A) $6$ cm

B) $3$ cm

C) $2$ cm

D) $4$ cm

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A $40^{\circ}$ angle can be drawn by drawing an angle bisector of a given $80^{\circ}$ angle.

A) True

B) False

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In what direction should a line be drawn through the point $(1,2)$ so that its point of intersection with the line $x+y=4$ will be at a distance of $\frac{\sqrt{6}}{3}$?

(A) $30^{\circ}$ (B) $45^{\circ}$ (C) $60^{\circ}$ (D) $75^{\circ}$

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Bisector of an angle divides the angle into:

Four parts

Three parts

Two parts

Two equal parts