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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Sign up nowIn 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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Sign up nowExercise 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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Sign up nowIn Fig. 6.15, $\angle \mathrm{PQR}=\angle \mathrm{PRQ}$, then prove that $\angle \mathrm{PQS}=\angle \mathrm{PRT}$.
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Sign up nowIn Fig. 6.16, if $x+y=w+z$, then prove that $\mathrm{AOB}$ is a line.
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Sign up nowIn 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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Sign up nowIt 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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Sign up nowExercise 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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Sign up nowIn 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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Sign up nowIn 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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Sign up nowIn 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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Sign up nowExtra 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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Sign up nowThe angles of depression of the top and the bottom of a $10$ $\mathrm{m}$ tall building from the top of a multi-storeyed building are $30^\circ$ and $45^\circ$, respectively. Find the height of the multi-storeyed building.
A) $10$ $\mathrm{m}$
B) $15$ $\mathrm{m}$
C) $5(\sqrt{3}+3)$ $\mathrm{m}$
D) $5$ $\mathrm{m}$
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Sign up nowCorrected:
The letters/variables used in algebra for writing rules and formulas in a general way, can represent:
A) Numbers
B) Angles
C) Both A & B
D) None of these
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Sign up nowIf in a right angle $\triangle ABC$, acute angles $A$ and $B$ satisfy $\tan A + \tan B + \tan^2 A + \tan^2 B = 10$, then which of the following is/are correct?
A) $\tan A = \frac{3 + \sqrt{5}}{2}$ B) $\tan A = \frac{3 - \sqrt{5}}{2}$ C) $\tan A = -2 - \sqrt{3}$ D) $\tan A = -2 + \sqrt{3}
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Sign up nowIn a regular polygon, if the ratio of the interior and exterior angles is 3:2, then the interior angle of the polygon is $\qquad$
A) $70^{\circ}$
B) $36^{\circ}$
C) $108^{\circ}$
D) $100^{\circ}$
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Sign up nowThe angle between the pair of straight lines $y^{2} \sin^{2} \theta - xy \sin^{2} \theta + x^{2}(\cos^{2} \theta - 1) = 1$ is
(A) $\frac{\pi}{3}$
(B) $\frac{\pi}{4}$ (C) $\frac{2 \pi}{3}$
(D) $\frac{\pi}{2}$
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Sign up now1 The angles between North and West and South and East are (a) complementary (b) supplementary (c) both are acute (d) both are obtuse
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