2 Marks Questions

Question 12 Marks

In figure, find the value of $x.$

Answer

$\angle\text{AOE}=\angle\text{BOF}=5\text{x}$ [vertically opposite angle] $\angle\text{COA}+\angle\text{AOE}+\angle\text{EOD}=180^\circ$ [linear pair]
$\Rightarrow 3x + 5x + 2x = 180^\circ $
$\Rightarrow 10x = 180^\circ $
$\Rightarrow x = 18^\circ$

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Question 22 Marks

Which pair of lines in the below figure is parallel? Give reasons.

Answer

$\angle A +\angle B =115+65=180^{\circ}$ Therefore, $AB \| BC$ [As sum of co interior angles are supplementary $]$ $\angle B +\angle C =65+115=180^{\circ}$ Therefore, $AB \| CD$ [As sum of interior angles are supplementary $]$

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Question 32 Marks

An angle is equal to $8$ times its complement. Determine its measure.

Answer

Let the measure of an angle be $x^\circ .$
Therefore, its complement will be $(90 - x)^\circ .$
It is given that
$\Rightarrow x^\circ = 8(90 - x)^\circ $
$\Rightarrow x = 720 - 8x$
$\Rightarrow 9x = 720$
$\Rightarrow x = 80$
Therefore, the measure of the angles is $80^\circ $

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Question 42 Marks

In figure, $AOC$ is a line, find $x.$

Answer

Since $\angle\text{AOB}$ and $\angle\text{BOC}$ are linear pairs,
$\angle\text{AOB}+\angle\text{BOC}=180^\circ$
$70 + 2x = 180$
$2x = 180 - 70$
$2x = 110$
$\text{x}=\frac{110}{2}$
$x = 55$
Hence, the value of $x$ is $55^\circ $

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Question 52 Marks

Write the supplement of an angle of measure $2y^\circ .$

Answer

Let the required angle measures $x^{\circ}$. It is given that two angles measuring $x^{\circ}$ and $2 y^{\circ}$ are supplementary. Therefore, their sum must be equal to $180^{\circ}$. Or, we can say that: $x+2 y=180 x=180-2 y$ Hence, the required angle measures $(180-2 y)^{\circ}$

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Question 62 Marks

In figure, $ACB$ is a line such that $\angle\text{DCA}=5\text{x}$ and $\angle\text{DCB}=4\text{x}.$ Find the values of $\angle\text{DCA}$ and $\angle\text{DCB}.$

Answer

It is given that $ACB$ is a line in the figure given below. Thus, $\angle\text{ACD}$ and $\angle\text{BCD}$ from a linear pair.

Therefore, their sum must be equal to $180^\circ $. Or,
we can say that $\angle\text{ACD}+\angle\text{BCD}=180^\circ$
Also, $\angle\text{ACD}=4\text{x}$ and $\angle\text{BCD}=5\text{x}.$
This further simplifies to: $4x + 5x = 180 9x = 180$
$\text{x}=\frac{180}{9}$
$x = 20$
$\therefore\angle\text{DCA}=5\text{x}=5\times20^\circ=100^\circ$
$\angle\text{DCB}=4\text{x}=4\times20^\circ=80^\circ$
Hence, the values of $\angle\text{DCA}$ and $\angle\text{DCB}$ are $100^\circ $ and $80^\circ $ respectively.

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Question 72 Marks

In figure, determine the value of $x.$

Answer

Since the sum of all the angles round a point is equal to $360^\circ 3x + 3x + 150 + x = 360 7x = 360 - 150 7x = 210 $$\text{x}=\frac{210}{7}$$ x = 30 $Value of x is $30^\circ $

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Question 82 Marks

In figure, lines $AB, CD$ and $EF$ intersect at $O$. Find the measures of $\angle\text{AOC},\angle\text{COF},\angle\text{DOE}$ and $\angle\text{BOF}.$

Answer

$\angle\text{AOE}+\angle\text{DOE}+\angle\text{BOD}=180^\circ$ [linear pair]
$\Rightarrow\ \angle\text{DOE}=180^\circ-40^\circ-35^\circ=105^\circ$
$\angle\text{DOE}=\angle\text{COF}=105^\circ$ [vertically opposite angles]
Now, $\angle\text{AOE}+\angle\text{AOF}=180^\circ$ [lineat pair]
$\Rightarrow\ \angle\text{AOE}+\angle\text{AOC}+\angle\text{COF}=180^\circ$
$\Rightarrow\ 40^\circ+\angle\text{AOC}+105^\circ=180^\circ $
$\Rightarrow\ \angle\text{AOC}=35^\circ$
Also, $\angle\text{BOF}=\angle\text{AOE}=40^\circ$ [vertically opposite angles]

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Question 92 Marks

In figure, $POS$ is a line, Find $x.$

Answer

Since $\angle\text{POQ}$ and $\angle\text{QOS}$ are linear pairs $\angle\text{POQ}+\angle\text{QOS}=180^\circ$
$\angle\text{POQ}+\angle\text{QOR}+\angle\text{SOR}=180^\circ$ $60 + 4x + 40 = 180 4x = 180 - 100 4x = 80 x = 20$
Hence, Value of $x = 20$

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Question 102 Marks

In figure, write all pairs of adjacent angles and all the linear pairs.

Answer

Adjacent angles are:
$i. \angle\text{AOD},\angle\text{COD}$
$ii. \angle\text{BOC},\angle\text{COD}$
$iii. \angle\text{AOD},\angle\text{BOD}$
$iv. \angle\text{AOC},\angle\text{BOC}$
Linear pairs:
$i. \angle\text{AOD},\angle\text{BOD}$
$ii. \angle\text{AOC},\angle\text{BOC}$

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Question 112 Marks

Two supplementary angles are in the ratio $4 : 5$. Find the angles.

Answer

Let the angles be $4x$ and $5x$. It is given that they are supplem entary angles. $\therefore 4x + 5x = 180 $
$\Rightarrow 9x = 180 $
$\Rightarrow x = 20^\circ $
Hence $4x = 80, 5x = 100$
$\therefore$ Angle are $80^\circ $ and $100^\circ .$

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Question 122 Marks

If the angles $(2x - 10)^\circ $ and $(x - 5)^\circ $ are complementary angles, find $x.$

Answer

Since the angles are complementry. Theefore their sum will be $90^\circ . $
$\Rightarrow (2x - 10)^\circ + (x - 5)^\circ = 90^\circ $
$\Rightarrow 2x - 10 + x - 5 = 90 $
$\Rightarrow 3x = 105^\circ $
$\Rightarrow x = 35^\circ $

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Question 132 Marks

Two supplementary angles differ by $48^\circ $. Find the angles.

Answer

Let the measure of an angle $x^{\circ} . \therefore$ Its supplem entary will be $(180- x )^{\circ}$. It is given that $\left(180- x ^{\circ}\right)- x ^{\circ}=48^{\circ} $
$\Rightarrow 180-2 x$
$=48 $
$\Rightarrow 132=2 x $
$\Rightarrow x=66^{\circ}$ Hence, $180-x=180-66=114^{\circ}$.
Therefore, angles are $66^{\circ}$ and $114^{\circ}$.

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Question 142 Marks

In figure, rays $AB$ and $CD$ intersect at $O$. Determine $x$ when $y = 40^\circ .$

Answer

Here, $\angle\text{AOC}+\angle\text{BOC}=180^\circ$ [linear pair]
$\Rightarrow 2x + y = 180^\circ $
$\Rightarrow 2x + 40 = 180^\circ $
$\Rightarrow 2x = 180 - 40 = 140^\circ $
$\Rightarrow 2x = 140^\circ $
$\Rightarrow x = 70^\circ $

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Question 152 Marks

In figure, $OA$ and $OB$ are opposite rays: If $y = 35°$, what is the value of $x?$

Answer

Given that, $y = 35^\circ $
$\angle\text{AOC}+\angle\text{BOC}=180^\circ$
$(2y + 5) + 3x = 180 (2(35) + 5) + 3x = 180 (70 + 5) + 3x = 180 3x = 180 - 75 3x = 105 x = 35$

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Question 162 Marks

If an angle differs from its complement by $10^\circ $, find the angle.

Answer

Let the measure of the angle be $x^\circ$
. $\therefore$ its complement will be $(90 - x)^\circ $.
It is given that $x^\circ - (90 - x)^\circ = 10^\circ $
$\Rightarrow x - 90 + x = 10 $
$\Rightarrow 2x = 100 $
$\Rightarrow x = 50$
$\therefore$ The measure of the angle will be $50^\circ .$

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Question 172 Marks

In figure, find the values of $x, y$ and $z.$

Answer

From the given figure: $\angle\text{y}=25^\circ$ [vertically opposite angle]
Now, $\angle\text{x}+\angle\text{y}=180^\circ$ [linear pair] $\Rightarrow\ \angle\text{x}=180^\circ-25^\circ$
$\Rightarrow\ \angle\text{x}=155^\circ$ Also, $\angle\text{z}=\angle\text{x}=155^\circ$ [vertically opposite angle]
$\therefore \angle\text{y}=25^\circ$
$\angle\text{z}=\angle\text{x}=155^\circ$

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Question 182 Marks

If an angle is $28^\circ $ less than its complement, find its measure.

Answer

Let the measure of the angle be $x^\circ $.
$\therefore$ Its complement will be $(90 - x)^\circ $ It is given that Angle = complement $- 28^\circ .$
$\therefore x^\circ = (90 - x)^\circ - 28^\circ $
$ 2x^\circ = 62^\circ $
$ x = 31^\circ $
$\therefore$ Angle measure dis $31^\circ .$

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Question 192 Marks

If the supplement of an angle is two-third of itself. Determine the angle and its supplement.

Answer

Let the measure of the angle be $x^\circ $.
$\therefore$ its supplement will be $(180 - x)^\circ .$ It is given that $(180-\text{x})^\circ=\frac{2}{3}\text{x}^\circ$
$\Rightarrow\ 180-\text{x}=\frac{2}{3}\text{x}$
$\Rightarrow\ \frac{2}{3}\text{x}+\text{x}=180$
$\Rightarrow\ \frac{5}{3}\text{x}=180$
$\Rightarrow\ \text{x}=108^\circ$ Hence, supplement $= 180 - 108 = 72^\circ $
$\therefore$ Angle will be $108^\circ $ and its supplement will be $72^\circ .$

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Question 202 Marks

If the supplement of an angle is three times its complement, find the angle.

Answer

Let the measure of the angle be $x^{\circ} .$
$ \therefore$ Its supplement will be $(180-x)^{\circ}$.
And its complement will be $(90- x )^{\circ}$. It is
given that $(180-x)^{\circ}=3(90-x)^{\circ} $
$\Rightarrow 180-x=270-3 x $
$\Rightarrow 2 x=90 $
$\Rightarrow x=45$
$\therefore$ The measure of the angle will be $45^{\circ}$.

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Question 212 Marks

If $l, m, n$ are three lines such that $l ∥ m$ and $\text{n}\perp\text{l},$ prove that $\text{n}\perp\text{m}.$

Answer

Given, $l || m$, $\text{n}\perp\text{l}$ To prove: $\text{n}\perp\text{m}$ Since $l || m$ and $n$ intersects $\therefore\ \angle{1}=\angle{2}$ [Corresponding angles] But,$ U = 90 \Rightarrow\ \angle{2}=90^\circ$ Hence n is perpendicular to m.

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Question 222 Marks

If an angle is $30^\circ $ more than one half of its complement, find the measure of the angle.

Answer

Let the measure of the angle be $x^{\circ} . $
$\therefore$ its complement will be $(90-x)^{\circ}$. It is
given that Angle $=30^{\circ}+\frac{1}{2}$ complement $$
$\Rightarrow x ^{\circ}+\frac{1}{2}(90- x )^{\circ} $
$\Rightarrow x =\frac{60^{\circ}+90- x }{2} $
$\Rightarrow 2 x =60^{\circ}+90- x $
$\Rightarrow 3 x =150 $
$\Rightarrow x =50$
$\therefore$ Angle is $50^{\circ}$.

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Question 232 Marks

In figure, three coplanar lines intersect at a point $O$, forming angles as shown in the figure. Find the values of $x, y, z$ and $u$.

Answer

$\text{z}=\angle\text{BOD}=90^\circ$ [vertically opposite angle]
$\text{y}=\angle\text{DOF}=50^\circ$ [vertically oppposite angle]
Now, $x + y + z = 180^\circ $
$\Rightarrow x + 90^\circ + 50^\circ = 180^\circ $
$\Rightarrow x = 180^\circ - 140^\circ = 40^\circ $

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Question 242 Marks

The measure of an angle is twice the measure of its supplementary angle. Find its measure.

Answer

Let the measure of angle is $x^{\circ} . $
$\therefore$ Its supplementary will be $(180-x)^{\circ}$. It is given that $x^{\circ}=2(180-x)^{\circ} $
$\Rightarrow x=360-2 x $
$\Rightarrow 3 x=360 $
$\Rightarrow x=120^{\circ} $
$\therefore$ The measure of angle is $120^{\circ}$.

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Question 252 Marks

Write the complement of an angle of measure $x^\circ .$

Answer

We have to write the complement of an angle which measures $x^{\circ}$.
Let the other angle be $y^{\circ}$.
We know that the sum of the complementary angles be $90^{\circ}$.
Therefore, $x^{\circ}+y^{\circ}=90^{\circ} y^{\circ}=(90-x)^{\circ}$

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Question 262 Marks

In figure, rays $AB$ and $CD$ intersect at $O$. Determine y when$ x = 60^\circ .$

Answer

Here, $\angle\text{AOC}+\angle\text{BOC}=180^\circ$ [linear pair]
$\Rightarrow 2x + y = 180^\circ $ [linear pair]
$\Rightarrow 2 \times 60^\circ + y = 180^\circ $
$\Rightarrow y = 180^\circ - 120^\circ = 60^\circ $
$\therefore y = 60^\circ $

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Question 272 Marks

How many pairs of adjacent angles are formed when two lines intersect in a point?

Answer

Four pairs of adjacent angles will be formed when two lines intersect at a point. Considering two lines $AB$ and $CD$ intersecting at $O$. The $4$ pairs are: $(\angle\text{AOD},\angle\text{DOB})$ $(\angle\text{DOB},\angle\text{BOC})$ $(\angle\text{COA},\angle\text{AOD})$ And $(\angle\text{BOC},\angle\text{COA})$ Hence, $4$ pairs of adjacent angles are formed when two lines intersect at a point.

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Question 282 Marks

An angle is $14^\circ $ more than its complementary angle. What is its measure?

Answer

Let the angle measures $x^{\circ}$. Therefore, the measure of its complement becomes $(90- x )^{\circ}$ According to the given statement, the angle is $14$ more than its complement. Thus, we have, $x=14+(90-x) x=104-x x+x=1042 x=$
$104 x=\frac{104}{2} x=52$ The measure of its complement becomes $90-x=90-52=38$ Hence, the required angle measures $52^{\circ}$ and its complement measures $38^{\circ}$.

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