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Maths 3 Marks Question

Lines And Angles · Jharkhand Board (JAC) STD 9 (Jharkhand - English Medium). 28 questions.

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Question 13 Marks
If a wheel has six spokes equally spaced, then find the measure of the angle between two adjacent spokes.
Answer
It is given that the six spokes are equally spaced, thus, two adjacent spokes subtend equal angle at the centre of the wheel.
Let that angle measures $x^\circ$ .
Also, The six spokes form a complete angle, that is $360^\circ$ .
Therefore, $x + x + x + x + x + x = 360^\circ$ $6x = 360^\circ$ $\text{x}=\frac{360^\circ}{6}$ $x = 60^\circ$
Hence, the measure of the angle between two adjacent spokes measures $60^\circ$ .
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Question 23 Marks
Prove that the straight lines perpendicular to the same straight line are parallel to one another.
Answer
Let $AB$ and $CD$ be drawn perpendicular to the Line $MN$ $\angle\text{ABD}=90^\circ$ $... (i)$
$[AB$ is perpendicular to $MN]$
$\angle\text{CON}=90^\circ$ $... (ii) [CD$ is perpendicular to $MN ]$
Now, $\angle\text{ABD}=\angle\text{CDN}=90^\circ$$[$From $(i)$ and $(ii)]$
Therefore,$ AB\ ||\ CD$, Since corresponding angles are equal.
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Question 33 Marks
In figure, arms $BA$ and $BC$ of $\angle\text{ABC}$ are respectively parallel to arms $ED$ and $EF$ of $\angle\text{DEF}.$ Prove that $\angle\text{ABC}=\angle\text{DEF}.$
Answer
Given $AB\ ||\ DE$ and $BC\ ||\ EF$
To Prove: $\angle\text{ABC}=\angle\text{DEF}$
 Construction: Produce $BC$ to $x$ such that it intersects $DE$ at $M$.
Proof: Since $AB\ ||\ DE$ and $BX$ is the transversal $ABC = DMX ...(1)$ [Corresponding angle]
Also, $BX\ ||\ EF$ and $DE$ is the transversal $DMX = DEF ...(2)$ [Corresponding angles] From $(1)$ and $(2)$ $\angle\text{ABC}=\angle\text{DEF}$
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Question 43 Marks
How many pairs of adjacent angles, in all, can you name in figure $10.36$.
Answer
Pairs of adjacent angles are:
$\angle\text{EOC},\angle\text{DOC}$
$\angle\text{EOD},\angle\text{DOB}$
$\angle\text{DOC},\angle\text{COB}$
$\angle\text{EOD},\angle\text{DOA}$
$\angle\text{DOC},\angle\text{COA}$
$\angle\text{BOC},\angle\text{BOA}$
$\angle\text{BOA},\angle\text{BOD}$
$\angle\text{BOA},\angle\text{BOE}$
$\angle\text{EOC},\angle\text{COA}$
$\angle\text{EOC},\angle\text{COB}$
 Hence, $10$ pair of adjacent angles.
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Question 53 Marks
In figure, $AB\ ||\ CD\ ||\ EF$ and $GH\ ||\ KL$. Find $\angle\text{HKL}.$
Answer
Produce $LK$ to meet $GF$ at $N$.
Now, alternative angles are equal $\angle\text{CHG}=\angle\text{HGN}=60^\circ$
$\angle\text{HGN}=\angle\text{KNF}=60^\circ$ [Corresponding angles]
Hence, $\angle\text{KNG}=180-60=120$
$\Rightarrow\ \angle\text{GNK}=\angle\text{AKL}=120^\circ$ [Corresponding angles]
$\angle\text{AKH}=\angle\text{KHD}=25^\circ$ [alternative angles]
Therefore, $\angle\text{HKL}=\angle\text{AKH}+\angle\text{AKL}=25+120=145^\circ.$
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Question 63 Marks
In figure, lines $AB$ and $CD$ are parallel and $P$ is any point as shown in the figure. Show that $\angle\text{ABP}+\angle\text{CDP}=\angle\text{DPB}.$
Answer
Given that $AB\ ||\ CD$ Let $EF$ be the parallel line to $AB$ and $CD$ which passes through $P$.
It can be seen from the figure. Alternative angles are equal $\angle\text{ABP}=\angle\text{BPF}$ Alternative angles are equal $\angle\text{CDP}=\angle\text{DPF}$ $\angle\text{ABP}+\angle\text{CDP}=\angle\text{BPF}+\angle\text{DPF}$ $\angle\text{ABP}+\angle\text{CDP}=\angle\text{DPB}$
Hence proved $AB$ parallel to $CD, P$ is any point
To prove: $\angle\text{ABP}+\angle\text{BPD}+\angle\text{CDP}=360^\circ$
Construction: Draw $EF\ ||\ AB$ passing through $P$.

Proof: Since $AB\ ||\ EF$ and $AB\ ||\ CD$, Therefore $EF\ ||\ CD$ [Lines parallel to the same line are parallel to each other] $\angle\text{ABP}+\angle\text{EPB}=180^\circ$[Sum of co-interior angles is $180$]
$\angle\text{EPD}+\angle\text{COP}=180^\circ\dots(1)$ [Sum of co-interior angles is $180$]
$\angle\text{EPD}+\angle\text{CDP}=180^\circ\dots(2)$ By adding $(1)$ end $(2)$
$\angle\text{ABP}+\angle\text{EPB}+\angle\text{EPD}+\angle\text{CDP}=(180+180)^\circ$
$\angle\text{ABP}+\angle\text{EPB}+\angle\text{COP}=360^\circ$
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Question 73 Marks
In figure, $\angle{1}=60^\circ$ and $\angle{2}=\Big(\frac{2}{3}\Big)^\text{rd}$ of a right angle. Prove that l∥m.
Answer
Given: $\angle{1}=60^\circ$ and $\angle{2}=\Big(\frac{2}{3}\Big)^\text{rd}$ of a right angle
To Prove: Parallel Drawn to m Proof: $\angle{1}=60^\circ$ $\angle{2}=\Big(\frac{2}{3}\Big)\times90=60$
Since $\angle{1}=\angle{1}=60^\circ$
Therefore, Parallel to m as pair of corresponding angles are equal.
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Question 83 Marks
Define supplementary angles.
Answer
supplementary Angles: Two angles, the sum of whose measures is $180^\circ $, are called suppplementary angles.
Thus,
angles $\angle\text{BAC}$ and $\angle\text{DAC}$ are aupplementary angles.
If $x + y = 180^\circ $.

Example $1$: Angles of measure $50^\circ$ and $130^\circ $ are supplementary angles, because
$50^\circ + 130^\circ = 180^\circ $
Example $2$: Angles of measure $60^\circ$ and $120^\circ $ are supplementary angles, because
$60^\circ + 120^\circ = 180^\circ $
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Question 93 Marks
In figure, state Which lines are parallel and why.
Answer
Vertically opposite angles are equal
$\Rightarrow\ \angle\text{EOC}=\angle\text{DOK}=100^\circ$
$\Rightarrow\ \angle\text{DOK}=\angle\text{ACO}=100^\circ$
Here two lines $EK$ and $CA$ cut by a third line and the corresponding angles to it are equal Therefore, $EK\ ||\  AC$.
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Question 103 Marks
Given $\angle\text{POR}=3\text{x}$ and $\angle\text{QOR}=2\text{x}+10,$ Find the value of $x$ for which $POQ$ will be a line. (Figure $10.41$).
Answer
For the case that $POR$ is a line $\angle\text{POR}$ and $\angle\text{QOR}$ are linear parts $\angle\text{POR}+\angle\text{QOR}=180^\circ$
Also, given that, $\angle\text{POR}=3\text{x}$ and $\angle\text{QOR}=2\text{x}+10$
 $2x + 10 + 3x = 180$
$5x + 10 = 180 $
$5x = 180 - 10 $
$5x = 170 x $
$​​​​​​​= 34$
Hence the value of $x$ is $34^\circ $
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Question 113 Marks
In figure, $AB\ ||\ CD$ and $P$ is any point shown in the figure. Prove that: $\angle\text{ABP}+\angle\text{BPD}+\angle\text{CDP}=360^\circ$
Answer
The given figure is as follows:

It is give that $AB || CD$ Let us draw a line $XY$ passing through point $P$ and parallel to $AB$ and $CD$.
We have $XY\ ||\ CD$, thus, $\angle\text{CDP}$ and $\angle{2}$ are consecutive interior angles.
Therefore, $\angle{2}+\angle\text{CDP}=180^\circ\dots(\text{i})$
Similarly, we have $XY\ ||\ AB$, thus, $\angle\text{ABP}$ and $\angle{1}$ are consecutive interior angles.
​​​​​​​Therefore, $\angle{1}+\angle\text{ABP}=180^\circ\dots(\text{ii})$ On adding equation $(i)$ and $(ii)$, we get: $\angle{2}+\angle\text{CDP}+\angle{1}+\angle\text{ABP}=180^\circ+180^\circ$ $(\angle{2}+\angle{1})+\angle\text{CDP}+\angle\text{ABP}=360^\circ$ $\angle\text{ABP}+\angle\text{BPD}+\angle\text{CDP}=360^\circ$ Hence proved.
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Question 123 Marks
In figure, $OA$ and $OB$ are opposite rays: If $x = 25^\circ $, what is the value of $y$?
Answer
Given that, $x = 25$ Since $\angle\text{AOC}$ and $\angle\text{BOC}$ form a linear pair $\angle\text{AOC}+\angle\text{BOC}=180^\circ$
Given that $\angle\text{AOC}=2\text{y}+5$ and $\angle\text{BOC}=3\text{x}$
$\angle\text{AOC}+\angle\text{BOC}=180^\circ$
$(2y + 5) + 3x $
$= 180 (2y + 5) + 3(25) $
$= 180 2y + 5 + 75$
$ = 180 2y + 80$
$ = 180 2y $
$= 180 - 80 $
$= 100$
$\text{y}=\frac{100}{2}$ $y = 50$
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Question 133 Marks
If figure, if $AB\ ||\ CD$ and $CD\ ||\ EF$, find $\angle\text{ACE}.$
Answer
Since $EF\ ||\ CD$
Therefore, $EFC + ECD = 180$ [Co-interior angles are supplementary]
$\Rightarrow ECD = 180 - 130 = 50$
Also $BA\ ||\ CD$
$\Rightarrow BAC = ACD = 70$ [alternative angles] But,
$ACE + ECD = 70 $
$\Rightarrow ACE = 70 - 50 = 20$
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Question 143 Marks
In figure $POQ$ is a line. Ray $OR$ is perpendicular to line $PQ$. $OS$ is another ray lying between rays $OP$ and $OR$. Prove that $\angle\text{ROS}=\frac{1}{2}(\angle\text{QOS}-\angle\text{POS}).$
Answer
Given that OR perpendicular $\therefore\angle\text{POR}=90^\circ$
$\angle\text{POS}+\angle\text{SOR}=90$
$[\therefore\angle\text{POR}=\angle\text{POS}+\angle\text{SOR}]$
$\angle\text{ROS}=90^\circ-\angle\text{POS}\dots(1)$
$\angle\text{QOR}=90$
$(\because\text{OR}\perp\text{PQ})$
 $\angle\text{QOS}-\angle\text{ROS}=90^\circ$
$\angle\text{ROS}=\angle\text{QOS}-90^\circ\dots(2)$ By adding $(1)$ and $(2)$ equations,
we get: $\therefore\ \angle\text{ROS}=\angle\text{QOS}-\angle\text{POS}$
$\angle\text{ROS}=\frac{1}{2}(\angle\text{QOS}-\angle\text{POS})$
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Question 153 Marks
In figure, $p$ is a transversal to lines $m$ and $n$, $\angle{2}=120^\circ$and $\angle{5}=60^\circ.$Prove that $m\ ||\ n$.
Answer
Given that
$\angle{2}=120^\circ$ and $\angle{5}=60^\circ.$
To prove,
$\angle{2}+\angle{1}=180^\circ$ [Linear pair]
$120+\angle{1}=180^\circ$
$\angle{1}=180-120$
$\angle{1}=60^\circ$
Since $\angle{1}=\angle{5}=60^\circ$
Therefore, $m\ ||\ n$. [As pair of corresponding angles are equal]
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Question 163 Marks
In figure transversal $l$ intersects two lines $m$ and $n$, $\angle{4}=110^\circ$ and $\angle{7}=65^\circ.$ Is $m\ ||\ n$?
Answer
Given: $\angle{4}=110^\circ$and $\angle{7}=65^\circ.$
To find: is $m\ ||\ n$ Here, $\angle{7}=\angle{5}=65^\circ$ [Vertically opposite angle]
Now, $\angle{4}+\angle{5}=110+65=175^\circ$
Therefore, $m$ is not parallel to $n$ as the pair of co interior angles is not supplementary.
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Question 173 Marks
If the complement of an angle is equal to the supplement of the thrice of it. Find the measure of the angle.
Answer
Let the measure of the angle be $x^\circ $.
Its complement will be $(90^\circ - x^\circ )$ and its supplement will be $(180^\circ - x^\circ )$.
Supplement of thrice of the angle $= (180^\circ - 3x^\circ )$
According to the given information:
$(90^\circ - x^\circ ) = (180^\circ - 3x^\circ )$
$3x - x = 180 - 90$
$2x = 90$
$x = 45$
Thus, the measure of the angle is $45^\circ $.
The measure of the angle is $45^\circ $.
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Question 183 Marks
In figure, $PQ\ ||\ AB$ and $PR\ ||\ BC$. If $\angle\text{QPR}=102^\circ,$ determine $\angle\text{ABC}.$ Give reasons.
Answer
$AB$ is produce to meet $PR$ at $K$ Since $PQ\ ||\ AB$
$\angle\text{QPR}=\angle\text{BKR}=102^\circ$[corresponding angles]
Since $PR\ ∥\ BC$ $\angle\text{RKB}+\angle\text{CBK}=180^\circ$[Since Corresponding angles are supplementary]
$\angle\text{CKB}=180-102=78$
$\therefore\ \angle\text{CKB}=78^\circ$
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Question 193 Marks
$AB, CD$ and $EF$ are three concurrent lines passing throught the point $O$ such that $OF$ bisects $\angle\text{BOD}.$ If $\angle\text{BOF}=35^\circ,$ find $\angle\text{BOC}$ and $\angle\text{AOD}.$
Answer

$\angle\text{BOF}=35^\circ$ $\therefore\angle\text{BOD}=2\angle\text{BOF}=70^\circ$
$[\because$ $OF$ bisects $\angle\text{BOD}]$
$\angle\text{BOD}=\angle\text{AOC}=70^\circ$ [vertically opposite angles]
Now, $\angle\text{BOC}+\angle\text{AOC}=180^\circ$ [linear pair]
$\Rightarrow\ \angle\text{BOC}+70^\circ=180^\circ$
$\Rightarrow\ \angle\text{BOC}=110^\circ$
$\therefore\ \angle\text{AOD}=\angle\text{BOC}=110^\circ$ [vertically opposite angles]
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Question 203 Marks
In figure if $l\ ||\ m, n∥p$ and $\angle\text{1}=85^\circ,$ find $\angle{2}.$
Answer
Corresponding angles are equal
$\Rightarrow\ \angle{1}=\angle{3}=85^\circ$
By using the property of co-interior angles are supplementary
$\angle{2}+\angle{3}=180^\circ$
$\angle{2}+55=180^\circ$
$\angle{2}=180-85$
$\angle{2}=95^\circ$
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Question 213 Marks
What value of $y$ would make $AOB$ a line in the below figure, If $\angle\text{AOC}=4\text{y}$ and $\angle\text{BOC}=(6\text{y}+30)?$
Answer
Since, $\angle\text{AOC}$ and $\angle\text{BOC}$ are linear pairs $\angle\text{AOC}+\angle\text{BOC}=180^\circ$
$6y + 30 + 4y = 180\ 10y + 30 = 180\ 10y = 180 - 30\ 10y = 150$ $\text{y}=\frac{150}{10}$ y = 15
Hence value of $y$ that will make $AOB$ a line is $15^\circ$ .
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Question 223 Marks
How many pairs of adjacent angles are formed when two lines intersect in a point?
Answer
Let us draw the following diagram showing two lines $AB$ and $CD$ intersecting at a point $O$.

We have the following pair of adjacent angles,
so formed: $\angle\text{AOC}$ and $\angle\text{BOC}$
$\angle\text{AOC}$ and $\angle\text{AOD}$
$\angle\text{BOD}$ and $\angle\text{BOC}$
$\angle\text{BOD}$ and $\angle\text{AOD}$
Hence, in total four pair of adjacent angles are formed.
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Question 233 Marks
An angle is equal to five times its complement. Determine its measure.
Answer
Let the complement of the required angle measures $x^\circ$ .
Therefore, the required angle becomes $5x^\circ $.
Since, the angles are complementary.
Thus, their sum must be equal to $90^\circ $.
Or we can say that : $x + 5x = 90^\circ 6x = 90^\circ $ $\text{x}=\frac{90^\circ}{6}$ $x = 15^\circ $
Hence, the required angle becomes: $5x = 5(15^\circ ) = 75^\circ $.
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Question 243 Marks
An angle is equal to its supplement. Determine its measure.
Answer
Let the supplement of the angle be $x^\circ$ .
According the given statement, the required angle is equal to its supplement, therefore, the required angle becomes $x^\circ$ .
 Sine both the angles are supplementary, therefore, their sum must be equal to $180^\circ$ .
Or we can say that: $x + x = 180^\circ$ $2x = 180^\circ$ $\text{x}=\frac{180^\circ}{2}$$ x = 90^\circ$
Hence, the required angle measures $90^\circ$ .
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Question 253 Marks
Two lines $AB$ and $CD$ intersect at $O$. If $\angle\text{AOC}+\angle\text{COB}+\angle\text{BOD}=270^\circ,$ find the measure of $\angle\text{AOC},\angle\text{COB},\angle\text{BOD}$ and $\angle\text{DOA}.$
Answer

Given: $\angle\text{AOC}+\angle\text{COB}+\angle\text{BOD}=270^\circ$
To find: $\angle\text{AOC},\angle\text{COB},\angle\text{BOD},\angle\text{DOA}$
Here, $\angle\text{AOC},\angle\text{COB},\angle\text{BOD}=270^\circ$[Complete angle]
$\Rightarrow 270 + AOD = 360 $
$\Rightarrow AOD = 360 - 270$
$ \Rightarrow AOD = 90$
Now, $AOD + BOD = 180$ [Linear pair]
$90 + BOD = 180 $
$\Rightarrow BOD = 180 - 90$
$ \Rightarrow BOD = 90 AOD = BOC = 90$ [Vertically opposite angles]
$BOD = AOC = 90$ [Vertically opposite angles]
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Question 263 Marks
In figure, $l, m$ and $n$ are parallel lines intersected by transversal $p$ at $X. Y$ and $Z$ respectively. Find $\angle\text{1},\angle\text{2}$ and $\angle\text{3}.$
Answer
From the given figure:
$\angle\text{3}+\angle\text{myz}=180^\circ$ [Linear pair]
$\Rightarrow\ \angle{3}=180-120$
$\Rightarrow\ \angle3=60^\circ$
Now line $I$ parallel to $m$
$\angle{1}=\angle{3}$ [Corresponding angles]
$\angle{1}=60^\circ$
Also $m$ parallel to $n$
$\Rightarrow\ \angle{2}=120^\circ$ [Alternative interior angle]
Hence,
$\angle{1}=\angle{3}=60^\circ$
$\angle{2}=120^\circ$
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Question 273 Marks
In figure, lines $AB$ and $CD$ intersect at $O$. If $\angle\text{AOC}+\angle\text{BOE}=70^\circ$ and $\angle\text{BOD}=40^\circ,$ find $\angle\text{BOE}$ and reflex $\angle\text{COE}.$
Answer
In the figure, $\angle\text{AOC},\angle\text{BOE}$ and $\angle\text{COE} $ from a linear pair.
Thus,
$\angle\text{AOC}+\angle\text{BOE}+\angle\text{COE}=180^\circ$
It is given that $\angle\text{AOC}+\angle\text{BOE}=70^\circ,$
on substituting this value, we get:
$70^\circ+\angle\text{COE}=180^\circ$
$\angle\text{COE}=180^\circ-70^\circ$
$\angle\text{COE}=110^\circ$
Thus, reflex $\angle\text{COE}=360^\circ-110^\circ$
Therefore, reflex $\angle\text{COE}=250^\circ.$
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Question 283 Marks
In figure, show that $AB\ ||\ EF$.
Answer
Produce $EF$ to intersect $AC$ at $K$.
Now, $\angle\text{DCE}+\angle\text{CEF}=35+145=180^\circ$
Therefore, $EF\ ||\ CD ... (1)$ [Since Sum of Co-interior angles is $180$]
Now, $\angle\text{BAC}=\angle\text{ACD}=57^\circ$ $\Rightarrow BA\ ||\ EF ... (2)$ [Alternative angles are equal]
From $(1)$ and $(2)$ $AB\ ||\ EF$ [Since, Lines parallel to the same line are parallel to each other] Hence proved.
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