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MATHS 3 Marks Question

Lines And Angles · Rajasthan Board (RBSE) STD 9 (Rajasthan - English Medium). 28 questions.

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Question 13 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°.
Its complement will be (90° - x°) and its supplement will be (180° - x°).
Supplement of thrice of the angle = (180° - 3x°)
According to the given information:
(90° - x°) = (180° - 3x°)
3x - x = 180 - 90
2x = 90
x = 45
Thus, the measure of the angle is 45°.
The measure of the angle is 45°.
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Question 23 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°.
Also,
The six spokes form a complete angle, that is 360°.
Therefore,
x + x + x + x + x + x = 360°
6x = 360°
$\text{x}=\frac{360^\circ}{6}$
x = 60°
Hence, the measure of the angle between two adjacent spokes measures 60°.
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Question 33 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 43 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 53 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 63 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 73 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 83 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 93 Marks
Define supplementary angles.
Answer
supplementary Angles: Two angles, the sum of whose measures is 180°, are called suppplementary angles.
Thus,
angles $\angle\text{BAC}$ and $\angle\text{DAC}$ are aupplementary angles.
If x + y = 180°

Example 1: Angles of measure 50° and 130° are supplementary angles, because
50° + 130° = 180°
Example 2: Angles of measure 60° and 120° are supplementary angles, because
60° + 120° = 180°
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Question 103 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 113 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°
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Question 123 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 133 Marks
In figure, OA and OB are opposite rays:
If x = 25°, 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 143 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]
⇒ ECD = 180 - 130 = 50
Also BA || CD
⇒ BAC = ACD = 70 [alternative angles]
But, ACE + ECD = 70
⇒ ACE = 70 - 50 = 20
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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
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 183 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 193 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 203 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°.
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Question 213 Marks
An angle is equal to five times its complement. Determine its measure.
Answer
Let the complement of the required angle measures x°.
Therefore, the required angle becomes 5x°.
Since, the angles are complementary.
Thus, their sum must be equal to 90°.
Or we can say that :
x + 5x = 90°
6x = 90°
$\text{x}=\frac{90^\circ}{6}$
x = 15°
Hence, the required angle becomes:
5x = 5(15°)
= 75°.
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Question 223 Marks
An angle is equal to its supplement. Determine its measure.
Answer
Let the supplement of the angle be x°.
According the given statement, the required angle is equal to its supplement, therefore, the required angle becomes x°.
Sine both the angles are supplementary, therefore, their sum must be equal to 180°.
Or we can say that:
x + x = 180°
2x = 180°
$\text{x}=\frac{180^\circ}{2}$
x = 90°
Hence, the required angle measures 90°.
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Question 233 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]
⇒ 270 + AOD = 360
⇒ AOD = 360 - 270
⇒ AOD = 90
Now,
AOD + BOD = 180 [Linear pair]
90 + BOD = 180
⇒ BOD = 180 - 90
⇒ BOD = 90
AOD = BOC = 90 [Vertically opposite angles]
BOD = AOC = 90 [Vertically opposite angles]
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Question 243 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 253 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 263 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,
$\angle\text{BAC}=\angle\text{ACD}=57^\circ$
⇒ 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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Question 273 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 283 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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