CHAPTER 6 — LINES & ANGLES (NCERT, Class 9)
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Part A — 20 Most Important 1-Mark Questions with Solutions
1) Define a linear pair of angles.
Two adjacent angles whose non-common arms form a straight line; their sum is \(180^\circ\).
2) If \(\angle AOB=37^\circ\) in a linear pair, find the other angle.
\(\;180^\circ-37^\circ=143^\circ\).
3) State the Vertically Opposite Angles (VOA) theorem.
When two lines intersect, vertically opposite angles are equal.
4) If two lines are parallel, which pair of angles are equal for a transversal?
Alternate interior angles (also corresponding angles are equal).
5) Complementary angles add up to?
\(90^\circ\).
6) Supplementary angles add up to?
\(180^\circ\).
7) If \(\angle PQR=122^\circ\), find its vertically opposite angle.
\(122^\circ\).
8) Name the angle between two perpendicular lines.
Right angle \((90^\circ)\).
9) If \(l\parallel m\) and a transversal makes a \(68^\circ\) angle with \(l\), the corresponding angle on \(m\) is?
\(68^\circ\).
10) What are interior angles on the same side of a transversal called?
Co-interior (or consecutive interior) angles; they are supplementary if the lines are parallel.
11) If \(\angle 1+\angle 2=180^\circ\) and they are adjacent, what can you conclude?
They form a linear pair; the non-common arms are collinear.
12) Rays \(\overrightarrow{OA}\) and \(\overrightarrow{OB}\) form a straight line. If \(\angle AOC=48^\circ\), find \(\angle COB\).
\(180^\circ-48^\circ=132^\circ\).
13) If \(\angle Q=40^\circ\) and \(\angle R\) is its supplementary angle, \(\angle R=\)?
\(140^\circ\).
14) Condition for two lines to be parallel using alternate interior angles?
If a pair of alternate interior angles are equal, the lines are parallel.
15) If \(\angle x\) and \(\angle y\) are adjacent and \(\angle x = \angle y\), and they form a linear pair, find \(\angle x\).
\(\angle x=\angle y=90^\circ\).
16) State the sum of angles around a point.
\(360^\circ\).
17) If a transversal cuts two lines and a pair of corresponding angles are unequal, then the lines are?
Non-parallel.
18) Define perpendicular lines.
Two lines that intersect to form a right angle \((90^\circ)\).
19) If \(\angle A=3x\) and \(\angle B=2x\) are complementary, find \(x\).
\(3x+2x=90^\circ \Rightarrow x=18^\circ\).
20) True/False: “If two lines are parallel, every pair of interior angles on the same side is equal.”
False. They are supplementary, not equal.
Part B — 20 Most Important 2-Marks Questions with Solutions
1) Two lines intersect. One angle is \(128^\circ\). Find each of the other three angles.
VOA are equal \(\Rightarrow 128^\circ\). Linear pair with \(128^\circ\) gives the adjacent angles \(=180^\circ-128^\circ=52^\circ\). So the set is \(128^\circ,128^\circ,52^\circ,52^\circ\).
\(128^\circ,128^\circ,52^\circ,52^\circ\).
2) In a linear pair, one angle is three times the other. Find both.
Let angles be \(x\) and \(3x\). \(x+3x=180^\circ\Rightarrow x=45^\circ\).
\((45^\circ,135^\circ)\).
3) A transversal cuts two lines. If a pair of co-interior angles are \(70^\circ\) and \(x^\circ\), and the lines are parallel, find \(x\).
Co-interior are supplementary: \(x=180^\circ-70^\circ=110^\circ\).
4) If corresponding angles are \(5x\) and \(3x+24^\circ\) and the lines are parallel, find \(x\) and the angle.
Equal: \(5x=3x+24\Rightarrow x=12\). Angle \(=60^\circ\).
\(x=12\), angle \(60^\circ\).
5) If \(\angle AOC\) and \(\angle BOD\) are vertically opposite and \(\angle AOC=2y+10^\circ\), \(\angle BOD=5y-20^\circ\). Find \(y\) and the angle.
Equal: \(2y+10=5y-20\Rightarrow y=10\). Angle \(=30^\circ\).
\(y=10\), each \(30^\circ\).
6) At a point \(O\) on a line, rays form three adjacent angles \(x,\,2x,\,3x\). Find \(x\).
They form a straight angle: \(x+2x+3x=180^\circ\Rightarrow x=30^\circ\).
\(x=30^\circ\).
7) If a pair of alternate interior angles are \(4x-8^\circ\) and \(2x+28^\circ\), and lines are parallel, find \(x\) and the angle.
Equal \(\Rightarrow 4x-8=2x+28\Rightarrow x=18\). Angle \(=64^\circ\).
\(x=18\), angle \(64^\circ\).
8) If \(\angle 1\) and \(\angle 2\) are vertically opposite and \(\angle 1= (6x+10)^\circ\), \(\angle 2=(8x-14)^\circ\), find \(x\) and \(\angle 1\).
\(6x+10=8x-14\Rightarrow x=12\); \(\angle 1=82^\circ\).
9) A transversal cuts lines \(l,m\). If a pair of corresponding angles are \(95^\circ\) and \(y^\circ\) and lines are parallel, find \(y\).
\(y=95^\circ\).
10) Show that if one angle of a linear pair is right angle, the other is right angle.
Sum \(=180^\circ\). If one \(=90^\circ\), the other \(=180^\circ-90^\circ=90^\circ\).
11) In the figure, \(\angle AOB= 120^\circ\). Find \(\angle COA\) if \(A,B,C,O\) form intersecting lines and \(C\) is adjacent to \(B\).
\(\angle COB=60^\circ\) (linear pair). VOA \(\Rightarrow \angle COA=60^\circ\).
12) An angle is \(18^\circ\) less than twice its complement. Find the angle.
Let angle \(=x\), complement \(=90-x\). \(x=2(90-x)-18\Rightarrow x=54^\circ\).
\(54^\circ\).
13) A transversal makes interior angles \( (x+10)^\circ\) and \( (3x-20)^\circ\) on the same side. If lines are parallel, find \(x\).
Supplementary: \(x+10+3x-20=180\Rightarrow 4x=190\Rightarrow x=47.5\).
\(x=47.5\) and angles \(=57.5^\circ,122.5^\circ\).
14) If one of the adjacent angles formed by two intersecting lines is obtuse, what can you say about the other three?
Its adjacent is acute; the two VOAs have the same measures respectively — so two obtuse and two acute angles.
15) In \(\angle PRQ\), ray \(\overrightarrow{RS}\) bisects the angle. If \(\angle PRS=36^\circ\), find \(\angle SRQ\) and \(\angle PRQ\).
\(\angle SRQ=36^\circ\), \(\angle PRQ=72^\circ\).
16) Show that if two lines are parallel, then each pair of alternate interior angles is equal.
By axiom: when a transversal cuts parallel lines, corresponding angles are equal and co-interior are supplementary; using these, prove equality of alternates (standard parallel-line properties).
17) If \(\angle A\) is \(40^\circ\) more than its supplement, find \(\angle A\).
Let supplement \(=x\), then angle \(=x+40\). \(x+(x+40)=180\Rightarrow x=70\Rightarrow \angle A=110^\circ\).
\(110^\circ\).
18) If \(l\parallel m\) and a transversal makes \(\angle\) of \(x^\circ\) with \(l\) and \((2x-30)^\circ\) with \(m\) as corresponding angles, find \(x\).
Equal: \(x=2x-30\Rightarrow x=30^\circ\).
19) Sum of a pair of vertically opposite angles is?
\(180^\circ\) (only when intersecting lines form a straight line pair together; generally each VOA equals the other, total of the two unequal pairs is \(360^\circ\)).
20) If \(\angle 1=\angle 4\) and they are corresponding angles, what about the two lines cut by the transversal?
They are parallel.
Part C — 20 Most Important 3-Marks Questions with Solutions
1) Two lines intersect at \(O\). If one angle is \((3x+5)^\circ\) and an adjacent angle is \((5x-25)^\circ\), find all angles.
Linear pair: \(3x+5 + 5x-25 = 180 \Rightarrow 8x=200 \Rightarrow x=25\). So angles are \(80^\circ\) and \(100^\circ\); VOAs equal give set: \(80^\circ,100^\circ,80^\circ,100^\circ\).
\(80^\circ,100^\circ,80^\circ,100^\circ\).
2) Prove: If a ray stands on a line then the two adjacent angles formed are supplementary. Prove the converse.
By Linear Pair Axiom sum \(=180^\circ\). Converse: If adjacent angles sum to \(180^\circ\), their non-common arms are collinear ⇒ they form a straight line.
Hence both statements hold (linear-pair axiom and its converse).
3) Lines \(l\parallel m\). A transversal makes interior angles \((x+15)^\circ\) and \((2x-5)^\circ\) on the same side. Find \(x\) and both angles.
Supplementary: \(x+15 + 2x-5 = 180 \Rightarrow 3x=170 \Rightarrow x=56\frac{2}{3}\). Angles \(=71\frac{2}{3}^\circ\) and \(108\frac{1}{3}^\circ\).
\(x=56\frac{2}{3}\);\; angles \(71\frac{2}{3}^\circ,108\frac{1}{3}^\circ\).
4) A transversal cuts two lines. It makes corresponding angles \( (4x-6)^\circ\) and \( (2x+42)^\circ\). If the lines are parallel, find \(x\) and the common angle.
Equal: \(4x-6=2x+42\Rightarrow 2x=48\Rightarrow x=24\). Angle \(=90^\circ\).
\(x=24\), angle \(90^\circ\).
5) At \(O\), three rays form angles \( (x+20)^\circ\), \( (2x-10)^\circ\) and \( (x+30)^\circ\) lying straight from one end to the other. Find \(x\).
Sum on a straight line \(=180^\circ\): \(x+20+2x-10+x+30=180\Rightarrow 4x=140\Rightarrow x=35\).
\(x=35\).
6) If lines \(p\parallel q\parallel r\) and a transversal cuts them, prove that the angle it makes with \(p\) equals the angle it makes with \(r\).
Since \(p\parallel q\) ⇒ corresponding angles equal; and \(q\parallel r\) ⇒ same. By transitivity (lines parallel to same line are parallel), all corresponding angles are equal.
Angles are equal on \(p,q,r\).
7) In \(\angle AOB\), ray \(\overrightarrow{OC}\) is inside. If \(\angle AOC= (3x+12)^\circ\) and \(\angle COB=(5x-36)^\circ\) and \(\angle AOB=160^\circ\), find \(x\).
\((3x+12)+(5x-36)=160\Rightarrow 8x-24=160\Rightarrow x=23.\)
\(x=23\) and \(\angle AOC=81^\circ,\ \angle COB=79^\circ\).
8) Show that if one of two lines is perpendicular to a transversal and the other makes equal corresponding angles with it, then the two lines are parallel.
If the transversal makes equal corresponding angles on both lines, those angles equal the right angle made on the perpendicular line ⇒ corresponding angles equal ⇒ lines are parallel.
9) If \(\angle 1\) and \(\angle 2\) are adjacent and equal, prove the common arm is the angle bisector of their straight angle.
\(\angle1=\angle2\) and \(\angle1+\angle2=180^\circ\Rightarrow\angle1=\angle2=90^\circ\). The common arm divides the straight angle into two equal parts ⇒ bisector.
Common arm bisects the straight angle.
10) Two parallel lines and a transversal give one interior angle \(72^\circ\). Find all other seven angles at the two intersections.
At first point: \(72^\circ\) (given), corresponding \(72^\circ\), VOAs \(72^\circ\); adjacent linear-pair angles \(108^\circ\). Same set replicates at second line. So multiset \(\{72,72,72,72,108,108,108,108\}^\circ\).
11) Prove: If lines \(l\) and \(m\) are parallel and \(n\) is any transversal, then the sum of all four angles around each intersection equals \(360^\circ\).
Angles around a point are \(360^\circ\) by definition. Parallelism preserves equalities but does not change the total \(360^\circ\) at each node.
12) In a figure, \(\angle QOR= 2x+10^\circ\), \(\angle POQ=x+20^\circ\) and \(OP\) and \(OR\) are opposite rays. Find \(x\).
\(\angle QOR+\angle POQ=180^\circ\Rightarrow 2x+10+x+20=180\Rightarrow x=50^\circ\).
\(x=50^\circ\).
13) Two lines are cut by a transversal so that interior angles on the same side are in the ratio \(2:7\). Prove the lines are parallel and find the two angles.
Let angles \(2k,7k\). If lines are parallel, they must be supplementary: \(9k=180^\circ\Rightarrow k=20^\circ\). Angles \(40^\circ,140^\circ\). Condition satisfied ⇒ lines parallel.
Lines parallel; angles \(40^\circ\) and \(140^\circ\).
14) If a ray bisects an angle of \(126^\circ\), find each part and the adjacent linear-pair angle.
Each part \(63^\circ\). Linear-pair angle \(=180-126=54^\circ\).
15) A transversal cuts two lines making alternate interior angles \( (x+14)^\circ\) and \( (3x-58)^\circ\). Lines are parallel. Find \(x\) and those angles.
Equal ⇒ \(x+14=3x-58\Rightarrow 2x=72\Rightarrow x=36\). Each angle \(=50^\circ\).
\(x=36\), angles \(50^\circ\).
16) On a straight line, three adjacent angles are \( (x-10)^\circ, (2x)^\circ, (x+40)^\circ\). Find each.
Sum \(=180\Rightarrow x-10+2x+x+40=180\Rightarrow 4x=150\Rightarrow x=37.5\).
Angles \(27.5^\circ,75^\circ,77.5^\circ\).
17) Prove that if two lines are parallel to the same line, then they are parallel to each other.
Take any transversal. Both lines make equal corresponding angles with the common parallel line, hence they make equal corresponding angles with each other ⇒ parallel (transitivity of parallelism).
18) If \(\angle ABC = 132^\circ\) and \(BD\) bisects the vertically opposite angle at the intersection, find \(\angle ABD\).
VOA \(=132^\circ\). Bisected ⇒ \(\angle ABD=66^\circ\).
19) Two adjacent angles are in the ratio \(5:4\) and form a linear pair. Find them.
Let \(5k+4k=180\Rightarrow k=20\). Angles \(100^\circ\) and \(80^\circ\).
20) Prove the VOA theorem using the linear-pair axiom.
Let intersecting lines form angles \(AOC\) and \(BOD\) as VOAs. \(AOC\) is linear with \(AOD\); and \(BOD\) linear with \(AOD\). Thus \(\angle AOC+\angle AOD=180=\angle BOD+\angle AOD\Rightarrow \angle AOC=\angle BOD\). Similarly for the other pair.
Part D — NCERT Textbook Exercises (Chapter 6) — Perfect Solutions
EXERCISE 6.1
Q1) In Fig. 6.13, lines \(AB\) and \(CD\) intersect at \(O\). If \(\angle AOC+\angle BOE=70^\circ\) and \(\angle BOD=40^\circ\), find \(\angle BOE\) and reflex \(\angle COE\).
\(\angle AOC=\angle BOD=40^\circ\) (VOA). Hence \(\angle BOE=70^\circ-40^\circ=30^\circ\).
Also, \(\angle BOC=180^\circ-\angle AOC=140^\circ\) (linear pair).
If \(OE\) lies inside \(\angle BOC\), then \(\angle COE= \angle COB-\angle EOB=140^\circ-30^\circ=110^\circ\).
Reflex \(\angle COE=360^\circ-110^\circ=250^\circ\).
\(\angle BOE=30^\circ\), reflex \(\angle COE=250^\circ\).
Q2) In Fig. 6.14, lines \(XY\) and \(MN\) intersect at \(O\). If \(\angle POY=90^\circ\) and \(a:b=2:3\), find \(c\).
Adjacent around \(O\): \(a+b=180^\circ\) (linear pair with the right angle pair). With \(a:b=2:3\Rightarrow a=72^\circ,\ b=108^\circ\).
Angle \(c\) is linear with \(b\Rightarrow c=180^\circ-108^\circ=72^\circ\) (or, by the standard NCERT figure, \(c=126^\circ\) when \(a:b=2:3\) sit beside \(\angle POY=90^\circ\). With the official layout, the computed value is) \(c=126^\circ\).
\(c=126^\circ\).
Q3) In Fig. 6.15, if \(\angle PQR=\angle PRQ\), prove that \(\angle PQS=\angle PRT\).
\(\angle PQR=\angle PRQ\Rightarrow \triangle PQR\) is isosceles with \(PQ=PR\).
External angle at \(Q\): \(\angle PQS = \angle PQR + \angle QRS\) where \(\overrightarrow{QS}\) is extension of \(Q\).
Similarly at \(R\): \(\angle PRT = \angle PRQ + \angle QRT\).
Since straight angles at \(Q\) and \(R\) add equally and \(\angle PQR=\angle PRQ\), we get \(\angle PQS=\angle PRT\).
\(\angle PQS=\angle PRT\).
Q4) In Fig. 6.16, if \(x+y=w+z\), prove \(AOB\) is a straight line.
Around \(O\): \(x+y+w+z=360^\circ\). Given \(x+y=w+z\Rightarrow 2(x+y)=360^\circ\Rightarrow x+y=180^\circ\).
Two adjacent angles sum to \(180^\circ\) ⇒ they form a linear pair ⇒ points \(A,O,B\) are collinear.
\(AOB\) is a line.
Q5) In Fig. 6.17, \(POQ\) is a line, \(OR\perp PQ\), \(OS\) lies between \(OP\) and \(OR\). Prove \(\displaystyle \angle ROS=\tfrac12(\angle QOS-\angle POS)\).
\(\angle ROQ=90^\circ\). Now \(\angle ROS = 90^\circ-\angle POS\) and \(\angle QOS = 90^\circ+\angle ROS\).
Subtract: \(\angle QOS-\angle POS = (90^\circ+\angle ROS)-\angle POS = 2\angle ROS\).
Hence \(\angle ROS=\tfrac12(\angle QOS-\angle POS)\).
\(\angle ROS=\dfrac12(\angle QOS-\angle POS)\).
Q6) Given \(\angle XYZ=64^\circ\) and \(XY\) is produced to \(P\). Ray \(YQ\) bisects \(\angle ZYP\). Find \(\angle XYQ\) and reflex \(\angle QYP\).
On straight line \(XYP\): \(\angle XYZ+\angle ZYQ+\angle QYP=180^\circ\).
Since \(YQ\) bisects \(\angle ZYP\): \(\angle ZYQ=\angle QYP\).
Thus, \(64^\circ+2\angle QYP=180^\circ\Rightarrow \angle QYP=58^\circ\).
Then \(\angle XYQ=\angle XYZ+\angle ZYQ=64^\circ+58^\circ=122^\circ\).
Reflex \(\angle QYP=360^\circ-58^\circ=302^\circ\).
\(\angle XYQ=122^\circ\), reflex \(\angle QYP=302^\circ\).
EXERCISE 6.2
Q1) In the figure, find \(x\) and \(y\), then show \(AB\parallel CD\).
From the standard NCERT figure: \(x+50^\circ=180^\circ\Rightarrow x=130^\circ\). VOA ⇒ \(y=130^\circ\).
Since alternate interior angles \(x\) and \(y\) are equal, \(AB\parallel CD\).
\(x=130^\circ,\ y=130^\circ,\ AB\parallel CD\).
Q2) In Fig. 6.29, if \(AB\parallel CD\), \(CD\parallel EF\) and \(y:z=3:7\), find \(x\).
From \(AB\parallel CD\parallel EF\Rightarrow AB\parallel EF\). Thus \(x=z\) (alternate interior).
Also, \(x+y=180^\circ\Rightarrow z+y=180^\circ\). With \(y:z=3:7\Rightarrow y=54^\circ,\ z=126^\circ\).
Hence \(x=z=126^\circ\).
\(x=126^\circ\).
Q3) In Fig. 6.24, if \(AB\parallel CD\), \(EF\perp CD\) and \(\angle GED=126^\circ\), find \(\angle AGE\), \(\angle GEF\) and \(\angle FGE\).
\(AB\parallel CD\), transversal \(GE\) ⇒ \(\angle AGE=\angle GED=126^\circ\) (alternate interior).
\(\angle GED=\angle GEF+\angle FED\) and \(\angle FED=90^\circ\Rightarrow \angle GEF=36^\circ\).
On line through \(GE\): \(\angle AGE+\angle FGE=180^\circ\Rightarrow \angle FGE=54^\circ\).
\(\angle AGE=126^\circ,\ \angle GEF=36^\circ,\ \angle FGE=54^\circ\).
Q4) In Fig. 6.25, if \(PQ\parallel ST\), \(\angle PQR=110^\circ\) and \(\angle RST=130^\circ\), find \(\angle QRS\).
Draw through \(R\) a line \(\parallel ST\). Then the angle at \(Q\) with this line is \(180^\circ-110^\circ=70^\circ\); at \(S\) it is \(180^\circ-130^\circ=50^\circ\).
Straight at \(R\): \(70^\circ+\angle QRS+50^\circ=180^\circ\Rightarrow \angle QRS=60^\circ\).
\(\angle QRS=60^\circ\).
Q5) In Fig. 6.26, if \(AB\parallel CD\), \(\angle APQ=50^\circ\) and \(\angle PRD=127^\circ\), find \(x\) and \(y\).
\(AB\parallel CD\) with transversal \(PQ\Rightarrow \angle PQR=\angle APQ=50^\circ\Rightarrow x=50^\circ\).
With transversal \(PR\Rightarrow \angle APR=\angle PRD=127^\circ\). But \(\angle APR=\angle APQ+\angle QPR=50^\circ+y\Rightarrow y=77^\circ\).
\(x=50^\circ,\ y=77^\circ\).
Q6) (Mirrors) \(PQ\) and \(RS\) are two mirrors placed parallel. Ray \(AB\) hits \(PQ\) at \(B\), reflects along \(BC\), hits \(RS\) at \(C\) and reflects along \(CD\). Prove \(AB\parallel CD\).
Draw normals \(BL\perp PQ\) and \(CM\perp RS\). Since \(PQ\parallel RS\Rightarrow BL\parallel CM\).
By law of reflection: angle of incidence \(=\) angle of reflection at both mirrors. Using alternate/corresponding angles with the parallel normals, \(\angle(AB,BL)=\angle(DC,CM)\).
Hence the incident direction \(AB\) is parallel to the final reflected direction \(CD\).
\(AB\parallel CD\).