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Lines and Angles
Notes
MODULE - 3
Geometry
Mathematics Secondary Course 261
10
LINES AND ANGLES
Observe the top of your desk or table. Now move your hand on the top of your table. It
gives an idea of a plane. Its edges give an idea of a line, its corner, that of a point and the
edges meeting at a corner give an idea of an angle.
OBJECTIVES
Afterstudyingthislesson,youwillbeableto
• illustrate the concepts of point, line, plane, parallel lines and interesecting lines;
• recognise pairs of angles made by a transversal with two or more lines;
• verify that when a ray stands on a line, the sum of two angles so formed is 1800
;
• verify that when two lines intersect, vertically opposite angles are equal;
• verify that if a transversal intersects two parallel lines then corresponding angles
in each pair are equal;
• verify that if a transversal intersects two parallel lines then
(a) alternate angles in each pair are equal
(b) interior angles on the same side of the transversal are supplementary;
• prove that the sum of angles of a triangle is 1800
• verify that the exterior angle of a triangle is equal to the sum of two interior
opposite angles; and
• explain the concept of locus and exemplify it through daily life situations.
• find the locus of a point equidistent from (a) two given points, (b) two intersecting
lines.
• solve problems based on starred result and direct numerical problems based on
unstarred results given in the curriculum.
Notes
MODULE - 3
Geometry
Lines and Angles
Mathematics Secondary Course262
EXPECTED BACKGROUND KNOWLEDGE
• point,line,plane,intersectinglines,raysandangles.
• parrallellines
10.1 POINT, LINE AND ANGLE
In earlier classes, you have studied about a point, a line, a plane and an angle. Let us
quicklyrecalltheseconcepts.
Point : If we press the tip of a pen or pencil on a piece of paper, we get a fine dot, which
is called a point.
.
. .
Fig. 10.1
Apoint is used to show the location and is represented by capital lettersA, B, C etc.
10.1.1 Line
Now mark two pointsAand B on your note book. Join them with the help of a ruler or a
scale and extend it on both sides. This gives us a straight line or simply a line.
Fig. 10.2
In geometry, a line is extended infinitely on both sides and is marked with arrows to give
thisidea.Alineisnamedusinganytwopointsonit,viz,ABorbyasinglesmallletterl,m
etc. (See fig. 10.3)
Fig. 10.3
ThepartofthelinebetweentwopointsAandBiscalledalinesegmentandwillbenamed
AB.
Observe that a line segment is the shortest path between two pointsAand B. (See Fig.
10.4)
A
B
C
Lines and Angles
Notes
MODULE - 3
Geometry
Mathematics Secondary Course 263
Fig. 10.4
10.1.2 Ray
Ifwemark apointXanddrawaline,startingfromitextendinginfinitelyinonedirection
only, then we get a ray XY.
Fig. 10.5
XiscalledtheinitialpointoftherayXY.
10.1.3 Plane
If we move our palm on the top of a table, we get an idea of a plane.
Fig. 10.6
Similarly, floor of a room also gives the idea of part of a plane.
Planealsoextendsinfintelylengthwiseandbreadthwise.
Mark a pointAon a sheet of paper.
Howmanylinescanyoudrawpassingthoughthispoint?Asmanyasyouwish.
Fig. 10.7
Notes
MODULE - 3
Geometry
Lines and Angles
Mathematics Secondary Course264
C
In fact, we can draw an infinite number of lines through a point.
Take another point B, at some distance fromA.We can again draw an infinite number of
linespassingthroughB.
Fig. 10.8
Out of these lines, how many pass through both the pointsAand B? Out of all the lines
passingthroughA,onlyonepassesthroughB.Thus,onlyonelinepassesthroughboththe
pointsAandB.Weconcludethat oneandonlyonelinecanbedrawnpassingthrough
two given points.
Now we take three points in plane.
Fig. 10.9
We observe that a line may or may not pass through the three given points.
If a line can pass through three or more points, then these points are said to be collinear.
For example the pointsA, B and C in the Fig. 10.9 are collinear points.
If a line cannot be drawn passing through all three points (or more points), then they are
said to be non-collinear. For example points P, Q and R, in the Fig. 10.9, are non-
collinearpoints.
Sincetwopointsalwayslieonaline,wetalkofcollinearpointsonlywhentheirnumberis
three or more.
Let us now take two distinct linesAB and CD in a plane.
Fig. 10.10
Howmanypointscantheyhaveincommon?Weobservethattheselinescanhave.either
(i) one point in common as in Fig. 10.10 (a) and (b). [In such a case they are called
Lines and Angles
Notes
MODULE - 3
Geometry
Mathematics Secondary Course 265
intersecting lines] or (ii) no points in common as in Fig. 10.10 (c). In such a case they are
called parrallel lines.
Now observe three (or more) distinct lines in plane.
(a) (b) (c) (d)
Fig. 10.11
Whatarethepossibilities?
(i) They may interest in more than one point as in Fig. 10.11 (a) and 10.11 (b).
or (ii) They may intesect in one point only as in Fig. 10.11 (c). In such a case they are
calledconcurrentlines.
or (iii) They may be non intersecting lines parrallel to each other as in Fig. 10.11 |(d).
10.1.4 Angle
Mark a point O and draw two rays OA and OB starting from O. The figure we get is
called an angle. Thus, an angle is a figure consisting of two rays starting from a common
point.
Fig. 10.11(A)
ThisanglemaybenamedasangleAOBorangleBOAorsimplyangleO;andiswrittenas
∠ΑΟΒ or ∠ΒΟΑ or ∠Ο. [see Fig. 10.11A]
An angle is measured in degrees. If we take any point O and draw two rays starting from
itinoppositedirectionsthenthemeasureofthisangleistakentobe1800
degrees,written
as 1800
.
Fig. 10.12
B O A
Notes
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Mathematics Secondary Course266
This measure divided into 180 equal parts is called one degree (written as 1o
).
Angle obtained by two opposite rays is called a straight angle.
An angle of 900
is called a right angle, for example ∠ΒΟΑ or ∠BOC is a right angle in
Fig. 10.13.
Fig. 10.13
Twolinesorraysmakingarightanglewitheachotherarecalled perpendicularlines.In
Fig. 10.13 we can say OA is perpendicular to OB or vice-versa.
An angle less than 900
is called an acute angle. For example ∠POQ is an acute angle in
Fig. 10.14(a).
An angle greater than 900
but less than 1800
is called an obtuse angle. For example,
∠XOY is an obtuse angle in Fig. 10.14(b).
Fig. 10.14
10.2 PAIRS OF ANGLES
Fig. 10.15
(a) (b)
Lines and Angles
Notes
MODULE - 3
Geometry
Mathematics Secondary Course 267
C
(a) (b)
(a)
(b)
Observe the two angles ∠1 and ∠2 in each of the figures in Fig. 10.15. Each pair has a
common vertex O and a common side OA in between OB and OC. Such a pair of angles
is called a ‘pair of adjacent angles’.
Fig. 10.16
Observe the angles in each pair in Fig. 10.16[(a) and (b)]. They add up to make a total of
90o
.
Apairofangles,whosesumis90o
,iscalledapairofcomplementaryangles.Eachangle
is called the complement of the other.
Fig. 10.17
Again observe the angles in each pair in Fig. 10.17[(a) and (b)].
These add up to make a total of 180o
.
A pair of angles whose sum is 1800
, is called a pair of supplementary angles.
Each such angle is called the supplement of the other.
Draw a lineAB. From a point C on it draw a ray CD making two angles ∠X and ∠Y.
Fig. 10.18
Notes
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Mathematics Secondary Course268
If we measure ∠X and ∠Y and add, we will always find the sum to be 1800
, whatever be
the position of the ray CD. We conclude
If a ray stands on a line then the sum of the two adjacent angles so formed
is 180o
.
The pair of angles so formed as in Fig. 10.18 is called a linear pair of angles.
Note that they also make a pair of supplementary angles.
Draw two intersecting linesAB and CD, intersecting each other at O.
Fig. 10.19
∠AOC and ∠DOB are angles opposite to each other. These make a pair of vertically
oppposite angles. Measure them.You will always find that
∠AOC = ∠DOB.
∠AOD and ∠BOC is another pair of vertically opposite angles. On measuring, you will
againfindthat
∠AOD = ∠BOC
We conclude :
If two lines intersect each other, the pair of vertically opposite angles are
equal.
Anactivityforyou.
Attach two strips with a nail or a pin as shown in the figure.
Fig. 10.20
Lines and Angles
Notes
MODULE - 3
Geometry
Mathematics Secondary Course 269
Rotate one of the strips, keeping the other in position and observe that the pairs of verti-
callyoppositeanglesthusformedarealwaysequal.
A line which intersects two or more lines at distinct points is called a transversal. For
example line l in Fig. 10.21 is a transversal.
Fig. 10.21
Whenatransversalintersectstwolines,eightanglesareformed.
Fig. 10.22
Theseanglesinpairsareveryimportantinthestudyofpropertiesofparallellines.Someof
theusefulpairsareasfollows:
(a) ∠1 and ∠5 is a pair of corresponding angles. ∠2 and ∠6, ∠3 and ∠7 and ∠4 and
∠8 are other pairs of corresponding angles.
(b) ∠3and ∠6isapairofalternateangles. ∠4and∠5isanotherpairofalternateangles.
(c) ∠3 and ∠5 is a pair of interior angles on the same side of the transversal.
∠4 and ∠6 is another pair of interior angles.
InFig.10.22above,linesmandnarenotparallel;assuch,theremaynotexistanyrelation
between the angles of any of the above pairs. However, when lines are parallel, there are
someveryusefulrelationsinthesepairs,whichwestudyinthefollowing:
Whenatransversalintersectstwoparallellines,eightanglesareformed,whateverbethe
positionofparallellinesorthetransversal.
Notes
MODULE - 3
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Lines and Angles
Mathematics Secondary Course270
Fig. 10.23
Ifwemeasuretheangles,weshallalwysfindthat
∠1 = ∠5, ∠2 = ∠6, ∠3 = ∠7 and ∠4 = ∠8
that is, angles in each pair of corresponding angles are equal.
Also ∠3 = ∠6 and ∠4 = ∠5
that is, angles in each pair of alternate angle are equal.
Also, ∠3 + ∠5 = 180o
and ∠4 + ∠6 = 180o
.
Hence we conclude :
When a transversal intersects two parallel lines, then angles in
(i) each pair of corresponding angles are equal
(ii) each pair of alternate angles are equal
(iii) each pair of interior angles on the same side of the transversal are supple-
mentary,
You may also verify the truth of these results by drawing a pair of parallel lines (using
paralleledgesofyourscale)andatransversalandmeasuringanglesineachofthesepairs.
Converse of each of these results is also true.To verify the truth of the first converse, we
draw a lineAB and mark two points C and D on it.
Fig. 10.24
At C and D, we construct two angles ACF and CDH equal to each other, say 50o
, as
shown in Fig. 10.24. On producing EF and GH on either side, we shall find that they do
not intersect each other, that is, they are parallel.
Lines and Angles
Notes
MODULE - 3
Geometry
Mathematics Secondary Course 271
In a similar way, we can verify the truth of the other two converses.
Hence we conclude that
When a transversal inersects two lines in such a way that angles in
(i) any pair of corresponding angles are equal
or (ii) any pair of alternate angles are equal
or (iii) any pair of interior angles on the same side of transversal are supple-
mentary then the two lines are parallel.
Example 10.1 : Choose the correct answwer out of the alternative options in the follow-
ingmultiplechoicequestions.
Fig. 10.25
(i) In Fig. 10.25, ∠FOD and ∠BOD are
(A)supplementaryangles (B)complementaryangles
(C)verticallyoppositeangles (D) a linear pair of angles Ans. (B)
(ii) In Fig. 10.25, ∠COE and ∠BOE are
(A)complementaryangles (B)supplementaryangles
(C) a linear pair (D) adjacent angles Ans. (D)
(iii) In Fig. 10.25, ∠BOD is equal to
(A) xo
(B) (90 + x)o
(C) (90 – x)o
(D) (180 – x)o
Ans (C)
(iv) Anangleis4timesitssupplement;theangleis
(A) 39o
(B) 72o
(C) 108o
(D) 144o
Ans (D)
Notes
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Lines and Angles
Mathematics Secondary Course272
(v) What value of x will makeACB a straight angle in Fig. 10.26
Fig. 10.26
(A) 30o
(B) 40o
(C) 50o
(D) 60o
Ans (C)
Fig. 10.27
In the above figure, l is parallel to m and p is parallel to q.
(vi) ∠3 and ∠5 form a pair of
(A)Alternateangles (B)interiorangles
(C)verticallyopposite (D)correspondingangles AAns (D)
(vii) In Fig. 10.27, if ∠1 = 80o
, then ∠6 is equal to
(A) 80o
(B) 90o
(C) 100o
(D) 110o
Ans (C)
Fig. 10.28
(viii)In Fig. 10.28, OA bisects ∠LOB, OC bisects ∠MOB and ∠AOC = 900
. Show that
the points L, O and M are collinear.
Lines and Angles
Notes
MODULE - 3
Geometry
Mathematics Secondary Course 273
Solution : ∠BOL = 2 ∠BOA ...(i)
and ∠BOM = 2 ∠BOC ...(ii)
Adding (i) and (ii), ∠BOL + ∠BOM = 2 ∠BOA + 2∠BOC
∴ ∠LOM = 2[∠BOA + ∠BOC]
= 2 ×90o
= 180o
= a straight angle
∴ L, O and M are collinear.
CHECK YOUR PROGRESS 10.1.
1. Choose the correct answer out of the given alternatives in the following multiple
choicequestions:
Fig. 10.29
In Fig. 10.29,AB || CD and PQ intersects them at R and S respectively.
(i) ∠ARS and ∠BRS form
(A) a pair of alternate angles
(B) alinearpair
(C) a pair of corresponding angles
(D) apairofverticallyoppositeangles
(ii) ∠ARS and ∠RSD form a pair of
(A)Alternateangles (B)Verticallyoppositeangles
(C)Correspondingangles (D)Interiorangles
(iii) If ∠PRB = 60o
, then ∠QSC is
(A) 120o
(B) 60o
Notes
MODULE - 3
Geometry
Lines and Angles
Mathematics Secondary Course274
(C) 30o
(D) 90o
Fig. 10.30
(iv) In Fig. 10.30 above,AB and CD intersect at O. ∠COB is equal to
(A) 36o
(B) 72o
(C) 108o
(D) 144o
Fig. 10.31
2. In Fig. 10.31 above,AB is a straight line. Find x
3. In Fig. 10.32 below, l is parallel to m. Find angles 1 to 7.
Fig. 10.32
10.3 TRIANGLE, ITS TYPES AND PROPERTIES
Triangle is the simplest polygon of all the closed figures formed in a plane by three line
segments.
(5x+10)o
5x
o
72o
Lines and Angles
Notes
MODULE - 3
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Mathematics Secondary Course 275
Fig. 10.33
Itisaclosedfigureformedbythreelinesegmentshavingsixelements,namelythreeangles
(i) ∠ABC or ∠B (ii) ∠ACB or ∠C (iii) ∠CAB or ∠Aand three sides : (iv)AB (v) BC
(vi)CA
It is named as Δ ABC or Δ BAC or Δ CBAand read as triangleABC or triangle BAC or
triangleCBA.
10.3.1 Types of Triangles
Trianglescanbeclassifiedintodifferenttypesintwoways.
(a) On the basis of sides
(i) (ii) (iii)
Fig. 10.34
(i) Equilateral triangle : a triangle in which all the three sides are equal is called an
equilateral trangle. [ΔABC in Fig. 10.34(i)]
(ii) Isosceles triangle : Atriangle in which two sides are equal is called an isosceles
triangle.[ΔDEFinFig.10.34(ii)]
(iii) Scalene triangle : Atriangle in which all sides are of different lengths, is called a
sclene triangle [ΔLMNinFig.10.34(iii)]
(b) On the basis of angles :
(i) (ii) (iii)
Fig. 10.35
Notes
MODULE - 3
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(i) Obtuseangledtriangle:Atriangleinwhichoneoftheanglesisanobtuseangleis
calledanobtuseangledtriangleorsimplyobtusetriangle[ΔPQRisFig.10.35(i)]
(ii) Rightangledtriangle:Atriangleinwhichoneoftheanglesisarightangleiscalled
a right angled triangle or right triangle. [Δ UVW in Fig. 10.35(ii)]
(iii) Acuteangledtriangle:Atriangleinwhichallthethreeanglesareacuteiscalledan
acute angled triangle or acute triangle [Δ XYZ in Fig. 10.35(iii)
Now we shall study some important properties of angles of a triangle.
10.3.2Angle Sum Property of a Triangle
We draw two triangles and measure their angles.
Fig. 10.36
In Fig. 10.36 (a), ∠A = 80o
, ∠B = 40o
and ∠C = 60o
∴ ∠A + ∠B + ∠C = 80o
+ 40o
+ 60o
= 180o
In Fig. 10.36(b), ∠P = 30o
, ∠Q = 40o
, ∠R = 110o
∴ ∠P + ∠Q + ∠R = 30o
+ 40o
+ 110o
= 180o
What do you observe? Sum of the angles of triangle in each case in 1800
.
Wewillprovethisresultinalogicalwaynamingitasatheorem.
Theorem : The sum of the three angles of triangle is 180o
.
Fig. 10.37
Given : AtriangleABC
To Prove : ∠A + ∠B + ∠C = 1800
Construction : ThroughA, draw a line DE parallel to BC.
Proof : Since DE is parallel to BC andAB is a transversal.
Lines and Angles
Notes
MODULE - 3
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Mathematics Secondary Course 277
∴ ∠B = ∠DAB (Pairofalternateangles)
Similarly ∠C=∠EAC (Pairofalternateangles)
∴ ∠B + ∠C = ∠DAB + ∠EAC ...(1)
Now adding ∠A to both sides of (1)
∠A + ∠B + ∠C = ∠A + ∠DAB + ∠EAC
= 180o
(Anglesmakingastraightangle)
10.3.3 ExteriorAngles of a Triangle
Let us produce the side BC of ΔABC to a point D.
Fig. 10.38
In Fig. 10.39, observe that there are six exterior angles of the ΔABC, namely ∠1, ∠2,
∠3, ∠4, ∠5 and ∠6.
Fig. 10.39
In Fig. 10.38, ∠ACD so obtained is called an exterior angle of the ΔABC. Thus,
The angle formed by a side of the triangle produced and another side of the
triangle is called an exterior angle of the triangle.
Corresponding to an exterior angle of a triangle, there are two interior opposite angles.
Interior opposite angles are the angles of the triangle not forming a linear
pair with the given exterior angle.
For example in Fig. 10.38, ∠A and ∠B are the two interior opposite angles correspond-
ing to the exterior angleACD of ΔABC. We measure these angles.
∠A = 60o
∠B = 50o
Notes
MODULE - 3
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Lines and Angles
Mathematics Secondary Course278
and ∠ACD = 110o
We observe that ∠ACD = ∠A + ∠B.
This observation is true in general.
Thus, we may conclude :
An exterior angle of a triangle is equal to the sum of the two interior
opposite angles.
Examples 10.3 : Choose the correct answer out of the given alternatives in the fol-
lowing multiple choice questions:
(i) Which of the following can be the angles of a triangle?
(A) 65o
, 45o
and 80o
(B) 90o
, 30o
and 61o
(C) 60o
, 60o
and 59o
(D) 60o
, 60o
and 60o
. Ans (D)
Fig. 10.40
(ii) In Fig. 10.40 ∠A is equal to
(A) 30o
(B) 35o
(C) 45o
(D) 75o
Ans (C)
(iii) In a triangle, one angle is twice the other and the third angle is 600
. Then the
largest angle is
(A) 60o
(B) 80o
(C) 100o
(D) 120o
Ans (B)
Example 10.4:
Fig. 10.41
In Fig. 10.41, bisctors of ∠PQR and ∠PRQ intersect each other at O. Prove that
∠QOR = 90o
+
2
1
∠P.
Lines and Angles
Notes
MODULE - 3
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Mathematics Secondary Course 279
Solution : ∠QOR = 180o
–
2
1
[∠PQR + ∠PRQ)]
= 180o
–
2
1
(∠PQR + ∠PRQ)
= 180o
–
2
1
(180ο
– ∠P)
= 180o
– 90o
+
2
1
∠P = 90o
+
2
1
∠P
CHECK YOUR PROGRESS 10.2
1. Choosethecorrectansweroutofgivenalternativesinthefollowingmultiplechoice
questions:
(i) Atrianglecanhave
(A)Tworightangles (B)Two obtuse angles
(C)At the most two acute angles (D)Allthreeacuteangles
(ii)Inarighttriangle,oneexterioranglesis1200
,Thesmallestangleofthetrianglesis
(A) 20o
(B) 300
(C) 40o
(D) 600
(iii)
Fig. 10.42
In Fig. 10.42, CD is parallel to BA. ∠ACB is equal to
(A) 55o
(B) 60o
(C) 65o
(D) 70o
2. The angles of a triangle are in the ratio 2 : 3 : 5, find the three angles.
3. Prove that the sum of the four angles of a quadrilateral is 360o
.
Notes
MODULE - 3
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Lines and Angles
Mathematics Secondary Course280
4. In Fig. 10.43, ABCD is a trapezium such that AB||DC. Find ∠D and ∠C and
verify that sum of the four angles is 360o
.
Fig. 10.43
5. Prove that if one angle of a triangle is equal to the sum of the other two angles,
then it is a right triangle.
6. In Fig. 10.44, ABC is triangle such that ∠ABC = ∠ACB. Find the angles of the
triangle.
Fig. 10.44
10.4 LOCUS
During the game of cricket, when a player hits the ball, it describes a path, before being
caughtortouchingtheground.
Fig. 10.44
The path described is called Locus.
A figure in geometry is a result of the path traced by a point (or a very small particle)
movingundercertainconditions.
Forexample:
(1) Given two parallel lines l and m, also a point P between them equidistant from both
thelines.
Lines and Angles
Notes
MODULE - 3
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Mathematics Secondary Course 281
Fig. 10.45
If the particle moves so that it is equidistant from both the lines, what will be its path?
Fig. 10.46
The path traced by P will be a line parallel to both the lines and exactly in the middle of
them as in Fig. 10.46.
(2) Given a fixed point O and a point P at a fixed distance d.
Fig. 10.47
If the point P moves in a plane so that it is always at a constant distance d from the
fixed point O, what will be its path?
Fig. 10.48
The path of the moving point P will be a circle as shown in Fig. 10.48.
(3) Place a small piece of chalk stick or a pebble on top of a table. Strike it hard with a
pencil or a stick so that it leaves the table with a certain speed and observe its path
after it leaves the table.
.
. . . .
.
.
Notes
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Mathematics Secondary Course282
Fig. 10.49
The path traced by the pebble will be a curve (part of what is known as a parabola) as
shown in Fig. 10.49.
Thus,locusofapointmovingundercertainconditionsisthepathorthegeometricalfigure,
everypointofwhichsatisfiesthegivenconditon(s).
10.4.1 Locus of a point equidistant from two given points
LetAand B be the two given points.
.
. .
Fig. 10.50
We have to find the locus of a point Psuch that PA= PB.
JointAB.MarkthemindpointofABasM.Clearly,Misapointwhichisequidistantfrom
Aand B. Mark another point Pusing compasses such that PA= PB. Join PM and extend
itonbothsides.Usingapairofdividerorascale,itcaneasilybeverifiedthateverypoint
onPMisequidistantfromthepointsAandB.Also,ifwetakeanyotherpointQnotlying
on line PM, then QA ≠ QB.
Also ∠AMP = ∠BMP = 90o
That is, PM is the perpendicular bisector ofAB.
Fig. 10.51
A
P
B
Q
Lines and Angles
Notes
MODULE - 3
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Mathematics Secondary Course 283
Thus,wemayconcludethefollowing:
The locus of a point equidistant from two given poitns is the perpendicular
bisector of the line segment joining the two points.
Activityforyou:
Mark two pointsAand B on a sheet of paper and join them. Fold the paper along mid-
point ofAB so thatAcoincides with B. Make a crease along the line of fold.This crease
isastraightline.ThisisthelocusofthepointequidistantfromthegivenpointsAandB.It
can be easily checked that very point on it is equidistant fromAand B.
10.4.2 Locus of a point equidistant from two lines intersecting at O
LetAB and CD be two given lines intersecting at O.
Fig. 10.52
We have to find the locus of a point Pwhich is equidistant from bothAB and CD.
Draw bisectors of ∠BOD and ∠BOC.
Fig. 10.53
IfwetakeanypointPonanybisector lorm,wewillfindperpendiculardistancesPLand
PM of Pfrom the linesAB and CD are equal.
thatis, PL = PM
If we take any other point, say Q, not lying on any bisector l or m, then QL will not be
equal to QM.
Thus,wemayconclude:
The locus of a point equidistant from two intersecting lines is the pair of
lines, bisecting the angles formed by the given lines.
D
B
Notes
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Mathematics Secondary Course284
Activity for you :
Draw two linesAB and CD intersecting at O, on a sheet of paper. Fold the paper through
O so thatAO falls on CO and OD falls on OB and mark the crease along the fold. Take
a piont P on this crease which is the bisector of ∠BOD and check using a set square that
PL = PM
Fig. 10.54
Inasimilarwayfindtheotherbisectorbyfoldingagainandgettingcrease2.Anypointon
this crease 2 is also equidistant from both the lines.
Example 10.5 : Find the locus of the centre of a circle passing through two given points.
Solution : Let the two given points beAand B.We have to find the position or positions
of centre O of a circle passing throughAand B.
.
. .
Fig. 10.55
Point O must be equidistant from both the pointsAand B.As we have already learnt, the
locus of the point O will be the perpendicular bisector ofAB.
Fig. 10.56
A
O
B
Lines and Angles
Notes
MODULE - 3
Geometry
Mathematics Secondary Course 285
CHECK YOU PROGRESS 10.3
1. Find the locus of the centre of a circle passing through three given points A, B
and C which are non-collinear.
2. There are two villages certain distance apart.Awell is to be dug so that it is equidis-
tant from the two villages such that its distance from each village is not more than
the distance between the two villages. Representing the villages by pointsAand B
and the well by point P. show in a diagram the locus of the point P.
3. Two straight roadsAB and CD are intersecting at a point O.An observation post
is to be constructred at a distance of 1 km from O and equidistant from the roads
AB and CD. Show in a diagram the possible locations of the post.
4. Find the locus of a point which is always at a distance 5 cm from a given lineAB.
LET US SUM UP
• A line extends to inifinity on both sides and a line segment is only a part of it
between two points.
• Two distinct lines in a plane may either be intersecting or parallel.
• If three or more lines intersect in one point only then they are called cocurrent lines.
• Two rays starting from a common point form an angle.
• A pair of angles, whose sum is 900
is called a pair of complementary angles.
• A pair of angles whose sum is 1800
is called a pair of supplementary angles.
• If a ray stands on a line then the sum of the two adjacent angles, so formed is 1800
• If two lines intersect each other the pairs of vertically opposite angles are equal
• When a transversal intersects two parallel lines, then
(i) corresponding angles in a pair are equal.
(ii) alternate angles are equal.
(iii) interior angles on the same side of the transversal are supplementary.
• The sum of the angles of a triangle is 1800
• An exterior angle of a triangle to equal to the sum of the two interior opposite angles
• Locus of a point equidistant from two given points is the perpendicular bisector
of the line segment joing the points.
Notes
MODULE - 3
Geometry
Lines and Angles
Mathematics Secondary Course286
• Thelocusofapointequidistantfromtheintersectinglinesisthepairoflines,bisecting
theangleformedbythegivenlines.
TERMINAL EXERCISE
1. In Fig. 10.57, if x = 42, then determine (a) y (b) ∠AOD
Fig. 10.57
2.
Fig. 10.58
In the above figure p, q and r are parallel lines intersected by a transversal l atA, B
and C respectively. Find ∠1 and ∠2.
3. The sum of two angles of a triangle is equal to its third angle. Find the third angle.
Whattypeoftriangleisit?
4.
Fig. 10.59
In Fig. 10.59, sides of Δ ABC have been produced as shown. Find the angles of the
triangle.
Lines and Angles
Notes
MODULE - 3
Geometry
Mathematics Secondary Course 287
5.
Fig. 10.60
In Fig. 10.60, sides AB, BC and CA of the triangle ABC have been produced as
shown. Show that the sum of the exterior angles so formed is 360o
.
6.
Fig. 10.61
In Fig. 10.61ABC is a triangle in which bisectors of ∠B and ∠C meet at O. Show
that ∠BOC = 125o
.
7.
Fig. 10.62
In Fig. 10.62 above, find the sum of the angles, ∠A , ∠F , ∠C , ∠D , ∠B and ∠E.
8.
Fig. 10.63
Notes
MODULE - 3
Geometry
Lines and Angles
Mathematics Secondary Course288
In Fig. 10.63 in Δ ABC,AD is perpendicular to BC andAE is bisector of ∠ΒAC.
Find∠DAE,
9.
Fig. 10.64
In Fig. 10.64 above, in Δ PQR, PT is bisector of ∠P and QR is produced to S.
Show that ∠PQR + ∠PRS = 2 ∠PTR.
10. Prove that the sum of the (interior) angles of a pentagon is 5400
.
11. Findthelocusofapointequidistantfromtwoparallellines l andmatadistanceof5
cm from each other.
12. Find the locus of a point equidistant from pointsAand B and also equidistant from
raysAB andAC of Fig. 10.65.
Fig. 10.65
ANSWERS TO CHECK YOUR PROGRESS
10.1
1. (i) (B) (ii) (A) (iii) (B) (iv) (C)
2. x = 170
.
3. ∠1 = ∠3 = ∠4 = ∠6 = 110o
and ∠2 = ∠5 = ∠7 = 70o
.
10.2
1. (i) (D) (ii) (B) (iii) (B)
2. 36o
, 54o
and 90o
4. ∠D = 140o
and ∠C = 110o
6. ∠ABC = 45o
, ∠ACB = 45o
and ∠A = 90o
.
..A
B
C
Lines and Angles
Notes
MODULE - 3
Geometry
Mathematics Secondary Course 289
10.3
1. Only a point, which is the point of intersection of perpendicular bisectors of AB
and BC.
2. Let the villages beAand B, then locus will be the line segment PQ, perpendicular
bisector ofAB such that
AP = BP = QA = QB = AB
Fig. 10.65
3. Possible locations will be four points two points P and Q on the bisector of ∠AOC
and two points R and S on the bisector of ∠BOC.
Fig. 10.66
4. Two on either side of AB and lines parallel to AB at a distance of 5 cm from AB.
ANSWERS TO TERMINAL EXERCISE
1. (a) y = 27 (b) = 126o
2. ∠1 = 48o
and ∠2 = 132o
3. Third angle = 90o
, Right triangle 4. ∠A = 35o
, ∠B = 75o
∠C = 70o
7. 360o
8. 12o
11. A line parallel to locus l and m at a distance of 2.5 cm from each.
12. Point of intersection of the perpendicular bisector of AB and bisector of ∠BAC.

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Lines and angles /GEOMETRY

  • 1. Lines and Angles Notes MODULE - 3 Geometry Mathematics Secondary Course 261 10 LINES AND ANGLES Observe the top of your desk or table. Now move your hand on the top of your table. It gives an idea of a plane. Its edges give an idea of a line, its corner, that of a point and the edges meeting at a corner give an idea of an angle. OBJECTIVES Afterstudyingthislesson,youwillbeableto • illustrate the concepts of point, line, plane, parallel lines and interesecting lines; • recognise pairs of angles made by a transversal with two or more lines; • verify that when a ray stands on a line, the sum of two angles so formed is 1800 ; • verify that when two lines intersect, vertically opposite angles are equal; • verify that if a transversal intersects two parallel lines then corresponding angles in each pair are equal; • verify that if a transversal intersects two parallel lines then (a) alternate angles in each pair are equal (b) interior angles on the same side of the transversal are supplementary; • prove that the sum of angles of a triangle is 1800 • verify that the exterior angle of a triangle is equal to the sum of two interior opposite angles; and • explain the concept of locus and exemplify it through daily life situations. • find the locus of a point equidistent from (a) two given points, (b) two intersecting lines. • solve problems based on starred result and direct numerical problems based on unstarred results given in the curriculum.
  • 2. Notes MODULE - 3 Geometry Lines and Angles Mathematics Secondary Course262 EXPECTED BACKGROUND KNOWLEDGE • point,line,plane,intersectinglines,raysandangles. • parrallellines 10.1 POINT, LINE AND ANGLE In earlier classes, you have studied about a point, a line, a plane and an angle. Let us quicklyrecalltheseconcepts. Point : If we press the tip of a pen or pencil on a piece of paper, we get a fine dot, which is called a point. . . . Fig. 10.1 Apoint is used to show the location and is represented by capital lettersA, B, C etc. 10.1.1 Line Now mark two pointsAand B on your note book. Join them with the help of a ruler or a scale and extend it on both sides. This gives us a straight line or simply a line. Fig. 10.2 In geometry, a line is extended infinitely on both sides and is marked with arrows to give thisidea.Alineisnamedusinganytwopointsonit,viz,ABorbyasinglesmallletterl,m etc. (See fig. 10.3) Fig. 10.3 ThepartofthelinebetweentwopointsAandBiscalledalinesegmentandwillbenamed AB. Observe that a line segment is the shortest path between two pointsAand B. (See Fig. 10.4) A B C
  • 3. Lines and Angles Notes MODULE - 3 Geometry Mathematics Secondary Course 263 Fig. 10.4 10.1.2 Ray Ifwemark apointXanddrawaline,startingfromitextendinginfinitelyinonedirection only, then we get a ray XY. Fig. 10.5 XiscalledtheinitialpointoftherayXY. 10.1.3 Plane If we move our palm on the top of a table, we get an idea of a plane. Fig. 10.6 Similarly, floor of a room also gives the idea of part of a plane. Planealsoextendsinfintelylengthwiseandbreadthwise. Mark a pointAon a sheet of paper. Howmanylinescanyoudrawpassingthoughthispoint?Asmanyasyouwish. Fig. 10.7
  • 4. Notes MODULE - 3 Geometry Lines and Angles Mathematics Secondary Course264 C In fact, we can draw an infinite number of lines through a point. Take another point B, at some distance fromA.We can again draw an infinite number of linespassingthroughB. Fig. 10.8 Out of these lines, how many pass through both the pointsAand B? Out of all the lines passingthroughA,onlyonepassesthroughB.Thus,onlyonelinepassesthroughboththe pointsAandB.Weconcludethat oneandonlyonelinecanbedrawnpassingthrough two given points. Now we take three points in plane. Fig. 10.9 We observe that a line may or may not pass through the three given points. If a line can pass through three or more points, then these points are said to be collinear. For example the pointsA, B and C in the Fig. 10.9 are collinear points. If a line cannot be drawn passing through all three points (or more points), then they are said to be non-collinear. For example points P, Q and R, in the Fig. 10.9, are non- collinearpoints. Sincetwopointsalwayslieonaline,wetalkofcollinearpointsonlywhentheirnumberis three or more. Let us now take two distinct linesAB and CD in a plane. Fig. 10.10 Howmanypointscantheyhaveincommon?Weobservethattheselinescanhave.either (i) one point in common as in Fig. 10.10 (a) and (b). [In such a case they are called
  • 5. Lines and Angles Notes MODULE - 3 Geometry Mathematics Secondary Course 265 intersecting lines] or (ii) no points in common as in Fig. 10.10 (c). In such a case they are called parrallel lines. Now observe three (or more) distinct lines in plane. (a) (b) (c) (d) Fig. 10.11 Whatarethepossibilities? (i) They may interest in more than one point as in Fig. 10.11 (a) and 10.11 (b). or (ii) They may intesect in one point only as in Fig. 10.11 (c). In such a case they are calledconcurrentlines. or (iii) They may be non intersecting lines parrallel to each other as in Fig. 10.11 |(d). 10.1.4 Angle Mark a point O and draw two rays OA and OB starting from O. The figure we get is called an angle. Thus, an angle is a figure consisting of two rays starting from a common point. Fig. 10.11(A) ThisanglemaybenamedasangleAOBorangleBOAorsimplyangleO;andiswrittenas ∠ΑΟΒ or ∠ΒΟΑ or ∠Ο. [see Fig. 10.11A] An angle is measured in degrees. If we take any point O and draw two rays starting from itinoppositedirectionsthenthemeasureofthisangleistakentobe1800 degrees,written as 1800 . Fig. 10.12 B O A
  • 6. Notes MODULE - 3 Geometry Lines and Angles Mathematics Secondary Course266 This measure divided into 180 equal parts is called one degree (written as 1o ). Angle obtained by two opposite rays is called a straight angle. An angle of 900 is called a right angle, for example ∠ΒΟΑ or ∠BOC is a right angle in Fig. 10.13. Fig. 10.13 Twolinesorraysmakingarightanglewitheachotherarecalled perpendicularlines.In Fig. 10.13 we can say OA is perpendicular to OB or vice-versa. An angle less than 900 is called an acute angle. For example ∠POQ is an acute angle in Fig. 10.14(a). An angle greater than 900 but less than 1800 is called an obtuse angle. For example, ∠XOY is an obtuse angle in Fig. 10.14(b). Fig. 10.14 10.2 PAIRS OF ANGLES Fig. 10.15 (a) (b)
  • 7. Lines and Angles Notes MODULE - 3 Geometry Mathematics Secondary Course 267 C (a) (b) (a) (b) Observe the two angles ∠1 and ∠2 in each of the figures in Fig. 10.15. Each pair has a common vertex O and a common side OA in between OB and OC. Such a pair of angles is called a ‘pair of adjacent angles’. Fig. 10.16 Observe the angles in each pair in Fig. 10.16[(a) and (b)]. They add up to make a total of 90o . Apairofangles,whosesumis90o ,iscalledapairofcomplementaryangles.Eachangle is called the complement of the other. Fig. 10.17 Again observe the angles in each pair in Fig. 10.17[(a) and (b)]. These add up to make a total of 180o . A pair of angles whose sum is 1800 , is called a pair of supplementary angles. Each such angle is called the supplement of the other. Draw a lineAB. From a point C on it draw a ray CD making two angles ∠X and ∠Y. Fig. 10.18
  • 8. Notes MODULE - 3 Geometry Lines and Angles Mathematics Secondary Course268 If we measure ∠X and ∠Y and add, we will always find the sum to be 1800 , whatever be the position of the ray CD. We conclude If a ray stands on a line then the sum of the two adjacent angles so formed is 180o . The pair of angles so formed as in Fig. 10.18 is called a linear pair of angles. Note that they also make a pair of supplementary angles. Draw two intersecting linesAB and CD, intersecting each other at O. Fig. 10.19 ∠AOC and ∠DOB are angles opposite to each other. These make a pair of vertically oppposite angles. Measure them.You will always find that ∠AOC = ∠DOB. ∠AOD and ∠BOC is another pair of vertically opposite angles. On measuring, you will againfindthat ∠AOD = ∠BOC We conclude : If two lines intersect each other, the pair of vertically opposite angles are equal. Anactivityforyou. Attach two strips with a nail or a pin as shown in the figure. Fig. 10.20
  • 9. Lines and Angles Notes MODULE - 3 Geometry Mathematics Secondary Course 269 Rotate one of the strips, keeping the other in position and observe that the pairs of verti- callyoppositeanglesthusformedarealwaysequal. A line which intersects two or more lines at distinct points is called a transversal. For example line l in Fig. 10.21 is a transversal. Fig. 10.21 Whenatransversalintersectstwolines,eightanglesareformed. Fig. 10.22 Theseanglesinpairsareveryimportantinthestudyofpropertiesofparallellines.Someof theusefulpairsareasfollows: (a) ∠1 and ∠5 is a pair of corresponding angles. ∠2 and ∠6, ∠3 and ∠7 and ∠4 and ∠8 are other pairs of corresponding angles. (b) ∠3and ∠6isapairofalternateangles. ∠4and∠5isanotherpairofalternateangles. (c) ∠3 and ∠5 is a pair of interior angles on the same side of the transversal. ∠4 and ∠6 is another pair of interior angles. InFig.10.22above,linesmandnarenotparallel;assuch,theremaynotexistanyrelation between the angles of any of the above pairs. However, when lines are parallel, there are someveryusefulrelationsinthesepairs,whichwestudyinthefollowing: Whenatransversalintersectstwoparallellines,eightanglesareformed,whateverbethe positionofparallellinesorthetransversal.
  • 10. Notes MODULE - 3 Geometry Lines and Angles Mathematics Secondary Course270 Fig. 10.23 Ifwemeasuretheangles,weshallalwysfindthat ∠1 = ∠5, ∠2 = ∠6, ∠3 = ∠7 and ∠4 = ∠8 that is, angles in each pair of corresponding angles are equal. Also ∠3 = ∠6 and ∠4 = ∠5 that is, angles in each pair of alternate angle are equal. Also, ∠3 + ∠5 = 180o and ∠4 + ∠6 = 180o . Hence we conclude : When a transversal intersects two parallel lines, then angles in (i) each pair of corresponding angles are equal (ii) each pair of alternate angles are equal (iii) each pair of interior angles on the same side of the transversal are supple- mentary, You may also verify the truth of these results by drawing a pair of parallel lines (using paralleledgesofyourscale)andatransversalandmeasuringanglesineachofthesepairs. Converse of each of these results is also true.To verify the truth of the first converse, we draw a lineAB and mark two points C and D on it. Fig. 10.24 At C and D, we construct two angles ACF and CDH equal to each other, say 50o , as shown in Fig. 10.24. On producing EF and GH on either side, we shall find that they do not intersect each other, that is, they are parallel.
  • 11. Lines and Angles Notes MODULE - 3 Geometry Mathematics Secondary Course 271 In a similar way, we can verify the truth of the other two converses. Hence we conclude that When a transversal inersects two lines in such a way that angles in (i) any pair of corresponding angles are equal or (ii) any pair of alternate angles are equal or (iii) any pair of interior angles on the same side of transversal are supple- mentary then the two lines are parallel. Example 10.1 : Choose the correct answwer out of the alternative options in the follow- ingmultiplechoicequestions. Fig. 10.25 (i) In Fig. 10.25, ∠FOD and ∠BOD are (A)supplementaryangles (B)complementaryangles (C)verticallyoppositeangles (D) a linear pair of angles Ans. (B) (ii) In Fig. 10.25, ∠COE and ∠BOE are (A)complementaryangles (B)supplementaryangles (C) a linear pair (D) adjacent angles Ans. (D) (iii) In Fig. 10.25, ∠BOD is equal to (A) xo (B) (90 + x)o (C) (90 – x)o (D) (180 – x)o Ans (C) (iv) Anangleis4timesitssupplement;theangleis (A) 39o (B) 72o (C) 108o (D) 144o Ans (D)
  • 12. Notes MODULE - 3 Geometry Lines and Angles Mathematics Secondary Course272 (v) What value of x will makeACB a straight angle in Fig. 10.26 Fig. 10.26 (A) 30o (B) 40o (C) 50o (D) 60o Ans (C) Fig. 10.27 In the above figure, l is parallel to m and p is parallel to q. (vi) ∠3 and ∠5 form a pair of (A)Alternateangles (B)interiorangles (C)verticallyopposite (D)correspondingangles AAns (D) (vii) In Fig. 10.27, if ∠1 = 80o , then ∠6 is equal to (A) 80o (B) 90o (C) 100o (D) 110o Ans (C) Fig. 10.28 (viii)In Fig. 10.28, OA bisects ∠LOB, OC bisects ∠MOB and ∠AOC = 900 . Show that the points L, O and M are collinear.
  • 13. Lines and Angles Notes MODULE - 3 Geometry Mathematics Secondary Course 273 Solution : ∠BOL = 2 ∠BOA ...(i) and ∠BOM = 2 ∠BOC ...(ii) Adding (i) and (ii), ∠BOL + ∠BOM = 2 ∠BOA + 2∠BOC ∴ ∠LOM = 2[∠BOA + ∠BOC] = 2 ×90o = 180o = a straight angle ∴ L, O and M are collinear. CHECK YOUR PROGRESS 10.1. 1. Choose the correct answer out of the given alternatives in the following multiple choicequestions: Fig. 10.29 In Fig. 10.29,AB || CD and PQ intersects them at R and S respectively. (i) ∠ARS and ∠BRS form (A) a pair of alternate angles (B) alinearpair (C) a pair of corresponding angles (D) apairofverticallyoppositeangles (ii) ∠ARS and ∠RSD form a pair of (A)Alternateangles (B)Verticallyoppositeangles (C)Correspondingangles (D)Interiorangles (iii) If ∠PRB = 60o , then ∠QSC is (A) 120o (B) 60o
  • 14. Notes MODULE - 3 Geometry Lines and Angles Mathematics Secondary Course274 (C) 30o (D) 90o Fig. 10.30 (iv) In Fig. 10.30 above,AB and CD intersect at O. ∠COB is equal to (A) 36o (B) 72o (C) 108o (D) 144o Fig. 10.31 2. In Fig. 10.31 above,AB is a straight line. Find x 3. In Fig. 10.32 below, l is parallel to m. Find angles 1 to 7. Fig. 10.32 10.3 TRIANGLE, ITS TYPES AND PROPERTIES Triangle is the simplest polygon of all the closed figures formed in a plane by three line segments. (5x+10)o 5x o 72o
  • 15. Lines and Angles Notes MODULE - 3 Geometry Mathematics Secondary Course 275 Fig. 10.33 Itisaclosedfigureformedbythreelinesegmentshavingsixelements,namelythreeangles (i) ∠ABC or ∠B (ii) ∠ACB or ∠C (iii) ∠CAB or ∠Aand three sides : (iv)AB (v) BC (vi)CA It is named as Δ ABC or Δ BAC or Δ CBAand read as triangleABC or triangle BAC or triangleCBA. 10.3.1 Types of Triangles Trianglescanbeclassifiedintodifferenttypesintwoways. (a) On the basis of sides (i) (ii) (iii) Fig. 10.34 (i) Equilateral triangle : a triangle in which all the three sides are equal is called an equilateral trangle. [ΔABC in Fig. 10.34(i)] (ii) Isosceles triangle : Atriangle in which two sides are equal is called an isosceles triangle.[ΔDEFinFig.10.34(ii)] (iii) Scalene triangle : Atriangle in which all sides are of different lengths, is called a sclene triangle [ΔLMNinFig.10.34(iii)] (b) On the basis of angles : (i) (ii) (iii) Fig. 10.35
  • 16. Notes MODULE - 3 Geometry Lines and Angles Mathematics Secondary Course276 (i) Obtuseangledtriangle:Atriangleinwhichoneoftheanglesisanobtuseangleis calledanobtuseangledtriangleorsimplyobtusetriangle[ΔPQRisFig.10.35(i)] (ii) Rightangledtriangle:Atriangleinwhichoneoftheanglesisarightangleiscalled a right angled triangle or right triangle. [Δ UVW in Fig. 10.35(ii)] (iii) Acuteangledtriangle:Atriangleinwhichallthethreeanglesareacuteiscalledan acute angled triangle or acute triangle [Δ XYZ in Fig. 10.35(iii) Now we shall study some important properties of angles of a triangle. 10.3.2Angle Sum Property of a Triangle We draw two triangles and measure their angles. Fig. 10.36 In Fig. 10.36 (a), ∠A = 80o , ∠B = 40o and ∠C = 60o ∴ ∠A + ∠B + ∠C = 80o + 40o + 60o = 180o In Fig. 10.36(b), ∠P = 30o , ∠Q = 40o , ∠R = 110o ∴ ∠P + ∠Q + ∠R = 30o + 40o + 110o = 180o What do you observe? Sum of the angles of triangle in each case in 1800 . Wewillprovethisresultinalogicalwaynamingitasatheorem. Theorem : The sum of the three angles of triangle is 180o . Fig. 10.37 Given : AtriangleABC To Prove : ∠A + ∠B + ∠C = 1800 Construction : ThroughA, draw a line DE parallel to BC. Proof : Since DE is parallel to BC andAB is a transversal.
  • 17. Lines and Angles Notes MODULE - 3 Geometry Mathematics Secondary Course 277 ∴ ∠B = ∠DAB (Pairofalternateangles) Similarly ∠C=∠EAC (Pairofalternateangles) ∴ ∠B + ∠C = ∠DAB + ∠EAC ...(1) Now adding ∠A to both sides of (1) ∠A + ∠B + ∠C = ∠A + ∠DAB + ∠EAC = 180o (Anglesmakingastraightangle) 10.3.3 ExteriorAngles of a Triangle Let us produce the side BC of ΔABC to a point D. Fig. 10.38 In Fig. 10.39, observe that there are six exterior angles of the ΔABC, namely ∠1, ∠2, ∠3, ∠4, ∠5 and ∠6. Fig. 10.39 In Fig. 10.38, ∠ACD so obtained is called an exterior angle of the ΔABC. Thus, The angle formed by a side of the triangle produced and another side of the triangle is called an exterior angle of the triangle. Corresponding to an exterior angle of a triangle, there are two interior opposite angles. Interior opposite angles are the angles of the triangle not forming a linear pair with the given exterior angle. For example in Fig. 10.38, ∠A and ∠B are the two interior opposite angles correspond- ing to the exterior angleACD of ΔABC. We measure these angles. ∠A = 60o ∠B = 50o
  • 18. Notes MODULE - 3 Geometry Lines and Angles Mathematics Secondary Course278 and ∠ACD = 110o We observe that ∠ACD = ∠A + ∠B. This observation is true in general. Thus, we may conclude : An exterior angle of a triangle is equal to the sum of the two interior opposite angles. Examples 10.3 : Choose the correct answer out of the given alternatives in the fol- lowing multiple choice questions: (i) Which of the following can be the angles of a triangle? (A) 65o , 45o and 80o (B) 90o , 30o and 61o (C) 60o , 60o and 59o (D) 60o , 60o and 60o . Ans (D) Fig. 10.40 (ii) In Fig. 10.40 ∠A is equal to (A) 30o (B) 35o (C) 45o (D) 75o Ans (C) (iii) In a triangle, one angle is twice the other and the third angle is 600 . Then the largest angle is (A) 60o (B) 80o (C) 100o (D) 120o Ans (B) Example 10.4: Fig. 10.41 In Fig. 10.41, bisctors of ∠PQR and ∠PRQ intersect each other at O. Prove that ∠QOR = 90o + 2 1 ∠P.
  • 19. Lines and Angles Notes MODULE - 3 Geometry Mathematics Secondary Course 279 Solution : ∠QOR = 180o – 2 1 [∠PQR + ∠PRQ)] = 180o – 2 1 (∠PQR + ∠PRQ) = 180o – 2 1 (180ο – ∠P) = 180o – 90o + 2 1 ∠P = 90o + 2 1 ∠P CHECK YOUR PROGRESS 10.2 1. Choosethecorrectansweroutofgivenalternativesinthefollowingmultiplechoice questions: (i) Atrianglecanhave (A)Tworightangles (B)Two obtuse angles (C)At the most two acute angles (D)Allthreeacuteangles (ii)Inarighttriangle,oneexterioranglesis1200 ,Thesmallestangleofthetrianglesis (A) 20o (B) 300 (C) 40o (D) 600 (iii) Fig. 10.42 In Fig. 10.42, CD is parallel to BA. ∠ACB is equal to (A) 55o (B) 60o (C) 65o (D) 70o 2. The angles of a triangle are in the ratio 2 : 3 : 5, find the three angles. 3. Prove that the sum of the four angles of a quadrilateral is 360o .
  • 20. Notes MODULE - 3 Geometry Lines and Angles Mathematics Secondary Course280 4. In Fig. 10.43, ABCD is a trapezium such that AB||DC. Find ∠D and ∠C and verify that sum of the four angles is 360o . Fig. 10.43 5. Prove that if one angle of a triangle is equal to the sum of the other two angles, then it is a right triangle. 6. In Fig. 10.44, ABC is triangle such that ∠ABC = ∠ACB. Find the angles of the triangle. Fig. 10.44 10.4 LOCUS During the game of cricket, when a player hits the ball, it describes a path, before being caughtortouchingtheground. Fig. 10.44 The path described is called Locus. A figure in geometry is a result of the path traced by a point (or a very small particle) movingundercertainconditions. Forexample: (1) Given two parallel lines l and m, also a point P between them equidistant from both thelines.
  • 21. Lines and Angles Notes MODULE - 3 Geometry Mathematics Secondary Course 281 Fig. 10.45 If the particle moves so that it is equidistant from both the lines, what will be its path? Fig. 10.46 The path traced by P will be a line parallel to both the lines and exactly in the middle of them as in Fig. 10.46. (2) Given a fixed point O and a point P at a fixed distance d. Fig. 10.47 If the point P moves in a plane so that it is always at a constant distance d from the fixed point O, what will be its path? Fig. 10.48 The path of the moving point P will be a circle as shown in Fig. 10.48. (3) Place a small piece of chalk stick or a pebble on top of a table. Strike it hard with a pencil or a stick so that it leaves the table with a certain speed and observe its path after it leaves the table. . . . . . . .
  • 22. Notes MODULE - 3 Geometry Lines and Angles Mathematics Secondary Course282 Fig. 10.49 The path traced by the pebble will be a curve (part of what is known as a parabola) as shown in Fig. 10.49. Thus,locusofapointmovingundercertainconditionsisthepathorthegeometricalfigure, everypointofwhichsatisfiesthegivenconditon(s). 10.4.1 Locus of a point equidistant from two given points LetAand B be the two given points. . . . Fig. 10.50 We have to find the locus of a point Psuch that PA= PB. JointAB.MarkthemindpointofABasM.Clearly,Misapointwhichisequidistantfrom Aand B. Mark another point Pusing compasses such that PA= PB. Join PM and extend itonbothsides.Usingapairofdividerorascale,itcaneasilybeverifiedthateverypoint onPMisequidistantfromthepointsAandB.Also,ifwetakeanyotherpointQnotlying on line PM, then QA ≠ QB. Also ∠AMP = ∠BMP = 90o That is, PM is the perpendicular bisector ofAB. Fig. 10.51 A P B Q
  • 23. Lines and Angles Notes MODULE - 3 Geometry Mathematics Secondary Course 283 Thus,wemayconcludethefollowing: The locus of a point equidistant from two given poitns is the perpendicular bisector of the line segment joining the two points. Activityforyou: Mark two pointsAand B on a sheet of paper and join them. Fold the paper along mid- point ofAB so thatAcoincides with B. Make a crease along the line of fold.This crease isastraightline.ThisisthelocusofthepointequidistantfromthegivenpointsAandB.It can be easily checked that very point on it is equidistant fromAand B. 10.4.2 Locus of a point equidistant from two lines intersecting at O LetAB and CD be two given lines intersecting at O. Fig. 10.52 We have to find the locus of a point Pwhich is equidistant from bothAB and CD. Draw bisectors of ∠BOD and ∠BOC. Fig. 10.53 IfwetakeanypointPonanybisector lorm,wewillfindperpendiculardistancesPLand PM of Pfrom the linesAB and CD are equal. thatis, PL = PM If we take any other point, say Q, not lying on any bisector l or m, then QL will not be equal to QM. Thus,wemayconclude: The locus of a point equidistant from two intersecting lines is the pair of lines, bisecting the angles formed by the given lines. D B
  • 24. Notes MODULE - 3 Geometry Lines and Angles Mathematics Secondary Course284 Activity for you : Draw two linesAB and CD intersecting at O, on a sheet of paper. Fold the paper through O so thatAO falls on CO and OD falls on OB and mark the crease along the fold. Take a piont P on this crease which is the bisector of ∠BOD and check using a set square that PL = PM Fig. 10.54 Inasimilarwayfindtheotherbisectorbyfoldingagainandgettingcrease2.Anypointon this crease 2 is also equidistant from both the lines. Example 10.5 : Find the locus of the centre of a circle passing through two given points. Solution : Let the two given points beAand B.We have to find the position or positions of centre O of a circle passing throughAand B. . . . Fig. 10.55 Point O must be equidistant from both the pointsAand B.As we have already learnt, the locus of the point O will be the perpendicular bisector ofAB. Fig. 10.56 A O B
  • 25. Lines and Angles Notes MODULE - 3 Geometry Mathematics Secondary Course 285 CHECK YOU PROGRESS 10.3 1. Find the locus of the centre of a circle passing through three given points A, B and C which are non-collinear. 2. There are two villages certain distance apart.Awell is to be dug so that it is equidis- tant from the two villages such that its distance from each village is not more than the distance between the two villages. Representing the villages by pointsAand B and the well by point P. show in a diagram the locus of the point P. 3. Two straight roadsAB and CD are intersecting at a point O.An observation post is to be constructred at a distance of 1 km from O and equidistant from the roads AB and CD. Show in a diagram the possible locations of the post. 4. Find the locus of a point which is always at a distance 5 cm from a given lineAB. LET US SUM UP • A line extends to inifinity on both sides and a line segment is only a part of it between two points. • Two distinct lines in a plane may either be intersecting or parallel. • If three or more lines intersect in one point only then they are called cocurrent lines. • Two rays starting from a common point form an angle. • A pair of angles, whose sum is 900 is called a pair of complementary angles. • A pair of angles whose sum is 1800 is called a pair of supplementary angles. • If a ray stands on a line then the sum of the two adjacent angles, so formed is 1800 • If two lines intersect each other the pairs of vertically opposite angles are equal • When a transversal intersects two parallel lines, then (i) corresponding angles in a pair are equal. (ii) alternate angles are equal. (iii) interior angles on the same side of the transversal are supplementary. • The sum of the angles of a triangle is 1800 • An exterior angle of a triangle to equal to the sum of the two interior opposite angles • Locus of a point equidistant from two given points is the perpendicular bisector of the line segment joing the points.
  • 26. Notes MODULE - 3 Geometry Lines and Angles Mathematics Secondary Course286 • Thelocusofapointequidistantfromtheintersectinglinesisthepairoflines,bisecting theangleformedbythegivenlines. TERMINAL EXERCISE 1. In Fig. 10.57, if x = 42, then determine (a) y (b) ∠AOD Fig. 10.57 2. Fig. 10.58 In the above figure p, q and r are parallel lines intersected by a transversal l atA, B and C respectively. Find ∠1 and ∠2. 3. The sum of two angles of a triangle is equal to its third angle. Find the third angle. Whattypeoftriangleisit? 4. Fig. 10.59 In Fig. 10.59, sides of Δ ABC have been produced as shown. Find the angles of the triangle.
  • 27. Lines and Angles Notes MODULE - 3 Geometry Mathematics Secondary Course 287 5. Fig. 10.60 In Fig. 10.60, sides AB, BC and CA of the triangle ABC have been produced as shown. Show that the sum of the exterior angles so formed is 360o . 6. Fig. 10.61 In Fig. 10.61ABC is a triangle in which bisectors of ∠B and ∠C meet at O. Show that ∠BOC = 125o . 7. Fig. 10.62 In Fig. 10.62 above, find the sum of the angles, ∠A , ∠F , ∠C , ∠D , ∠B and ∠E. 8. Fig. 10.63
  • 28. Notes MODULE - 3 Geometry Lines and Angles Mathematics Secondary Course288 In Fig. 10.63 in Δ ABC,AD is perpendicular to BC andAE is bisector of ∠ΒAC. Find∠DAE, 9. Fig. 10.64 In Fig. 10.64 above, in Δ PQR, PT is bisector of ∠P and QR is produced to S. Show that ∠PQR + ∠PRS = 2 ∠PTR. 10. Prove that the sum of the (interior) angles of a pentagon is 5400 . 11. Findthelocusofapointequidistantfromtwoparallellines l andmatadistanceof5 cm from each other. 12. Find the locus of a point equidistant from pointsAand B and also equidistant from raysAB andAC of Fig. 10.65. Fig. 10.65 ANSWERS TO CHECK YOUR PROGRESS 10.1 1. (i) (B) (ii) (A) (iii) (B) (iv) (C) 2. x = 170 . 3. ∠1 = ∠3 = ∠4 = ∠6 = 110o and ∠2 = ∠5 = ∠7 = 70o . 10.2 1. (i) (D) (ii) (B) (iii) (B) 2. 36o , 54o and 90o 4. ∠D = 140o and ∠C = 110o 6. ∠ABC = 45o , ∠ACB = 45o and ∠A = 90o . ..A B C
  • 29. Lines and Angles Notes MODULE - 3 Geometry Mathematics Secondary Course 289 10.3 1. Only a point, which is the point of intersection of perpendicular bisectors of AB and BC. 2. Let the villages beAand B, then locus will be the line segment PQ, perpendicular bisector ofAB such that AP = BP = QA = QB = AB Fig. 10.65 3. Possible locations will be four points two points P and Q on the bisector of ∠AOC and two points R and S on the bisector of ∠BOC. Fig. 10.66 4. Two on either side of AB and lines parallel to AB at a distance of 5 cm from AB. ANSWERS TO TERMINAL EXERCISE 1. (a) y = 27 (b) = 126o 2. ∠1 = 48o and ∠2 = 132o 3. Third angle = 90o , Right triangle 4. ∠A = 35o , ∠B = 75o ∠C = 70o 7. 360o 8. 12o 11. A line parallel to locus l and m at a distance of 2.5 cm from each. 12. Point of intersection of the perpendicular bisector of AB and bisector of ∠BAC.