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Question (1):
Write complement of 42o, 87o, 20o.
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Answer:
Angles 48o, 3o, 70o are the complements of the angles 42o, 87o, 20o respectively.
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Question (2):
Write the supplement of 108o, 93o, 170o.
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Answer:
Angles 72o, 87o, 10o are the supplements of the angles 108o, 93o, 170o respectively.
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Question (3):
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Answer:
2x+112=360
2x=360-112=248
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Question (4):
In the adjoining figure, OA and OB are opposite rays.
(i) If x=75o, what is the value of y?
(ii) If y=110o, what is the value of x?
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Answer:
(i) If x=75o then
x+y=180o ( Linear pair)
75o+y=180o
y=180o-75o
y=105o
(ii) If y=110o, then
x+y=180o (  Linear pair)
x+110o=180o
x=180o-110o
x=70o
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Question (5):
Determine the value of x.
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Answer:
4x+2x=180o ( Linear pair)
6x=180o
x = 30o
 4x = 4 x 30o
= 120o
2x = 2 x 30o
= 60o
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Question (6):
In the given figure,  form a linear pair. If a - b = 80o, find the values of 'a' and 'b'.
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Answer:
Adding (i) and (ii),
a + b + a - b = 1800 + 800
2a = 260o
a = 130o
Substituting a = 130o in (i)
a + b = 180o
130 + b = 180o
b = 180o - 130o
b = 50o
 a = 130o and b = 50o
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Question (7):
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Answer:
 
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Question (8):
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Answer:
 
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Question (9):
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Answer:
 ( straight angle)
Divide both sides by 2,
 
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Question (10):
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Answer:
 ...(i) ( they form a linear pair)
 ...(ii) ( Linear pair)
Equating (i) and (ii),
 
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Question (11):
In the given figure, each of the angles AOC and AOB are right angles. Show that BOC is a line.
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Answer:
=180o
Since the sum of angles formed by them is 180o, BOC is a line.
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Question (12):
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Answer:
Multiplying by 2,
Hence A, O and B are collinear points.
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Question (13):
Prove that the bisectors of a pair of consecutive angles of two parallel lines are at right angles to one another.
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Answer:
AB||CD and EFGH is a transversal.
 ...(i) (consecutive interior angles)
= 90o
i.e., the bisectors of a pair of consecutive interior angles are at right angles.
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Question (14):
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Answer:
= 60o
Therefore m and n are parallel.
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Question (15):
Determine all the angles from 1 to 8.
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Answer:
 
 
= 36o
 
=108o
=72o
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Question (16):
If two lines are intersected in such a way that bisectors of a pair of corresponding angles are parallel, show that the two lines are parallel.
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Answer:
iii) 
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Question (17):
A transversal intersects two given lines in such a way that the interior angles on the same side of transversal are equal. Show by means of a figure that the lines need not be parallel. State the condition under which the two lines will be parallel.
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Answer:
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Question (18):
If the angles of a triangle are in the ratio 2:3:4, determine the three angles.
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Answer:
Let the three angles of the triangle be 2xo, 3xo and 4xo.
 2xo+3xo+4xo= 180 o (Angle sum property )
9xo= 180o
 
xo= 20o
 The angles measure 2xo= 2 x 20o
= 40o
3xo= 3 x 20o
= 60o
4xo= 4 x 20o
= 80o
The three angles of the triangle in the ratio 2:3:4 are 40o, 60o, 80o.
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Question (19):
Can a triangle have
i) two right angles?
ii) two obtuse angles?
iii) two acute angles?
iv) all angles more than 60o?
v) all angles less than 60o?
vi) all angles equal to 60o?
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Answer:
i) No. The sum of those two angles itself becomes 180o.
ii) No. The sum of two obtuse angles will always be greater than 180o.
iii) Yes. The sum of two acute angles will always be less than 180o.
iv) No. The sum of all three angles will be more than 180o .
v) No. The sum of 3 angles will be less than 180o .
Vi) Yes. The sum of 3 angles, each of 60o amounts to 180o.
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Question (20):
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Answer:
(Exterior angle = sum of interior opposite angles)
(Exterior angle = sum of interior opposite angles)
Adding (i) and (ii),
 
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Question (21):
In the adjoining figure, m and n are two plane mirrors perpendicular to each other. Show that the incident ray CA is parallel to the reflected ray BD.
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Answer:
m and n are mirrors placed perpendicular to each other.
Similarly,
In right angled triangle BOA,
Multiplying by 2, we get
From (i) and (ii),
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Question (22):
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Answer:
Adding (i) and (ii)
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Question (23):
In the adjoining figure, l || m and p || q. Show that  and  are supplementary.
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Answer:
l || m and p || q
l || m
 ...(i) ( corresponding angles)
p || q
  ...(ii) (Interior angles are supplementary)
From (i) and (ii) we have,
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Question (24):
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Answer:
AE||BF
 
But these are alternate angles
 AE||BF.
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Question (25):
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Answer:
 ( sum of angles around a point)
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Question (26):
If lines AB, AC, AD and AE are parallel to a line l. What can be said about the points A, B, C, D and E?
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Answer:
AB, AC, AD, AE are parallel to l
l is a line, A is a fixed point not on the line l.
According to parallel axiom there is only one line through A which is parallel to l.
 A, B, C, D and E are collinear points.
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Question (27):
In the adjoining figure, rays OA, OB, OC, OD and OE have the common point O. Show that
 .
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Answer:
OA, OB, OC, OD, OE are rays with the common end point O.
By adding (i) and (ii), we get
 
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