Exercise 14.1
Question 1:
Write down each pair of adjacent angles shown in Fig.
Answer 1:
Adjacent angles are the angles that have a common vertex and a common arm.
Following are the adjacent angles in the given figure:
∠DOC and ∠BOC∠COB and ∠BOA
Question 2:
In Fig., name all the pairs of adjacent angles.
Answer 2:
In figure (i), the adjacent angles are:
∠EBA and∠ABC∠ACB and ∠BCF∠BAC and ∠CAD
In figure (ii), the adjacent angles are:
∠BAD and ∠DAC
∠BDA and ∠CDA
Question 3:
In figure, write down: (i) each linear pair (ii) each pair of vertically opposite angles.
Answer 3:
(i) Two adjacent angles are said to form a linear pair of angles if their non-common arms are two opposite rays.
∠1 and ∠3
∠1 and ∠2
∠4 and ∠3
∠4 and ∠2
∠5 and ∠6
∠5 and ∠7
∠6 and ∠8
∠7 and ∠8
(ii) Two angles formed by two intersecting lines having no common arms are called vertically opposite angles.
∠1 and ∠4
∠2 and ∠3
∠5 and ∠8
∠6 and ∠7
Question 4:
Are the angles 1 and 2 given in Fig. adjacent angles?
Answer 4:
No, because they have no common vertex.
Question 5:
Find the complement of each of the following angles:
(i) 35°
(ii) 72°
(iii) 45°
(iv) 85°
Answer 5:
Two angles are called complementary angles if the sum of those angles is 90°.
Complementary angles of the following angles are:
(i) 90°-35°=55°(ii) 90°-72°=18°(iii) 90°-45°=45°(iv) 90°-85°=5°
Question 6:
Find the supplement of each of the following angles:
(i) 70°
(ii) 120°
(iii) 135°
(iv) 90°
Answer 6:
Two angles are called supplementary angles if the sum of those angles is 180°.
Supplementary angles of the following angles are:
(i) 180° − 70° = 110°
(ii) 180° − 120° = 60°
(iii) 180° − 135° = 45°
(iv) 180° − 90° = 90°
Question 7:
Identify the complementary and supplementary pairs of angles from the following pairs:
(i) 25°, 65°
(ii) 120°, 60°
(iii) 63°, 27°
(iv) 100°, 80°
Answer 7:
Since
(i) 25°+65°=90° , therefore this is complementary pair of angle. (ii) 120°+ 60°= 180°, therefore this is supplementary pair of angle.(iii) 63°+27°= 90°, therefore this is complementary pair of angle.(iv) 100°+ 80°= 180° , therefore this is supplementary pair of angle.
Therefore, (i) and (iii) are the pairs of complementary angles and (ii) and (iv) are the pairs of supplementary angles.
Question 8:
Can two angles be supplementary, if both of them be
(i) obtuse?
(ii) right?
(iii) acute?
Answer 8:
(i) No, two obtuse angles cannot be supplementary.
(ii) Yes, two right angles can be supplementary. (∵∠90°+∠90°=∠180°)
(iii) No, two acute angles cannot be supplementary.
Question 9:
Name the four pairs of supplementary angles shown in Fig.
Answer 9:
Following are the supplementary angles:
∠AOC and ∠COB
∠BOC and ∠DOB
∠BOD and ∠DOA
∠AOC and ∠DOA
Question 10:
In Fig., A, B, C are collinear points and ∠DBA = ∠EBA.
(i) Name two linear pairs
(ii) Name two pairs of supplementary angles.
Answer 10:
(i) Linear pairs:
∠ABD and ∠DBC
∠ABE and ∠EBC
Because every linear pair forms supplementary angles, these angles are:
∠ABD and ∠DBC
∠ABE and ∠EBC
Question 11:
If two supplementary angles have equal measure, what is the measure of each angle?
Answer 11:
Let x and y be two supplementary angles that are equal.
∠x=∠y
According to the question,
∠x+∠y=180°⇒∠x+∠x=180°⇒2∠x=180°⇒∠x=180°2=90°∴∠x=∠y=90°
Question 12:
If the complement of an angle is 28°, then find the supplement of the angle.
Answer 12:
Let x be the complement of the given angle 28°.
∴ ∠x+28°=90°⇒∠x=90°-28°=62°
So, supplement of the angle = 180°-62°=118°
Question 13:
In Fig. 19, name each linear pair and each pair of vertically opposite angles:
Answer 13:
Two adjacent angles are said to form a linear pair of angles if their non-common arms are two opposite rays.
∠1 and ∠2∠2 and ∠3
∠3 and ∠4
∠1 and ∠4
∠5 and ∠6
∠6 and ∠7
∠7 and ∠8
∠8 and ∠5
∠9 and ∠10
∠10 and ∠11
∠11 and ∠12
∠12 and ∠9
Two angles formed by two intersecting lines having no common arms are called vertically opposite angles.
∠1 and ∠3
∠4 and ∠2
∠5 and ∠7
∠6 and ∠8
∠9 and ∠11
∠10 and ∠12
Question 14:
In Fig., OE is the bisector of ∠BOD. If ∠1 = 70°, find the magnitudes of ∠2, ∠3 and ∠4.
Answer 14:
Since OE is the bisector of ∠BOD,
∴∠DOE=∠EOB∠2+∠1+∠EOB=180° (Linear Pair)∠2+2∠1=180° (∠1=∠EOB)⇒∠2=180°-2∠1=180°-2×70°=180°-140°=40°
∠4=∠2=40° (Vertically opposite angles)∠3=∠DOB=∠1+∠EOB=70°+70°=140° [∠3=∠DOB (Vertically opposite angles)]
Question 15:
One of the angles forming a linear pair is a right angle. What can you say about its other angle?
Answer 15:
One angle of a linear pair is the right angle, i.e., 90°.
∴ The other angle = 180° - 90° = 90°
Question 16:
One of the angles forming a linear pair is an obtuse angle. What kind of angle is the other?
Answer 16:
If one of the angles of a linear pair is obtuse, then the other angle should be acute; only then can their sum be 180°.
Question 17:
One of the angles forming a linear pair is an acute angle. What kind of angle is the other?
Answer 17:
In a linear pair, if one angle is acute, then the other angle should be obtuse. Only then their sum can be 180°.
Question 18:
Can two acute angles form a linear pair?
Answer 18:
No, two acute angles cannot form a linear pair because their sum is always less than 180°.
Question 19:
If the supplement of an angle is 65°; then find its complement.
Answer 19:
Let x be the required angle.
Then, we have:
x + 65° = 180°
⇒x = 180° - 65° = 115°
The complement of angle x cannot be determined.
Question 20:
Find the value of x in each of the following figures.
Answer 20:
(i)
Since ∠BOA+∠BOC=180° (Linear pair)
∴ ∠x=180°-∠BOA=180°-60°=120°
(ii)
Since ∠QOP+∠QOR=180° (Linear pair)∴2x+3x=180°⇒5x=180°⇒x=180°5=36°
(iii)
Since ∠LOP+∠PON+∠NOM=180° (Linear pair)∴∠PON=180°-∠LOP-∠NOM⇒x=180°-35°-60°⇒x=180°-95°=85°
(iv)
Since ∠COD+∠DOE+∠EOA+∠AOB+∠BOC=360° (Sum of all angles at a point)∴83°+92°+75°+47°+x=360°⇒297°+x=360°⇒x=360°-297°=63°
(v)
2x°+x°+2x°+3x°=180°⇒8x=180⇒x=1808=22.5°
(vi)
3x°=105°⇒x=1053=35°
Question 21:
In Fig. 22, it being given that ∠1 = 65°, find all other angles.
Answer 21:
∠1=∠3 (Vertically opposite angles)
∴∠3=65°
Since ∠1+∠2=180° (Linear pair)
∴∠2=180°-65°=115°
∠2=∠4 (Vertically opposite angles)
∴∠4=∠2=115° and ∠3=65°
Question 22:
In Fig., OA and OB are opposite rays:
(i) If x = 25°, what is the value of y?
(ii) If y = 35°, what is the value of x?
Answer 22:
∠AOC + ∠BOC = 180° (Linear pair)
⇒(2y+5)+3x=180°⇒3x+2y=175°
(i) If x = 25°, then
3×25°+2y=175°⇒75°+2y=175°⇒2y=175°-75°=100°⇒y=100°2=50°
(ii) If y = 35°, then
3x+2×35°=175°⇒3x+70°=175°⇒3x=175°-70°=105°⇒x=105°3=35°
Question 23:
In Fig., write all pairs of adjacent angles and all the linear pairs.
Answer 23:
Adjacent angles:
∠DOA and ∠DOC∠DOC and ∠BOC
∠AOD and ∠DOB∠BOC and ∠AOC
Linear pairs of angles:
∠AOD and ∠DOB∠BOC and ∠AOC
Question 24:
In Fig. 25, find ∠x. Further find ∠BOC, ∠COD and ∠AOD.
Answer 24:
∠AOD+∠DOC+∠COB=180°(Linear pair)(x+10)°+x°+(x+20)°=180°3x+30°=180°3x=180°-30°3x=150°x=150°3=50°
∠BOC=x+20°=50°+20°=70°∠COD=x=50°∠AOD=x+10°=50°+10°=60°
Question 25:
How many pairs of adjacent angles are formed when two lines intersect in a point?
Answer 25:
If two lines intersect at a point, then four adjacent pairs are formed, and those pairs are linear as well.
Question 26:
How many pairs of adjacent angles, in all, can you name in Fig.?
Answer 26:
There are 10 adjacent pairs in the given figure; they are:
∠EOD and ∠DOC∠COD and ∠BOC∠COB and ∠BOA
∠AOB and ∠BOD∠BOC and ∠COE∠COD and ∠COA∠DOE and ∠DOB
∠EOD and ∠DOA∠EOC and ∠AOC∠AOB and ∠BOE
Question 27:
In Fig., determine the value of x.
Answer 27:
∠AOB+∠BOC=180° (Linear pair)⇒3x+3x=180°⇒6x=180°⇒x=180°6=30°
Question 28:
In Fig., AOC is a line, find x.
Answer 28:
∠AOB+∠BOC=180° (Linear pair)⇒70°+2x=180°⇒2x=180°-70°=110°⇒x=110°2=55°
Question 29:
In Fig., POS is a line, find x.
Answer 29:
∠QOP+∠QOR+∠ROS=180° (Angles on a straight line)
⇒60°+4x+40°=180°⇒100°+4x=180°⇒4x=180°-100°=80°⇒x=80°4=20°
Question 30:
In Fig., lines l1 and l2 intersect at O, forming angles as shown in the figure. If x = 45°, find the values of y, z and u.
Answer 30:
∠z=∠x=45° (Vertically opposite angles)Now,∠x+∠y=180° (Linear pair)⇒∠y=180°-45°=135°∠u=∠y=135° (Vertically opposite angles)
Question 31:
In Fig., three coplanar lines intersect at a point O, forming angles as shown in the figure. Find the values of x, y, z and u.
Answer 31:
∠BOD + ∠DOF + ∠FOA = 180° (Linear pair)
∴ ∠FOA = ∠u = 180°-90°-50°=40°
∠FOA=∠x=40° (Vertically opposite angles)
∠BOD=∠z=90° (Vertically opposite angles)
∠EOC=∠y=50° (Vertically opposite angles)
Question 32:
In Fig., find the values of x, y and z.
Answer 32:
∠y=25° (Vertically opposite angles)Since ∠x+∠y=180° (Linear pair)∴∠x=180°-25°=155°∠z=∠x=155° (Vertically opposite angles)
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