Chapter 8: Lines And Angles Exercise – 8.2
Question: 1
In the below 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?
Solution:
(i) Given that,
x = 25
Since ∠AOC and ∠BOC form a linear pair
∠AOC + ∠BOC = 180°
Given that ∠AOC = 2y + 5 and ∠BOC = 3x
∠AOC + ∠BOC = 180°
(2y + 5) + 3x = 180
(2y +5) + 3(25) = 180
2y + 5 + 75 = 180
2y + 80 = 180
2y = 180 - 80 = 100
y = 100/2 = 50
y = 50
(ii) Given that,
y = 35
∠AOC + ∠BOC = 180°
(2y + 5) + 3x = 180
(2(35) + 5) + 3x = 180
(70 + 5) + 3x = 180
3x = 180 - 75
3x = 105
x = 35
Question: 2
In the below figure, write all pairs of adjacent angles and all the linear pairs.
Solution:
Adjacent angles are:
(i) ∠AOC, ∠COB
(ii) ∠AOD ∠BOD
(i) ∠AOD, ∠COD
(i) ∠BOC, ∠COD
Linear pairs: ∠AOD, ∠BOD, ∠AOC, ∠BOC
Question: 3
In the given below figure, find x. Further find ∠COD, ∠AOD, ∠BOC
Solution:
Since ∠AOD and ∠BOD form a line pair,
∠AOD + ∠BOD = 180°
∠AOD + ∠BOC + ∠COD = 180°
Given that,
∠AOD = (x + 10)°, ∠COD = x°, ∠BOC = (x + 20)°
(x + 10) + x + (x + 20) = 180
3x + 30 = 180
3x = 180 - 30
3x = 150/3
x = 50
Therefore, ∠AOD = (x + 10)
= 50 + 10 = 60
∠COD = x = 50°
∠COD = (x + 20)
= 50 + 20 = 70
∠AOD = 60°∠COD = 50°∠BOC = 70°
Question: 4
In the given below figure rays OA, OB, OC, OP and OE have the common end point O. Show that ∠AOB + ∠BOC + ∠COD + ∠DOE + ∠EOA = 360°
Solution:
Given that OA, OB, OD and OE have the common end point O.
A ray opposite to OA is drawn
Since ∠AOB, ∠BOF are linear pairs,
∠AOB + ∠BOF = 180°
∠AOB + ∠BOC + ∠COF = 180° .... (1)
Also,
∠AOE and ∠EOF are linear pairs
∠AOE + ∠EOF = 180°
∠AOE + ∠DOF + ∠DOE = 180° .... (2)
By adding (1) and (2) equations we get
∠AOB + ∠BOC + ∠COF + ∠AOE + ∠DOF + ∠DOE = 180°
∠AOB + ∠BOC + ∠COD + ∠DOE + ∠EOA = 180°
Hence proved.
Question: 5
In the below figure, ∠AOC and ∠BOC form a linear pair. If a - 2b = 30°, find a and b?
Solution:
Given that,
∠AOC and ∠BOC form a linear pair
If a - b = 30
∠AOC = a°, ∠BOC = b°
Therefore, a + b = 180 ... (1)
Given a - 2b = 30 ... (2)
By subtracting (1) and (2)
a + b - a + 2b = 180 - 30
3b = 150
b = 150/3
b = 50
Since a - 2b = 30
a - 2(50) = 30
a = 30 + 100
a = 130
Hence, the values of a and b are 130° and 50° respectively.
Question: 6
How many pairs of adjacent angles are formed when two lines intersect at a point?
Solution:
Four pairs of adjacent angles will be formed when two lines intersect at a point.
Considering two lines AB and CD intersecting at O
The 4 pairs are:
(∠AOD, ∠DOB), (∠DOB, ∠BOC), (∠COA, ∠AOD) and (∠BOC, ∠COA)
Hence, 4 pairs of adjacent angles are formed when two lines intersect at a point.
Question: 7
How many pairs of adjacent angles, in all, can you name in the figure below?
Solution:
Pairs of adjacent angles are:
∠EOC, ∠DOC
∠EOD, ∠DOB
∠DOC, ∠COB
∠EOD, ∠DOA
∠DOC, ∠COA
∠BOC, ∠BOA
∠BOA, ∠BOD
∠BOA, ∠BOE
∠EOC, ∠COA
∠EOC, ∠COB
Hence, 10 pair of adjacent angles.
Question: 8
In the below figure, find value of x?
Solution:
Since the sum of all the angles round a point is equal to 360°
3x + 3x + 150 + x = 360
7x = 360 - 150
7x = 210
x = 210/7
x = 30
Value of x is 30°
Question: 9
In the below figure, AOC is a line, find x.
Solution:
Since ∠AOB and ∠BOC are linear pairs,
∠AOB + ∠BOC = 180°
70 + 2x = 180
2x = 180 - 70
2x = 110
x = 110/2
x = 55
Hence, the value of x is 55°
Question: 10
In the below figure, POS is a line, Find x?
Solution:
Since ∠POQ and ∠QOS are linear pairs
∠POQ + ∠QOS = 180°
∠POQ + ∠QOR + ∠SOR = 180°
60 + 4x + 40 = 180
4x = 180 - 100
4x = 80
x = 20
Hence, Value of x = 20
Question: 11
In the below figure, ACB is a line such that ∠DCA = 5x and ∠DCB = 4x. Find the value of x?
Solution:
Here, ∠ACD + ∠BCD = 180°
[Since they are linear pairs]
∠DCA = 5x and ∠DCB = 4x
5x + 4x = 180
9x = 180
x = 20
Hence, the value of x is 20°
Question: 12
In the given figure, Given ∠POR = 3x and ∠QOR = 2x + 10, Find the value of x for which POQ will be a line?
Solution:
For the case that POR is a line
∠POR and ∠QOR are linear parts
∠POR + ∠QOR = 180°
Also, given that,
∠POR = 3x and ∠QOR = 2x + 10
2x + 10 + 3x = 180
5x + 10 = 180
5x = 180 – 10
5x = 170
x = 34
Hence the value of x is 34°
Question: 13
In Fig: a is greater than b by one third of a right angle. Find the value of a and b?
Solution:
Since a and b are linear
a + b = 180
a = 180 – b ... (1)
From given data, a is greater than b by one third of a right angle
a = b + 90/3
a = b + 30
a – b = 30 ... (2)
Equating (1) and (2)
180 – b = b + 30
180 – 30 = 2b
b = 150 / 2
b = 75
From (1)
a = 180 – b
a = 180 – 75
a = 105
Hence the values of a and b are 105° and 75° respectively.
Question: 14
What value of y would make AOB a line in the below figure, If ∠AOB = 4y and ∠BOC = (6y + 30)?
Solution:
Since, ∠AOC and ∠BOC are linear pairs
∠AOC + ∠BOC = 180°
6y + 30 + 4y = 180
10y + 30 = 180
10y = 180 - 30
10y = 150
y = 150/10
y = 15
Hence value of y that will make AOB a line is 15°
Question: 15
If the figure below forms a linear pair,
∠EOB = ∠FOC = 90 and ∠DOC = ∠FOG = ∠AOB = 30
Find the measure of ∠FOE, ∠COB and ∠DOE
Name all the right angles
Name three pairs of adjacent complementary angles
Name three pairs of adjacent supplementary angles
Name three pairs of adjacent angles
Solution:
(i) ∠FOE = x, ∠DOE = y and ∠BOC = z
Since ∠AOF, ∠FOG is a linear pair
∠AOF + 30 = 180
∠AOF = 180 - 30
∠AOF = 150
∠AOB + ∠BOC + ∠COD + ∠DOE + ∠EOF = 150
30 + z + 30 + y + x = 150
x + y + z =150 - 30 - 30
x + y + z = 90 .... (1)
∠FOC = 90°
∠FOE + ∠EOD + ∠DOC = 90°
x + y + 30 = 90
x + y = 90 - 30
x + y =60 ... (2)
Substituting (2) in (1)
x + y + z = 90
60 + z = 90
z = 90 - 60 = 30
Given BOE = 90
∠BOC + ∠COD + ∠DOE = 90°
30 + 30 + DOE = 90
DOE = 90 - 60 = 30
DOE = x = 30
We also know that,
x + y = 60
y = 60 - x
y = 60 - 30
y = 30
Thus we have ∠FOE = 30, ∠COB = 30 and ∠DOE = 30
(ii) Right angles are ∠DOG, ∠COF, ∠BOF, ∠AOD
(iii) Adjacent complementary angles are (∠AOB, ∠BOD); ( ∠AOC, ∠COD); ( ∠BOC, ∠COE)
(iv) Adjacent supplementary angles are (∠AOB, ∠BOG); (∠AOC, ∠COG); (∠AOD, ∠DOG)
(v) Adjacent angles are (∠BOC, ∠COD); (∠COD, ∠DOE); (∠DOE, ∠EOF)
Question: 16
In below fig. OP, OQ, OR and OS are four rays. Prove that: ∠POQ + ∠QOR + ∠SOR + ∠POS = 360°
Solution:
Given that
OP, OQ, OR and OS are four rays
You need to produce any of the ray OP, OQ, OR and OS backwards to a point in the figure.
Let us produce ray OQ backwards to a point T
So that TOQ is a line
Ray OP stands on the TOQ
Since ∠TOP, ∠POQ is a linear pair
∠TOP + ∠POQ = 180° ... (1)
Similarly,
Ray OS stands on the line TOQ
∠TOS + ∠SOQ = 180° ... (2)
But ∠SOQ = ∠SOR + ∠QOR ... (3)
So, eqn (2) becomes
∠TOS + ∠SOR + ∠OQR = 180°
Now, adding (1) and (3) you get ∠TOP + ∠POQ + ∠TOS + ∠SOR + ∠QOR = 360° ... (4)
∠TOP + ∠TOS = ∠POS
Eqn: (4) becomes
∠POQ + ∠QOR + ∠SOR + ∠POS = 360°
Question: 17
In below fig, ray OS stand on a line POQ. Ray OR and ray OT are angle bisectors of ∠POS and ∠SOQ respectively. If ∠POS = x, find ∠ROT?
Solution:
Given,
Ray OS stand on a line POQ
Ray OR and Ray OT are angle bisectors of ∠POS and ∠SOQ respectively
∠POS = x
∠POS and ∠SOQ is linear pair
∠POS + ∠QOS = 180°
x + QOS = 180
QOS = 180 - x
Now, ray or bisector POS
∠ROS = 1/2 ∠POS
x/2
ROS = x/2 [Since POS = x]
Similarly ray OT bisector QOS
∠TOS = 1/2 ∠QOS
= (180 - x)/2 [QOS = 180 - x]
= 90 - x/2
Hence, ∠ROT = ∠ROS + ∠ROT
= x/2 + 90 - x/2
= 90
∠ROT = 180°
Question: 18
In the below fig, lines PQ and RS intersect each other at point O. If ∠POR: ∠ROQ = 5: 7. Find all the angles.
Solution:
Given
∠POR and ∠ROP is linear pair
∠POR + ∠ROP = 180°
Given that
∠POR: ∠ROQ = 5:7
Hence, POR = (5/12) × 180 = 75
Similarly ROQ = (7/7 + 5) × 180 = 105
Now POS = ROQ = 105° [Vertically opposite angles]
Also, SOQ = POR = 75° [Vertically opposite angles]
Question: 19
In the below fig. POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that ∠ROS = 1/2(∠QOS − ∠POS).
Solution:
Given that
OR perpendicular
∴ ∠POR = 90°
∠POS + ∠SOR = 90 [∴ ∠POR = ∠POS + ∠SOR]
∠ROS = 90° − ∠POS ... (1)
∠QOR = 90 (∵ OR ⊥ PQ)
∠QOS − ∠ROS = 90°
∠ROS = ∠QOS − 90°
By adding (1) and (2) equations, we get
∴ ∠ROS = ∠QOS − ∠POS
∠ROS = 1/2(∠QOS − ∠POS)