RD Sharma Solutions Class 9 Chapter 8 Ex 8.2
Q 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?
(i) Given to us here;
the value of x = 25
As you can see, ∠AOC and ∠BOC form a linear pair
∠AOC and ∠BOC = 180°
From the figure, ∠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
Therefore, y = 50
(ii) Given to us here;
the value of y = 35
Since, ∠AOC + ∠BOC = 180° (Linear pair angles)
(2y+5)+3x = 180
(2(35) + 5) + 3x = 180
(70+5) + 3x = 180
3x = 180 – 75
3x = 105
Therefore, x = 35
Q 2 : In the below figure, write all pairs of adjacent angles and all the linear pairs.
Solution: From the given figure, the Adjacent angles are :
And Linear pair of angles are ∠AOD,∠BOD,∠AOC,∠BOC.
Q 3 : In the given below the figure, find x. Further find ∠COD,∠AOD,∠BOC.
Solution: As,∠AOD and ∠BOD forms a line pair,
Therefore, ∠AOD+ ∠BOD=180°
In the figure,
∠AOD=(x+10)° ,∠COD=x° ,∠BOC=(x+20)°
hence, ( x + 10 ) + x + ( x + 20 ) = 180
3x + 30 = 180
3x = 180 – 30
3x = 150/3
x = 50
∠AOD=(x+10) =50 + 10 = 60
∠COD = x = 50°
∠BOC = (x+20) = 50 + 20 = 70
Hence, ∠AOD=60°, ∠COD=50°, ∠BOC=70°
Q 4 : In the Given below figure rays OA, OB, OC, OD and OE have the common end point 0. Show that ∠AOB+∠BOC+∠COD+∠DOE+∠EOA=360°.
Solution: Given to us here, ray OA, OB, OC, OD and OE have the common endpoint O.
A ray OX is drawn opposite to OA, as shown in the figure.
Since ∠AOB and ∠BOX are linear pair angles, therefore,
∠AOB+∠BOX = 180°
Or, ∠AOB + ∠BOC + ∠COF = 180° —–—–(1)
∠AOE and ∠EOX are linear pair angles, therefore,
Or, ∠AOE+∠DOF+∠DOE=180° —–(2)
By adding equations, (1) and (2), we get;
∠AOB + ∠BOC + ∠COF + ∠AOE + ∠DOF + ∠DOE = 180°
∠AOB + ∠BOC + ∠COD + ∠DOE + ∠EOA = 180°
Q 5 : In the Below figure, ∠AOC and ∠BOC form a linear pair. If a – 2b = 30°, find a and b?
Solution: Given to us here;
∠AOC and ∠BOC form a linear pair
Now, if a – 2b = 30°, we have to find the value of a and b.
From the figure, we can see,
∠AOC = a° ,∠BOC = b°
Thus, a + b = 180° —–(1)
Already given, a – 2b = 30° —–(2)
On subtracting equation (1) and (2) here, we get;
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°
Therefore, the values of a and b are 130° and 50° respectively.
Q 6: How many pairs of adjacent angles are formed when two lines intersect at a point?
Solution: Four pairs of adjacent angles are formed when two lines intersect each other at a single point.
Let, two lines AB and CD intersect at point O.
The 4 pair of adjacent angles are :
(∠AOD,∠DOB),(∠DOB,∠BOC),(∠COA, ∠AOD) and (∠BOC,∠COA).
Q 7: How many pairs of adjacent angles, in all, can you name in the figure below?
Solution: Number of Pairs of adjacent angles, from the figure, are :
∠EOC and ∠DOC
∠EOD and ∠DOB
∠DOC and ∠COB
∠EOD and ∠DOA
∠DOC and ∠COA
∠BOC and ∠BOA
∠BOA and ∠BOD
∠BOA and ∠BOE
∠EOC and ∠COA
∠EOC and ∠COB
Hence, there are 10 pair of adjacent angles.
Q.8: In the below figure, find the value of x?
Solution: As we know, the sum of all the angles around a point O is equal to 360°. Therefore,
3x + 3x + 150 + x = 360°
7x = 360° – 150°
7x = 210°
x = 210/7
x = 30°
Hence, the value of x is 30°.
Q.9: In the below figure, AOC is a line, find x.
Solution: Since, from the figure we can see, ∠AOB and ∠BOC are linear pairs,
∠AOB +∠BOC =180°
70 + 2x = 180
2x = 180 – 70
2x = 110
x = 110/2
x = 55
Therefore, the value of x is 55°.
Q.10: In the below figure, POS is a line, Find x?
Solution: As from the figure, we can see, ∠POQ and ∠QOS are linear pairs. Therefore,
∠POQ + ∠QOS=180°
∠POQ + ∠QOR+∠SOR=180°
60 + 4x +40 = 180
4x = 180 -100
4x = 80
x = 20
Hence, the value of x is 20.
Q 11: In the below figure, ACB is a line such that ∠DCA = 5x and ∠DCB = 4x. Find the value of x?
Solution: As you can see, ∠ACD and ∠BCD form a linear pair.
Therefore, ∠ACD + ∠BCD = 180°
∠DCA = 5x and ∠DCB = 4x
5x + 4x = 180°
9x = 180°
x = 20
Hence, the value of unknown variable, x is 20.
Q 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: As you can see, from the figure, POQ is a straight line. Therefore,∠POR and ∠QOR are linear parts.
So, ∠POR + ∠QOR = 180°
Also, given to us here;
∠POR =3x and ∠QOR = 2x + 10
2x + 10 + 3x = 180
5x + 10 = 180
5x = 180 – 10
5x = 170
x = 34
Hence the value of unknown variable, x is 34°.
Q 13: In Fig: a is greater than b by one-third of a right angle. Find the value of a and b?
Solution: Since angle a and b are linear pair, from the figure, therefore,
a + b = 180
a = 180 – b —–(1)
As per given question, a is greater than b by one third of a right angle, therefore;
a = b + 90/3
a = b + 30
a – b = 30 —–(2)
From equation (1) and (2), we get;
180 – b = b + 30
180 – 30 = 2b
b = 150 / 2
b = 75
From eq. (1)
a = 180 – b
a = 180 – 75
a = 105
Thus, the values of angle a and b are 105° and 75° respectively.
Q 14: What value of y would make AOB a line in the below figure, If ∠AOB=4y and ∠BOC=(6y+30)?
Solution: From the figure, ∠AOC and ∠BOC are linear pairs. Therefore;
Putting the values of both the angles, as per the given data;
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°.
Q 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) Let ∠FOE = a, ∠DOE = b and ∠BOC = c
Since ∠AOF and ∠FOG are linear pair, therefore;
∠AOF + ∠FOG = 180
∠AOF + 30 = 180 [Given, ∠FOG = 30]
∠AOF = 180 – 30
Or, ∠AOB +∠BOC+∠COD+∠DOE+∠EOF=150
30 + c + 30 + b + a = 150 [Given, ∠AOB =∠COD=30]
a + b + c =150 – 30 – 30
a + b + c =90 ——————–(1)
Given, ∠FOC = 90°
So, ∠FOE +∠EOD+∠DOC =90°
a + b +30 = 90
a + b = 90 – 30
a + b =60 —–(2)
Putting the value from eq (2) in (1), we get;
a + b + c = 90
60 + c = 90
c = 90 – 60 = 30°
Given, ∠BOE = 90°
∠BOC + ∠COD + ∠DOE = 90°
30 + 30 + ∠DOE = 90°
∠DOE = 90 – 60 = 30°
∠DOE = a = 30°
From eq.2 we know,
a + b = 60
b = 60 – a
b = 60 – 30
b = 30°
Therefore, ∠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)
Q16: In below fig. OP, OQ, OR and OS are four rays. Prove that: ∠POQ + ∠QOR+∠SOR + ∠POS = 360°
Solution: Given to us here;
The four rays OP, OQ, OR and OS, meet at a single point O.
Now, produce ray OQ backwards to a point T, to form a ray OT and TOQ is a straight line.
As you can see, ray OP stands on the TOQ.
Since ∠TOP and ∠POQ is a linear pair angle, therefore;
∠TOQ+∠POQ =180° —–(1)
In the same way, ray OS stands on the line TOQ. Thus,
∠TOS +∠SOQ =180° —–(2)
∠SOQ = ∠SOR + ∠QOR —–(3)
Now, equation (2) can be written as;
∠TOS+ ∠SOR +∠OQR =180°
If we add equation (1) and (3), we get;
∠TOP +∠POQ +∠TOS + ∠SOR + ∠QOR = 360° —–(4)
∠TOP + ∠TOS = ∠POS
Therefore, eqaution 4 becomes;
∠POQ + ∠QOR + ∠SOR + ∠POS = 360°
Q 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: As per the given question,
Ray OS stand on the line POQ.
And Ray OR and Ray OT are the angle bisectors of ∠POS and ∠SOQ, respectively.
Given, ∠POS = x
Since, ∠POS and ∠SOQ is linear pair
Therefore, ∠POS + ∠QOS =180°
x + ∠QOS = 180
Or, ∠QOS = 180 – x
Now, ray OR is the bisector ∠POS, therefore;
∠ROS = 1/2 ∠POS
∠ROS = x/2 …………..1 [Since ∠POS = x ]
In the same way, ray OT bisects ∠QOS, therefore,
∠TOS = 1/2 ∠QOS
∠TOS = (180 – x)/2 [ ∠QOS = 180 – x ]
∠TOS = 90 – x/2 ……………..2
Therefore, ∠ROT = ∠ROS + ∠SOT
= x/2 + 90 – x/2 [From eq.1 and eq.2]
Therefore, ∠ROT = 90°
Q 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 to us here;
From the figure,
∠POR and ∠ROQ are a linear pair of angles.
Therefore, ∠POR + ∠ROQ = 180°
As per the question is given;
∠POR: ∠ROQ = 5:7
So, ∠POR = (5/12) × 180=75
In the same way, ∠ROQ = (7/7+5) 180 = 105
Therefore, ∠POS = ∠ROQ = 105° [Both are Vertically opposite angles]
Also, ∠SOQ = ∠POR = 75° [Both are Vertically opposite angles]
Q 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 to us here;
Since, OR perpendicular to line PQ.
Therefore, ∠POR = 90°
∠POS + ∠SOR = 90 [ Since, ∠POR = ∠POS+∠SOR]
∠ROS = 90°- ∠POS ———–(1)
∠QOR = 90 (∵ OR ⊥ PQ)
∠QOS – ∠ROS = 90°
∠ROS = ∠QOS -90°
By adding equation (1) and (2), we get here;
2∠ROS = ∠QOS – ∠POS
∠ROS = 1/2(∠QOS – ∠POS)