Selina Solutions Concise Mathematics Class 6 Chapter 24 Angles Exercise 24(A) provides comprehensive answers on each and every question discussed under this exercise. The solutions are designed with the purpose of improving students’ knowledge in the respective subject. Those who aim to obtain skills in problem solving are advised to practice Selina Solutions frequently. The solutions can be referred by the students to score more marks in the annual exam. For a better hold on these concepts, students can download Selina Solutions Concise Mathematics Class 6 Chapter 24 Angles Exercise 24(A) free PDF, from the links provided below.
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1. For each angle given below, write the name of the vertex, the names of the arms and the name of the angle.
(iv) Name the angles marked by letters a, b, c, x and y.
Solution:
(i) In the given figure,
Vertex = O
Arms = OA and OB
Angle = ∠AOB or ∠BOA or ∠O
(ii) In the given figure,
Vertex = Q
Arms = QP and QR
Angle = ∠PQR or ∠RQP or ∠Q
(iii) In the given figure,
Vertex = M
Arms = MN and ML
Angle = ∠LMN or ∠NML or ∠M
(iv) a = ∠AOE
b = ∠AOB
c = ∠BOC
d = ∠COD
e = ∠DOE
2. Name the points:
(i) in the interior of the angle PQR,
(ii) in the exterior of the angle PQR.
Solution:
(i) The points in the interior of the angle = a, b and x
(ii) The points in the exterior of the angle = d, m, n, s and t
3. In the given figure, figure out the number of angles formed within the arms OA and OE.
Solution:
In the given figure, the angles within the arms OA and OE are as follows:
(i) ∠AOE
(ii) ∠AOD
(iii) ∠AOC
(iv) ∠AOB
(v) ∠BOC
(vi) ∠BOD
(vii) ∠BOE
(viii) ∠COD
(ix) ∠COE and
(x) ∠DOE
4. Add:
(i) 290 16’23” and 80 27’12”
(ii) 90 45’56” and 730 8’ 15”
(iii) 560 38’ and 270 42’ 30”
(iv) 470 and 610 17’4”
Solution:
(i) 290 16’23” + 80 27’12”
290 16’23”
80 27’12” +
_____________
370 43’ 35”
_____________
Hence, addition of 290 16’23” and 80 27’12” = 370 43’ 35”
(ii) 90 45’56” + 730 8’ 15”
90 45’56”
730 8’ 15” +
____________
820 53’ 71”
____________
Hence, addition of 90 45’56” and 730 8’ 15” = 820 53’ 71”
(iii) 560 38’ + 270 42’ 30”
560 38’
270 42’ 30” +
_____________
830 80’ 30”
_____________
Hence, addition of 560 38’ and 270 42’ 30” = 830 80’ 30”
(iv) 470 + 610 17’4”
470
610 17’4” +
____________
1080 17’ 4”
_____________
Hence, addition of 470 and 610 17’4” = 1080 17’ 4”
5. In the figure, given below name:
(i) three pairs of adjacent angles
(ii) two acute angles
(iii) two obtuse angles
(iv) two reflex angles
Solution:
(i) In the given figure, the three pairs of adjacent angles are as follows:
∠AOB and ∠BOC ,
∠BOC and ∠COD
∠COD and ∠DOA
(ii) In the given figure, the two acute angles are
∠AOB and ∠AOD
(iii) In the given figure, the two obtuse angles are
∠BOC and ∠COD
(iv) In the given figure, the two reflex angles are
∠AOB and ∠COB
6. In the given figure; PQR is a straight line. If:
(i) ∠SQR = 750; find ∠PQS.
(ii) ∠PQS = 1100; find ∠RQS
Solution:
(i) Given PQR is a straight line
In the given figure,
∠PQS + ∠SQR = 1800 {linear pair of angles}
∠PQ S + 75 = 180
On further calculation, we get,
∠PQS = 180 – 75
∠PQS = 1050
(ii) Given PQR is a straight line
∠PQS + ∠RQS = 1800
1100 + ∠RQS = 1800
On further calculation, we get,
∠RQS = 1800 – 1100
∠RQS = 700
7. In the given figure; AOC is a straight line. If angle AOB = 500, angle AOE = 900 and angle COD = 250; find the measure of:
(i) angle BOC
(ii) angle EOD
(iii) obtuse angle BOD
(iv) reflex angle BOD
(v) reflex angle COE
Solution:
(i) Given
AOC is a straight line
∠AOB = 500
∠AOE = 900
∠COD = 250
To find the measure of ∠BOC
∠AOB + ∠BOC = 1800 (Linear pairs of angle)
500 + ∠BOC = 1800
On further calculation, we get,
∠BOC = 1800 – 500
∠BOC = 1300
(ii) Given
AOC is a straight line
∠AOB = 500
∠AOE = 900
∠COD = 250
To find the measure ∠EOD
∠EOD + ∠COD = 900 (Since ∠AOE = 900)
∠EOD + 250 = 900
∠EOD = 900 – 250
We get,
∠EOD = 650
(iii) Given
AOC is a straight line
∠AOB = 500
∠AOE = 900
∠COD = 250
To find the measure of obtuse angle BOD,
∠BOD = ∠BOC + ∠COD
Substituting the value of ∠BOC and ∠COD, we get,
∠BOD = 1300 + 250
We get,
∠BOD = 1550
(iv) Given
AOC is a straight line
∠AOB = 500
∠AOE = 900
∠COD = 250
To find the measure of reflex angel BOD
∠BOD = 3600 – ∠BOD
= 3600 – 1550
We get,
∠BOD = 2050
(iv) Given
AOC is a straight line
∠AOB = 500
∠AOE = 900
∠COD = 250
To find the measure of reflex angle COE
∠COE = 3600 – ∠COE
= 3600 – (∠COD + ∠EOD)
= 3600 – (250 + 650)
We get,
= 3600 – 900
∠COE = 2700
8. In the given figure if:
(i) a = 1300; find b
(ii) b = 200; find a
(iii) a = 5 / 3 right angle, find b
Solution:
(i) From figure,
a + b = 3600
Substitute the value of a in above equation
1300 + b = 3600
b = 3600 – 1300
We get,
b = 2300
(ii) From figure,
a + b = 3600
a + 2000 = 3600
a = 3600 – 2000
We get,
a = 1600
(iii) From figure,
a = 5 / 3 right angle
= 5 / 3 × 900
We get,
a = 1500
Now,
a + b = 3600
Substitute the value of a in above equation
1500 + b = 3600
b = 3600 – 1500
b = 2100
9. In the given diagram, ABC is a straight line
(i) If x = 530, find y
(ii) If y =
right angles; find x.
Solution:
(i) From the figure,
Given that ABC is a straight line
Hence,
∠ABD + ∠DBC = 1800 {Linear pair of angles}
x + y = 1800
530 + y = 1800
y = 1800 – 530
y = 1270
(ii) From given figure,
x + y = 1800
x +
right angles = 1800
x + 3 / 2 × 900 = 1800
On further calculation, we get
x + 1350 = 1800
x = 1800 – 1350
x = 450
10. In the given figure, AOB is a straight line. Find the value of x and also answer each of the following:
(i) ∠AOP = …….
(ii) ∠BOP = …….
(iii) which angle is obtuse?
(iv) which angle is acute?
Solution:
Given from figure,
∠AOP = x + 300
∠BOP = x – 300
Now,
∠AOP + ∠BOP = 1800 (Since ∠AOB is a straight line)
(x + 300)+ (x – 300) = 1800
2x = 1800
We get,
x = 900
(i) ∠AOP = x + 300
= 900 + 300
= 1200
(ii) ∠BOP = x – 300
= 900 – 300
We get,
= 600
(iii) The obtuse angle is ∠AOP
(iv) The acute angle is ∠BOP
11.In the given figure, PQR is a straight line. Find x. Then complete the following:
(i) ∠AQB = ……………
(ii) ∠BQP = ……………
(iii) ∠AQR = …………….
Solution:
Given
PQR is a straight line
∠AQP = x + 200
∠AQB = 2x + 100
∠BQR = x – 100
Since PQR is a straight line
∠AQP + ∠AQB + ∠BQR = 1800
(x + 200) + (2x + 100) + (x – 100) = 1800
We get,
4x + 200 = 1800
4x = 1800 – 200
4x = 1600
x = 1600 / 4
We get,
x = 400
(i) ∠AQB = 2x + 100
= 2 × 400 + 100
= 800 + 100
= 900
Similarly,
∠AQP = x + 200
∠AQP = 400 + 200
∠AQP = 600
∠BQR = x – 100
∠BQR = 400 – 100
∠BQR = 300
(ii) ∠BQP = ∠AQP + ∠AQB
= 600 + 900
= 1500
(iii) ∠AQR = ∠AQB + ∠BQR
= 900 + 300
= 1200
12. In the given figure, lines AB and CD intersect at point O.
(i) Find the value of ∠a.
(ii) Name all the pairs of vertically opposite angles.
(iii) Name all the pairs of adjacent angles
(iv) Name all the reflex angles formed and write the measure of each.
Solution:
Given
AB and CD intersect each other at point C
∠AOC = 680
(i) Here, AOB is a line
Hence,
∠AOC + ∠BOC = 1800 (Linear pairs of angles)
680 + a = 1800
a = 1800 – 680
a = 1120
(ii) The pairs of vertically opposite angles are,
∠AOC and ∠BOD and ∠BOC and ∠AOD
(iii) The pairs of adjacent angles are,
∠AOC and ∠BOC, ∠BOC and ∠BOD, ∠BOD and ∠DOA, ∠DOA and ∠AOC
(iv) The reflex angles in the given figure are,
∠BOC and ∠DOA,
Reflex angle BOC = 1800 + 680
= 2480
Reflex angle DOA = 1800 + 680
= 2480
13. In the given figure:
(i) If ∠AOB = 450, ∠BOC = 300 and ∠AOD = 1100;
Find: angles COD and BOD
(ii) If ∠BOC = ∠DOC = 340 and ∠AOD = 1200;
Find: angle AOB and angle AOC
(iii) If ∠AOB = ∠BOC = ∠COD = 380
Find: reflex angle AOC and reflex angle AOD
Solution:
(i) ∠COD = ∠AOD – ∠AOC
= ∠AOD – (∠AOB + ∠BOC)
= 1100 – (450 + 300)
We get,
= 1100 – 750
= 350
Hence,
∠COD = 350
∠BOD = ∠BOC + ∠COD
= 300 + 350
= 650
Hence,
∠BOD = 650
(ii) ∠AOB = ∠AOD – ∠BOD
= ∠AOD – (∠BOC + ∠COD)
= 1200 – (340 + 340)
We get,
= 1200 – 680
= 520
Hence,
∠AOB = 520
∠AOC = ∠AOB + ∠BOC
= 520 + 340
= 860
Hence,
∠AOC = 860
(iii) Reflex angel AOC = 3600 – ∠AOC
= 3600 – (∠AOB + ∠BOC)
= 3600 – (380 + 380)
= 3600 – 760
We get,
= 2840
Hence,
∠AOC = 2840
Reflex angle AOD = 3600 – ∠AOD
= 3600 – (∠AOB + ∠BOC + ∠COD)
= 3600 – (380 + 380 + 380)
= 3600 – 1140
We get,
= 2460
Hence,
∠AOC = 2460
