{"id":599545,"date":"2022-05-06T04:13:29","date_gmt":"2022-05-06T04:13:29","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=599545"},"modified":"2022-05-10T05:18:28","modified_gmt":"2022-05-10T05:18:28","slug":"selina-class-6-icse-solutions-mathematics-chapter-24-angles","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/selina-class-6-icse-solutions-mathematics-chapter-24-angles\/","title":{"rendered":"Selina Class 6 ICSE Solutions Mathematics : Chapter 24-\u00a0Angles"},"content":{"rendered":"\n

Class 6: Maths Chapter 24 solutions. Complete Class 6 Maths Chapter 24 Notes.<\/p>\n\n\n\n

Selina Class 6 ICSE Solutions Mathematics : Chapter 24- Angles<\/h2>\n\n\n\n

Selina 6th Maths Chapter 24, Class 6 Maths Chapter 24 solutions<\/p>\n\n\n\n

Exercise 24(A)<\/h3>\n\n\n\n

1. For each angle given below, write the name of the vertex, the names of the arms and the name of the angle.<\/strong><\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n
\"Selina<\/figure>\n\n\n\n
\"Selina<\/figure>\n\n\n\n

(iv) Name the angles marked by letters a, b, c, x and y.<\/strong><\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) In the given figure,<\/p>\n\n\n\n

Vertex = O<\/p>\n\n\n\n

Arms = OA and OB<\/p>\n\n\n\n

Angle = \u2220AOB or \u2220BOA or \u2220O<\/p>\n\n\n\n

(ii) In the given figure,<\/p>\n\n\n\n

Vertex = Q<\/p>\n\n\n\n

Arms = QP and QR<\/p>\n\n\n\n

Angle = \u2220PQR or \u2220RQP or \u2220Q<\/p>\n\n\n\n

(iii) In the given figure,<\/p>\n\n\n\n

Vertex = M<\/p>\n\n\n\n

Arms = MN and ML<\/p>\n\n\n\n

Angle = \u2220LMN or \u2220NML or \u2220M<\/p>\n\n\n\n

(iv) a = \u2220AOE<\/p>\n\n\n\n

b = \u2220AOB<\/p>\n\n\n\n

c = \u2220BOC<\/p>\n\n\n\n

d = \u2220COD<\/p>\n\n\n\n

e = \u2220DOE<\/p>\n\n\n\n

2. Name the points:<\/strong><\/p>\n\n\n\n

(i) in the interior of the angle PQR,<\/strong><\/p>\n\n\n\n

(ii) in the exterior of the angle PQR.<\/strong><\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) The points in the interior of the angle = a, b and x<\/p>\n\n\n\n

(ii) The points in the exterior of the angle = d, m, n, s and t<\/p>\n\n\n\n

3. In the given figure, figure out the number of angles formed within the arms OA and OE.<\/strong><\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

In the given figure, the angles within the arms OA and OE are as follows:<\/p>\n\n\n\n

(i) \u2220AOE<\/p>\n\n\n\n

(ii) \u2220AOD<\/p>\n\n\n\n

(iii) \u2220AOC<\/p>\n\n\n\n

(iv) \u2220AOB<\/p>\n\n\n\n

(v) \u2220BOC<\/p>\n\n\n\n

(vi) \u2220BOD<\/p>\n\n\n\n

(vii) \u2220BOE<\/p>\n\n\n\n

(viii) \u2220COD<\/p>\n\n\n\n

(ix) \u2220COE and<\/p>\n\n\n\n

(x) \u2220DOE<\/p>\n\n\n\n

4. Add:<\/strong><\/p>\n\n\n\n

(i) 290 <\/sup>16\u201923\u201d and 80 <\/sup>27\u201912\u201d<\/strong><\/p>\n\n\n\n

(ii) 90<\/sup> 45\u201956\u201d and 730<\/sup> 8\u2019 15\u201d<\/strong><\/p>\n\n\n\n

(iii) 560<\/sup> 38\u2019 and 270<\/sup> 42\u2019 30\u201d<\/strong><\/p>\n\n\n\n

(iv) 470<\/sup> and 610<\/sup> 17\u20194\u201d<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) 290 <\/sup>16\u201923\u201d + 80 <\/sup>27\u201912\u201d<\/p>\n\n\n\n

290 <\/sup>16\u201923\u201d<\/p>\n\n\n\n

80 <\/sup>27\u201912\u201d +<\/p>\n\n\n\n

_____________<\/p>\n\n\n\n

370<\/sup> 43\u2019 35\u201d<\/p>\n\n\n\n

_____________<\/p>\n\n\n\n

Hence, addition of 290 <\/sup>16\u201923\u201d and 80 <\/sup>27\u201912\u201d = 370<\/sup> 43\u2019 35\u201d<\/p>\n\n\n\n

(ii) 90<\/sup> 45\u201956\u201d + 730<\/sup> 8\u2019 15\u201d<\/p>\n\n\n\n

90<\/sup> 45\u201956\u201d<\/p>\n\n\n\n

730<\/sup> 8\u2019 15\u201d +<\/p>\n\n\n\n

____________<\/p>\n\n\n\n

820<\/sup> 53\u2019 71\u201d<\/p>\n\n\n\n

____________<\/p>\n\n\n\n

Hence, addition of 90<\/sup> 45\u201956\u201d and 730<\/sup> 8\u2019 15\u201d = 820<\/sup> 53\u2019 71\u201d<\/p>\n\n\n\n

(iii) 560<\/sup> 38\u2019 + 270<\/sup> 42\u2019 30\u201d<\/p>\n\n\n\n

560<\/sup> 38\u2019<\/p>\n\n\n\n

270<\/sup> 42\u2019 30\u201d +<\/p>\n\n\n\n

_____________<\/p>\n\n\n\n

830<\/sup> 80\u2019 30\u201d<\/p>\n\n\n\n

_____________<\/p>\n\n\n\n

Hence, addition of 560<\/sup> 38\u2019 and 270<\/sup> 42\u2019 30\u201d = 830<\/sup> 80\u2019 30\u201d<\/p>\n\n\n\n

(iv) 470<\/sup> + 610<\/sup> 17\u20194\u201d<\/p>\n\n\n\n

470<\/sup><\/p>\n\n\n\n

610<\/sup> 17\u20194\u201d +<\/p>\n\n\n\n

____________<\/p>\n\n\n\n

1080<\/sup> 17\u2019 4\u201d<\/p>\n\n\n\n

_____________<\/p>\n\n\n\n

Hence, addition of 470<\/sup> and 610<\/sup> 17\u20194\u201d = 1080<\/sup> 17\u2019 4\u201d<\/p>\n\n\n\n

5. In the figure, given below name:<\/strong><\/p>\n\n\n\n

(i) three pairs of adjacent angles<\/strong><\/p>\n\n\n\n

(ii) two acute angles<\/strong><\/p>\n\n\n\n

(iii) two obtuse angles<\/strong><\/p>\n\n\n\n

(iv) two reflex angles<\/strong><\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) In the given figure, the three pairs of adjacent angles are as follows:<\/p>\n\n\n\n

\u2220AOB and \u2220BOC ,<\/p>\n\n\n\n

\u2220BOC and \u2220COD<\/p>\n\n\n\n

\u2220COD and \u2220DOA<\/p>\n\n\n\n

(ii) In the given figure, the two acute angles are<\/p>\n\n\n\n

\u2220AOB and \u2220AOD<\/p>\n\n\n\n

(iii) In the given figure, the two obtuse angles are<\/p>\n\n\n\n

\u2220BOC and \u2220COD<\/p>\n\n\n\n

(iv) In the given figure, the two reflex angles are<\/p>\n\n\n\n

\u2220AOB and \u2220COB<\/p>\n\n\n\n

6. In the given figure; PQR is a straight line. If:<\/strong><\/p>\n\n\n\n

(i) \u2220SQR = 750<\/sup>; find \u2220PQS.<\/strong><\/p>\n\n\n\n

(ii) \u2220PQS = 1100<\/sup>; find \u2220RQS<\/strong><\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) Given PQR is a straight line<\/p>\n\n\n\n

In the given figure,<\/p>\n\n\n\n

\u2220PQS + \u2220SQR = 1800<\/sup> {linear pair of angles}<\/p>\n\n\n\n

\u2220PQ S + 75 = 180<\/p>\n\n\n\n

On further calculation, we get,<\/p>\n\n\n\n

\u2220PQS = 180 \u2013 75<\/p>\n\n\n\n

\u2220PQS = 1050<\/sup><\/p>\n\n\n\n

(ii) Given PQR is a straight line<\/p>\n\n\n\n

\u2220PQS + \u2220RQS = 1800<\/sup><\/p>\n\n\n\n

1100<\/sup> + \u2220RQS = 1800<\/sup><\/p>\n\n\n\n

On further calculation, we get,<\/p>\n\n\n\n

\u2220RQS = 1800<\/sup> \u2013 1100<\/sup><\/p>\n\n\n\n

\u2220RQS = 700<\/sup><\/p>\n\n\n\n

7. In the given figure; AOC is a straight line. If angle AOB = 500<\/sup>, angle AOE = 900<\/sup> and angle COD = 250<\/sup>; find the measure of:<\/strong><\/p>\n\n\n\n

(i) angle BOC<\/strong><\/p>\n\n\n\n

(ii) angle EOD<\/strong><\/p>\n\n\n\n

(iii) obtuse angle BOD<\/strong><\/p>\n\n\n\n

(iv) reflex angle BOD<\/strong><\/p>\n\n\n\n

(v) reflex angle COE<\/strong><\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) Given<\/p>\n\n\n\n

AOC is a straight line<\/p>\n\n\n\n

\u2220AOB = 500<\/sup><\/p>\n\n\n\n

\u2220AOE = 900<\/sup><\/p>\n\n\n\n

\u2220COD = 250<\/sup><\/p>\n\n\n\n

To find the measure of \u2220BOC<\/p>\n\n\n\n

\u2220AOB + \u2220BOC = 1800<\/sup> (Linear pairs of angle)<\/p>\n\n\n\n

500<\/sup> + \u2220BOC = 1800<\/sup><\/p>\n\n\n\n

On further calculation, we get,<\/p>\n\n\n\n

\u2220BOC = 1800<\/sup> \u2013 500<\/sup><\/p>\n\n\n\n

\u2220BOC = 1300<\/sup><\/p>\n\n\n\n

(ii) Given<\/p>\n\n\n\n

AOC is a straight line<\/p>\n\n\n\n

\u2220AOB = 500<\/sup><\/p>\n\n\n\n

\u2220AOE = 900<\/sup><\/p>\n\n\n\n

\u2220COD = 250<\/sup><\/p>\n\n\n\n

To find the measure \u2220EOD<\/p>\n\n\n\n

\u2220EOD + \u2220COD = 900<\/sup> (Since \u2220AOE = 900<\/sup>)<\/p>\n\n\n\n

\u2220EOD + 250<\/sup> = 900<\/sup><\/p>\n\n\n\n

\u2220EOD = 900<\/sup> \u2013 250<\/sup><\/p>\n\n\n\n

We get,<\/p>\n\n\n\n

\u2220EOD = 650<\/sup><\/p>\n\n\n\n

(iii) Given<\/p>\n\n\n\n

AOC is a straight line<\/p>\n\n\n\n

\u2220AOB = 500<\/sup><\/p>\n\n\n\n

\u2220AOE = 900<\/sup><\/p>\n\n\n\n

\u2220COD = 250<\/sup><\/p>\n\n\n\n

To find the measure of obtuse angle BOD,<\/p>\n\n\n\n

\u2220BOD = \u2220BOC + \u2220COD<\/p>\n\n\n\n

Substituting the value of \u2220BOC and \u2220COD, we get,<\/p>\n\n\n\n

\u2220BOD = 1300<\/sup> + 250<\/sup><\/p>\n\n\n\n

We get,<\/p>\n\n\n\n

\u2220BOD = 1550<\/sup><\/p>\n\n\n\n

(iv) Given<\/p>\n\n\n\n

AOC is a straight line<\/p>\n\n\n\n

\u2220AOB = 500<\/sup><\/p>\n\n\n\n

\u2220AOE = 900<\/sup><\/p>\n\n\n\n

\u2220COD = 250<\/sup><\/p>\n\n\n\n

To find the measure of reflex angel BOD<\/p>\n\n\n\n

\u2220BOD = 3600<\/sup> \u2013 \u2220BOD<\/p>\n\n\n\n

= 3600<\/sup> \u2013 1550<\/sup><\/p>\n\n\n\n

We get,<\/p>\n\n\n\n

\u2220BOD = 2050<\/sup><\/p>\n\n\n\n

(iv) Given<\/p>\n\n\n\n

AOC is a straight line<\/p>\n\n\n\n

\u2220AOB = 500<\/sup><\/p>\n\n\n\n

\u2220AOE = 900<\/sup><\/p>\n\n\n\n

\u2220COD = 250<\/sup><\/p>\n\n\n\n

To find the measure of reflex angle COE<\/p>\n\n\n\n

\u2220COE = 3600<\/sup> \u2013 \u2220COE<\/p>\n\n\n\n

= 3600<\/sup> \u2013 (\u2220COD + \u2220EOD)<\/p>\n\n\n\n

= 3600<\/sup> \u2013 (250<\/sup> + 650<\/sup>)<\/p>\n\n\n\n

We get,<\/p>\n\n\n\n

= 3600<\/sup> \u2013 900<\/sup><\/p>\n\n\n\n

\u2220COE = 2700<\/sup><\/p>\n\n\n\n

8. In the given figure if:<\/strong><\/p>\n\n\n\n

(i) a = 1300<\/sup>; find b<\/strong><\/p>\n\n\n\n

(ii) b = 200; find a<\/strong><\/p>\n\n\n\n

(iii) a = 5 \/ 3 right angle, find b<\/strong><\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) From figure,<\/p>\n\n\n\n

a + b = 3600<\/sup><\/p>\n\n\n\n

Substitute the value of a in above equation<\/p>\n\n\n\n

1300<\/sup> + b = 3600<\/sup><\/p>\n\n\n\n

b = 3600<\/sup> \u2013 1300<\/sup><\/p>\n\n\n\n

We get,<\/p>\n\n\n\n

b = 2300<\/sup><\/p>\n\n\n\n

(ii) From figure,<\/p>\n\n\n\n

a + b = 3600<\/sup><\/p>\n\n\n\n

a + 2000<\/sup> = 3600<\/sup><\/p>\n\n\n\n

a = 3600<\/sup> \u2013 2000<\/sup><\/p>\n\n\n\n

We get,<\/p>\n\n\n\n

a = 1600<\/sup><\/p>\n\n\n\n

(iii) From figure,<\/p>\n\n\n\n

a = 5 \/ 3 right angle<\/p>\n\n\n\n

= 5 \/ 3 \u00d7 900<\/sup><\/p>\n\n\n\n

We get,<\/p>\n\n\n\n

a = 1500<\/sup><\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

a + b = 3600<\/sup><\/p>\n\n\n\n

Substitute the value of a in above equation<\/p>\n\n\n\n

1500<\/sup> + b = 3600<\/sup><\/p>\n\n\n\n

b = 3600<\/sup> \u2013 1500<\/sup><\/p>\n\n\n\n

b = 2100<\/sup><\/p>\n\n\n\n

9. In the given diagram, ABC is a straight line<\/strong><\/p>\n\n\n\n

(i) If x = 530<\/sup>, find y<\/strong><\/p>\n\n\n\n

(ii) If y =<\/strong>
\"Selina right angles; find x.<\/strong><\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) From the figure,<\/p>\n\n\n\n

Given that ABC is a straight line<\/p>\n\n\n\n

Hence,<\/p>\n\n\n\n

\u2220ABD + \u2220DBC = 1800<\/sup> {Linear pair of angles}<\/p>\n\n\n\n

x + y = 1800<\/sup><\/p>\n\n\n\n

530<\/sup> + y = 1800<\/sup><\/p>\n\n\n\n

y = 1800<\/sup> \u2013 530<\/sup><\/p>\n\n\n\n

y = 1270<\/sup><\/p>\n\n\n\n

(ii) From given figure,<\/p>\n\n\n\n

x + y = 1800<\/sup><\/p>\n\n\n\n

x +
\"Selinaright angles = 1800<\/sup><\/p>\n\n\n\n

x + 3 \/ 2 \u00d7 900<\/sup> = 1800<\/sup><\/p>\n\n\n\n

On further calculation, we get<\/p>\n\n\n\n

x + 1350<\/sup> = 1800<\/sup><\/p>\n\n\n\n

x = 1800<\/sup> \u2013 1350<\/sup><\/p>\n\n\n\n

x = 450<\/sup><\/p>\n\n\n\n

10. In the given figure, AOB is a straight line. Find the value of x and also answer each of the following:<\/strong><\/p>\n\n\n\n

(i) \u2220AOP = \u2026\u2026.<\/strong><\/p>\n\n\n\n

(ii) \u2220BOP = \u2026\u2026.<\/strong><\/p>\n\n\n\n

(iii) which angle is obtuse?<\/strong><\/p>\n\n\n\n

(iv) which angle is acute?<\/strong><\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given from figure,<\/p>\n\n\n\n

\u2220AOP = x + 300<\/sup><\/p>\n\n\n\n

\u2220BOP = x \u2013 300<\/sup><\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

\u2220AOP + \u2220BOP = 1800<\/sup> (Since \u2220AOB is a straight line)<\/p>\n\n\n\n

(x + 300<\/sup>)+ (x \u2013 300<\/sup>) = 1800<\/sup><\/p>\n\n\n\n

2x = 1800<\/sup><\/p>\n\n\n\n

We get,<\/p>\n\n\n\n

x = 900<\/sup><\/p>\n\n\n\n

(i) \u2220AOP = x + 300<\/sup><\/p>\n\n\n\n

= 900<\/sup> + 300<\/sup><\/p>\n\n\n\n

= 1200<\/sup><\/p>\n\n\n\n

(ii) \u2220BOP = x \u2013 300<\/sup><\/p>\n\n\n\n

= 900<\/sup> \u2013 300<\/sup><\/p>\n\n\n\n

We get,<\/p>\n\n\n\n

= 600<\/sup><\/p>\n\n\n\n

(iii) The obtuse angle is \u2220AOP<\/p>\n\n\n\n

(iv) The acute angle is \u2220BOP<\/p>\n\n\n\n

11.In the given figure, PQR is a straight line. Find x. Then complete the following:<\/strong><\/p>\n\n\n\n

(i) \u2220AQB = \u2026\u2026\u2026\u2026\u2026<\/strong><\/p>\n\n\n\n

(ii) \u2220BQP = \u2026\u2026\u2026\u2026\u2026<\/strong><\/p>\n\n\n\n

(iii) \u2220AQR = \u2026\u2026\u2026\u2026\u2026.<\/strong><\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given<\/p>\n\n\n\n

PQR is a straight line<\/p>\n\n\n\n

\u2220AQP = x + 200<\/sup><\/p>\n\n\n\n

\u2220AQB = 2x + 100<\/sup><\/p>\n\n\n\n

\u2220BQR = x \u2013 100<\/sup><\/p>\n\n\n\n

Since PQR is a straight line<\/p>\n\n\n\n

\u2220AQP + \u2220AQB + \u2220BQR = 1800<\/sup><\/p>\n\n\n\n

(x + 200<\/sup>) + (2x + 100<\/sup>) + (x \u2013 100<\/sup>) = 1800<\/sup><\/p>\n\n\n\n

We get,<\/p>\n\n\n\n

4x + 200<\/sup> = 1800<\/sup><\/p>\n\n\n\n

4x = 1800<\/sup> \u2013 200<\/sup><\/p>\n\n\n\n

4x = 1600<\/sup><\/p>\n\n\n\n

x = 1600<\/sup> \/ 4<\/p>\n\n\n\n

We get,<\/p>\n\n\n\n

x = 400<\/sup><\/p>\n\n\n\n

(i) \u2220AQB = 2x + 100<\/sup><\/p>\n\n\n\n

= 2 \u00d7 400<\/sup> + 100<\/sup><\/p>\n\n\n\n

= 800<\/sup> + 100<\/sup><\/p>\n\n\n\n

= 900<\/sup><\/p>\n\n\n\n

Similarly,<\/p>\n\n\n\n

\u2220AQP = x + 200<\/sup><\/p>\n\n\n\n

\u2220AQP = 400<\/sup> + 200<\/sup><\/p>\n\n\n\n

\u2220AQP = 600<\/sup><\/p>\n\n\n\n

\u2220BQR = x \u2013 100<\/sup><\/p>\n\n\n\n

\u2220BQR = 400<\/sup> \u2013 100<\/sup><\/p>\n\n\n\n

\u2220BQR = 300<\/sup><\/p>\n\n\n\n

(ii) \u2220BQP = \u2220AQP + \u2220AQB<\/p>\n\n\n\n

= 600<\/sup> + 900<\/sup><\/p>\n\n\n\n

= 1500<\/sup><\/p>\n\n\n\n

(iii) \u2220AQR = \u2220AQB + \u2220BQR<\/p>\n\n\n\n

= 900<\/sup> + 300<\/sup><\/p>\n\n\n\n

= 1200<\/sup><\/p>\n\n\n\n

12. In the given figure, lines AB and CD intersect at point O.<\/strong><\/p>\n\n\n\n

(i) Find the value of \u2220a.<\/strong><\/p>\n\n\n\n

(ii) Name all the pairs of vertically opposite angles.<\/strong><\/p>\n\n\n\n

(iii) Name all the pairs of adjacent angles<\/strong><\/p>\n\n\n\n

(iv) Name all the reflex angles formed and write the measure of each.<\/strong><\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given<\/p>\n\n\n\n

AB and CD intersect each other at point C<\/p>\n\n\n\n

\u2220AOC = 680<\/sup><\/p>\n\n\n\n

(i) Here, AOB is a line<\/p>\n\n\n\n

Hence,<\/p>\n\n\n\n

\u2220AOC + \u2220BOC = 1800<\/sup> (Linear pairs of angles)<\/p>\n\n\n\n

680<\/sup> + a = 1800<\/sup><\/p>\n\n\n\n

a = 1800<\/sup> \u2013 680<\/sup><\/p>\n\n\n\n

a = 1120<\/sup><\/p>\n\n\n\n

(ii) The pairs of vertically opposite angles are,<\/p>\n\n\n\n

\u2220AOC and \u2220BOD and \u2220BOC and \u2220AOD<\/p>\n\n\n\n

(iii) The pairs of adjacent angles are,<\/p>\n\n\n\n

\u2220AOC and \u2220BOC, \u2220BOC and \u2220BOD, \u2220BOD and \u2220DOA, \u2220DOA and \u2220AOC<\/p>\n\n\n\n

(iv) The reflex angles in the given figure are,<\/p>\n\n\n\n

\u2220BOC and \u2220DOA,<\/p>\n\n\n\n

Reflex angle BOC = 1800<\/sup> + 680<\/sup><\/p>\n\n\n\n

= 2480<\/sup><\/p>\n\n\n\n

Reflex angle DOA = 1800<\/sup> + 680<\/sup><\/p>\n\n\n\n

= 2480<\/sup><\/p>\n\n\n\n

13. In the given figure:<\/strong><\/p>\n\n\n\n

(i) If \u2220AOB = 450<\/sup>, \u2220BOC = 300<\/sup> and \u2220AOD = 1100<\/sup>;<\/strong><\/p>\n\n\n\n

Find: angles COD and BOD<\/strong><\/p>\n\n\n\n

(ii) If \u2220BOC = \u2220DOC = 340<\/sup> and \u2220AOD = 1200<\/sup>;<\/strong><\/p>\n\n\n\n

Find: angle AOB and angle AOC<\/strong><\/p>\n\n\n\n

(iii) If \u2220AOB = \u2220BOC = \u2220COD = 380<\/sup><\/strong><\/p>\n\n\n\n

Find: reflex angle AOC and reflex angle AOD<\/strong><\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) \u2220COD = \u2220AOD \u2013 \u2220AOC<\/p>\n\n\n\n

= \u2220AOD \u2013 (\u2220AOB + \u2220BOC)<\/p>\n\n\n\n

= 1100<\/sup> \u2013 (450<\/sup> + 300<\/sup>)<\/p>\n\n\n\n

We get,<\/p>\n\n\n\n

= 1100<\/sup> \u2013 750<\/sup><\/p>\n\n\n\n

= 350<\/sup><\/p>\n\n\n\n

Hence,<\/p>\n\n\n\n

\u2220COD = 350<\/sup><\/p>\n\n\n\n

\u2220BOD = \u2220BOC + \u2220COD<\/p>\n\n\n\n

= 300<\/sup> + 350<\/sup><\/p>\n\n\n\n

= 650<\/sup><\/p>\n\n\n\n

Hence,<\/p>\n\n\n\n

\u2220BOD = 650<\/sup><\/p>\n\n\n\n

(ii) \u2220AOB = \u2220AOD \u2013 \u2220BOD<\/p>\n\n\n\n

= \u2220AOD \u2013 (\u2220BOC + \u2220COD)<\/p>\n\n\n\n

= 1200<\/sup> \u2013 (340<\/sup> + 340<\/sup>)<\/p>\n\n\n\n

We get,<\/p>\n\n\n\n

= 1200<\/sup> \u2013 680<\/sup><\/p>\n\n\n\n

= 520<\/sup><\/p>\n\n\n\n

Hence,<\/p>\n\n\n\n

\u2220AOB = 520<\/sup><\/p>\n\n\n\n

\u2220AOC = \u2220AOB + \u2220BOC<\/p>\n\n\n\n

= 520<\/sup> + 340<\/sup><\/p>\n\n\n\n

= 860<\/sup><\/p>\n\n\n\n

Hence,<\/p>\n\n\n\n

\u2220AOC = 860<\/sup><\/p>\n\n\n\n

(iii) Reflex angel AOC = 3600<\/sup> \u2013 \u2220AOC<\/p>\n\n\n\n

= 3600<\/sup> \u2013 (\u2220AOB + \u2220BOC)<\/p>\n\n\n\n

= 3600<\/sup> \u2013 (380<\/sup> + 380<\/sup>)<\/p>\n\n\n\n

= 3600<\/sup> \u2013 760<\/sup><\/p>\n\n\n\n

We get,<\/p>\n\n\n\n

= 2840<\/sup><\/p>\n\n\n\n

Hence,<\/p>\n\n\n\n

\u2220AOC = 2840<\/sup><\/p>\n\n\n\n

Reflex angle AOD = 3600<\/sup> \u2013 \u2220AOD<\/p>\n\n\n\n

= 3600<\/sup> \u2013 (\u2220AOB + \u2220BOC + \u2220COD)<\/p>\n\n\n\n

= 3600<\/sup> \u2013 (380<\/sup> + 380<\/sup> + 380<\/sup>)<\/p>\n\n\n\n

= 3600<\/sup> \u2013 1140<\/sup><\/p>\n\n\n\n

We get,<\/p>\n\n\n\n

= 2460<\/sup><\/p>\n\n\n\n

Hence,<\/p>\n\n\n\n

\u2220AOC = 2460<\/sup><\/p>\n\n\n\n

\n\n\n\n

Exercise 24(B)<\/h3>\n\n\n\n

1. Write the complement angle of:<\/strong><\/p>\n\n\n\n

(i) 450<\/sup><\/strong><\/p>\n\n\n\n

(ii) x0<\/sup><\/strong><\/p>\n\n\n\n

(iii) (x \u2013 10)0<\/sup><\/strong><\/p>\n\n\n\n

(iv) 200<\/sup> + y0<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) The complement angle of 450<\/sup> is,<\/p>\n\n\n\n

= 900<\/sup> \u2013 450<\/sup><\/p>\n\n\n\n

= 450<\/sup><\/p>\n\n\n\n

Therefore, the complement angle of 450<\/sup> is 450<\/sup><\/p>\n\n\n\n

(ii) x0<\/sup><\/p>\n\n\n\n

The complement angle of x0<\/sup> is,<\/p>\n\n\n\n

= 900<\/sup> \u2013 x0<\/sup><\/p>\n\n\n\n

= (90 \u2013 x)0<\/sup><\/p>\n\n\n\n

Therefore, the complement angle of x0<\/sup> is (90 \u2013 x)0<\/sup><\/p>\n\n\n\n

(iii) The complement angle of (x \u2013 10)0<\/sup> is,<\/p>\n\n\n\n

= 900<\/sup> \u2013 (x \u2013 10)0<\/sup><\/p>\n\n\n\n

= 900<\/sup> \u2013 x0<\/sup> + 100<\/sup><\/p>\n\n\n\n

= 1000<\/sup> \u2013 x0<\/sup><\/p>\n\n\n\n

= (100 \u2013 x)0<\/sup><\/p>\n\n\n\n

Therefore, the complement of (x \u2013 10)0<\/sup> is (100 \u2013 x)0<\/sup><\/p>\n\n\n\n

(iv) The complement angle of 200<\/sup> + y0<\/sup> is,<\/p>\n\n\n\n

= 900<\/sup> \u2013 (200<\/sup> + y0<\/sup>)<\/p>\n\n\n\n

= 900<\/sup> \u2013 200<\/sup> \u2013 y0<\/sup><\/p>\n\n\n\n

We get,<\/p>\n\n\n\n

= 700<\/sup> \u2013 y0<\/sup><\/p>\n\n\n\n

= (70 \u2013 y)0<\/sup><\/p>\n\n\n\n

2. Write the supplement angle of:<\/strong><\/p>\n\n\n\n

(i) 490<\/sup><\/strong><\/p>\n\n\n\n

(ii) 1110<\/sup><\/strong><\/p>\n\n\n\n

(iii) (x \u2013 30)0<\/sup><\/strong><\/p>\n\n\n\n

(iv) 200<\/sup> + y0<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) The supplement angle of 490<\/sup> is,<\/p>\n\n\n\n

= 1800<\/sup> \u2013 490<\/sup><\/p>\n\n\n\n

= 1310<\/sup><\/p>\n\n\n\n

Hence, the supplement angle of 490<\/sup> is 1310<\/sup><\/p>\n\n\n\n

(ii) The supplement angle of 1110<\/sup> is,<\/p>\n\n\n\n

= 1800<\/sup> \u2013 1110<\/sup><\/p>\n\n\n\n

= 690<\/sup><\/p>\n\n\n\n

Hence, the supplement angle of 1110<\/sup> is 690<\/sup><\/p>\n\n\n\n

(iii) The supplement angle of (x \u2013 30)0<\/sup> is,<\/p>\n\n\n\n

= 1800<\/sup> \u2013 (x \u2013 30)0<\/sup><\/p>\n\n\n\n

= 1800<\/sup> \u2013 x0<\/sup> + 300<\/sup><\/p>\n\n\n\n

= 2100<\/sup> \u2013 x0<\/sup><\/p>\n\n\n\n

= (210 \u2013 x)0<\/sup><\/p>\n\n\n\n

Hence, the supplement angle of (x \u2013 30)0<\/sup> is (210 \u2013 x)0<\/sup><\/p>\n\n\n\n

(iv) The supplement angle of 200<\/sup> + y0<\/sup> is,<\/p>\n\n\n\n

= 1800<\/sup> \u2013 (200<\/sup> + y0<\/sup>)<\/p>\n\n\n\n

= 1800<\/sup> \u2013 200<\/sup> \u2013 y0<\/sup><\/p>\n\n\n\n

= 1600<\/sup> \u2013 y0<\/sup><\/p>\n\n\n\n

= (160 \u2013 y)0<\/sup><\/p>\n\n\n\n

Hence, the supplement angle of 200<\/sup> + y0<\/sup> is (160 \u2013 y)0<\/sup><\/p>\n\n\n\n

3. Write the complement angle of:<\/strong><\/p>\n\n\n\n

(i) 1 \/ 2 of 600<\/sup><\/strong><\/p>\n\n\n\n

(ii) 1 \/ 5 of 1600<\/sup><\/strong><\/p>\n\n\n\n

(iii) 2 \/ 5 of 700<\/sup><\/strong><\/p>\n\n\n\n

(iv) 1 \/ 6 of 900<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) The complement angle of (1 \/ 2 of 600<\/sup>) is,<\/p>\n\n\n\n

= 900<\/sup> \u2013 (1 \/ 2 \u00d7 600<\/sup>)<\/p>\n\n\n\n

We get,<\/p>\n\n\n\n

= 900<\/sup> \u2013 300<\/sup><\/p>\n\n\n\n

= 600<\/sup><\/p>\n\n\n\n

Therefore, the complement angle of (1 \/ 2 of 600<\/sup>) is 600<\/sup><\/p>\n\n\n\n

(ii) The complement angle of (1 \/ 5 of 1600<\/sup>) is,<\/p>\n\n\n\n

= 900<\/sup> \u2013 (1 \/ 5 \u00d7 1600<\/sup>)<\/p>\n\n\n\n

We get,<\/p>\n\n\n\n

= 900<\/sup> \u2013 320<\/sup><\/p>\n\n\n\n

= 580<\/sup><\/p>\n\n\n\n

Therefore, the complement angle of (1 \/ 5 of 1600<\/sup>) is 580<\/sup><\/p>\n\n\n\n

(iii) The complement angle of (2 \/ 5 of 700<\/sup>) is,<\/p>\n\n\n\n

= 900<\/sup> \u2013 (2 \/ 5 \u00d7 700<\/sup>)<\/p>\n\n\n\n

We get,<\/p>\n\n\n\n

= 900<\/sup> \u2013 280<\/sup><\/p>\n\n\n\n

= 620<\/sup><\/p>\n\n\n\n

Therefore, the complement of (2 \/ 5 of 700<\/sup>) is 620<\/sup><\/p>\n\n\n\n

(iv) The complement angle of (1 \/ 6 of 900<\/sup>) is,<\/p>\n\n\n\n

= 900<\/sup> \u2013 (1 \/ 6 \u00d7 900<\/sup>)<\/p>\n\n\n\n

We get,<\/p>\n\n\n\n

= 900<\/sup> \u2013 150<\/sup><\/p>\n\n\n\n

= 750<\/sup><\/p>\n\n\n\n

Therefore, the complement of (1 \/ 6 of 900<\/sup>) is 750<\/sup><\/p>\n\n\n\n

4.<\/strong><\/p>\n\n\n\n

(i) 50% of 1200<\/sup><\/strong><\/p>\n\n\n\n

(ii) 1 \/ 3 of 1500<\/sup><\/strong><\/p>\n\n\n\n

(iii) 60% of 1000<\/sup><\/strong><\/p>\n\n\n\n

(iv) 3 \/ 4 of 1600<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) Supplement angle of 50% of 1200<\/sup> is,<\/p>\n\n\n\n

= 1800<\/sup> \u2013 (50% of 1200<\/sup>)<\/p>\n\n\n\n

= 1800<\/sup> \u2013 [(1200<\/sup> \u00d7 50) \/ 100]<\/p>\n\n\n\n

We get,<\/p>\n\n\n\n

= 1800<\/sup> \u2013 600<\/sup><\/p>\n\n\n\n

= 1200<\/sup><\/p>\n\n\n\n

Hence, supplement angle of 50% of 1200<\/sup> is 1200<\/sup><\/p>\n\n\n\n

(ii) Supplement angle of (1 \/ 3 of 1500<\/sup>) is,<\/p>\n\n\n\n

= 1800<\/sup> \u2013 (1 \/ 3 \u00d7 1500<\/sup>)<\/p>\n\n\n\n

We get,<\/p>\n\n\n\n

= 1800<\/sup> \u2013 500<\/sup><\/p>\n\n\n\n

= 1300<\/sup><\/p>\n\n\n\n

Hence, supplement angle of (1 \/ 3 of 1500<\/sup>) is 1300<\/sup><\/p>\n\n\n\n

(iii) Supplement angle of 60% of 1000<\/sup> is,<\/p>\n\n\n\n

= 1800<\/sup> \u2013 (60% of 1000<\/sup>)<\/p>\n\n\n\n

= 1800<\/sup> \u2013 [(60 \u00d7 100) \/ 100]<\/p>\n\n\n\n

We get,<\/p>\n\n\n\n

= 1800<\/sup> \u2013 600<\/sup><\/p>\n\n\n\n

= 1200<\/sup><\/p>\n\n\n\n

Hence, the supplement angle of (60% of 1000<\/sup>) is 1200<\/sup><\/p>\n\n\n\n

(iv) Supplement angle of 3 \/ 4 of 1600<\/sup><\/p>\n\n\n\n

= 1800<\/sup> \u2013 (3 \/ 4 of 1600<\/sup>)<\/p>\n\n\n\n

We get,<\/p>\n\n\n\n

= 1800<\/sup> \u2013 1200<\/sup><\/p>\n\n\n\n

= 600<\/sup><\/p>\n\n\n\n

Hence, the supplement angle of (3 \/ 4 of 1600<\/sup>) is 600<\/sup><\/p>\n\n\n\n

5. Find the angle:<\/strong><\/p>\n\n\n\n

(i) that is equal to its complement?<\/strong><\/p>\n\n\n\n

(ii) that is equal to its supplement?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) The angle equal to its complement is 450<\/sup><\/p>\n\n\n\n

(ii) The angle equal to its supplement is 900<\/sup><\/p>\n\n\n\n

6. Two complementary angles are in the ratio 7: 8. Find the angles<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given<\/p>\n\n\n\n

Two complementary angles are in the ratio 7: 8<\/p>\n\n\n\n

Let the two complementary angles be 7x and 8x<\/p>\n\n\n\n

Hence,<\/p>\n\n\n\n

7x + 8x = 900<\/sup><\/p>\n\n\n\n

15x = 900<\/sup><\/p>\n\n\n\n

We get,<\/p>\n\n\n\n

x = 900<\/sup> \/ 15<\/p>\n\n\n\n

x = 60<\/sup><\/p>\n\n\n\n

So, two complementary angles are<\/p>\n\n\n\n

7x = 7 \u00d7 60<\/sup><\/p>\n\n\n\n

= 420<\/sup><\/p>\n\n\n\n

8x = 8 \u00d7 60<\/sup><\/p>\n\n\n\n

= 480<\/sup><\/p>\n\n\n\n

Therefore, two complementary angles are 420<\/sup> and 480<\/sup><\/p>\n\n\n\n

7. Two supplementary angles are in the ratio 7: 11. Find the angles<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given<\/p>\n\n\n\n

Two supplementary angles are in the ratio 7: 11<\/p>\n\n\n\n

Let the two supplementary angles be 7x and 11x<\/p>\n\n\n\n

Hence,<\/p>\n\n\n\n

7x + 11x = 1800<\/sup><\/p>\n\n\n\n

18x = 1800<\/sup><\/p>\n\n\n\n

x = 1800<\/sup> \/ 18<\/p>\n\n\n\n

We get,<\/p>\n\n\n\n

x = 100<\/sup><\/p>\n\n\n\n

So, two supplementary angles are<\/p>\n\n\n\n

7x = 7 \u00d7 100<\/sup><\/p>\n\n\n\n

= 700<\/sup><\/p>\n\n\n\n

11x = 11 \u00d7 100<\/sup><\/p>\n\n\n\n

= 1100<\/sup><\/p>\n\n\n\n

Therefore, two supplementary angles are 700<\/sup> and 1100<\/sup><\/p>\n\n\n\n

8. The measures of two complementary angles are (2x \u2013 7)0<\/sup> and (x + 4)0<\/sup>. Find x.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given<\/p>\n\n\n\n

(2x \u2013 7)0<\/sup> and (x + 4)0<\/sup> are two complementary angles<\/p>\n\n\n\n

We know that,<\/p>\n\n\n\n

Sum of two complementary angles = 900<\/sup><\/p>\n\n\n\n

Hence,<\/p>\n\n\n\n

(2x \u2013 7)0<\/sup> + (x + 4)0<\/sup> = 900<\/sup><\/p>\n\n\n\n

2x \u2013 7 + x + 4 = 900<\/sup><\/p>\n\n\n\n

3x \u2013 3 = 900<\/sup><\/p>\n\n\n\n

3x = 900<\/sup> + 30<\/sup><\/p>\n\n\n\n

3x = 930<\/sup><\/p>\n\n\n\n

x = 930<\/sup> \/ 3<\/p>\n\n\n\n

We get,<\/p>\n\n\n\n

x = 310<\/sup><\/p>\n\n\n\n

Therefore, the value of x = 310<\/sup><\/p>\n\n\n\n

9. The measures of two supplementary angles are (3x + 15)0<\/sup> and (2x + 5)0<\/sup>. Find x.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given<\/p>\n\n\n\n

(3x + 15)0<\/sup> and (2x + 5)0<\/sup> are two supplementary angles<\/p>\n\n\n\n

We know that,<\/p>\n\n\n\n

Sum of two supplementary angles = 1800<\/sup><\/p>\n\n\n\n

Hence,<\/p>\n\n\n\n

(3x + 15)0<\/sup> + (2x + 5)0<\/sup> = 1800<\/sup><\/p>\n\n\n\n

3x + 15 + 2x + 5 = 1800<\/sup><\/p>\n\n\n\n

5x + 200<\/sup> = 1800<\/sup><\/p>\n\n\n\n

5x = 1800<\/sup> \u2013 200<\/sup><\/p>\n\n\n\n

5x = 1600<\/sup><\/p>\n\n\n\n

x = 1600<\/sup> \/ 5<\/p>\n\n\n\n

We get,<\/p>\n\n\n\n

x = 320<\/sup><\/p>\n\n\n\n

Therefore, the value of x is 320<\/sup><\/p>\n\n\n\n

10. For an angle x0<\/sup>, find:<\/strong><\/p>\n\n\n\n

(i) the complementary angle<\/strong><\/p>\n\n\n\n

(ii) the supplementary angle<\/strong><\/p>\n\n\n\n

(iii) the value of x0<\/sup> if its supplementary angle is three times its complementary angle.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

For an angle x0<\/sup><\/p>\n\n\n\n

(i) Complementary angle of x0<\/sup> is,<\/p>\n\n\n\n

= (900<\/sup> \u2013 x)<\/p>\n\n\n\n

(ii) Supplementary angle of x0<\/sup> is,<\/p>\n\n\n\n

= (1800<\/sup> \u2013 x)<\/p>\n\n\n\n

(iii) As per the statement,<\/p>\n\n\n\n

Supplementary angle = 3 (Complementary angle)<\/p>\n\n\n\n

1800<\/sup> \u2013 x = 3 (900<\/sup> \u2013 x)<\/p>\n\n\n\n

1800<\/sup> \u2013 x = 2700<\/sup> \u2013 3x<\/p>\n\n\n\n

\u2013 x + 3x = 2700<\/sup> \u2013 1800<\/sup><\/p>\n\n\n\n

2x = 900<\/sup><\/p>\n\n\n\n

x = 900<\/sup> \/ 2<\/p>\n\n\n\n

We get,<\/p>\n\n\n\n

x = 450<\/sup><\/p>\n\n\n\n

Therefore, the value of x is 450<\/sup><\/p>\n\n\n\n

Download PDF<\/strong><\/h2>\n\n\n\n

Selina Class 6 ICSE Solutions Mathematics : Chapter 24- Angles<\/p>\n\n\n\n

Download PDF<\/strong>: Selina Class 6 ICSE Solutions Mathematics : Chapter 24-\u00a0Angles PDF<\/a><\/p>\n\n\n\n

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