{"id":546728,"date":"2021-10-07T11:05:20","date_gmt":"2021-10-07T11:05:20","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=546728"},"modified":"2021-10-09T07:45:02","modified_gmt":"2021-10-09T07:45:02","slug":"rd-sharma-solutions-for-class-7-maths-chapter-14-lines-and-angles","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-7-maths-chapter-14-lines-and-angles\/","title":{"rendered":"RD Sharma Solutions for Class 7 Maths Chapter 14\u2013Lines And Angles"},"content":{"rendered":"\n

Class 7: Maths Chapter 14 solutions. Complete Class 7 Maths Chapter 14 Notes.<\/p>\n\n\n\n

RD Sharma Solutions for Class 7 Maths Chapter 14\u2013Lines And Angles<\/h2>\n\n\n\n

RD Sharma 7th Maths Chapter 14, Class 7 Maths Chapter 14 solutions<\/p>\n\n\n\n

Exercise 14.1 Page No: 14.6<\/p>\n\n\n\n

1. Write down each pair of adjacent angles shown in fig. 13.<\/strong><\/p>\n\n\n\n

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

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

The angles that have common vertex and a common arm are known as adjacent angles<\/p>\n\n\n\n

Therefore the adjacent angles in given figure are:<\/p>\n\n\n\n

\u2220DOC and \u2220BOC<\/p>\n\n\n\n

\u2220COB and \u2220BOA<\/p>\n\n\n\n

2. In Fig. 14, name all the pairs of adjacent angles.<\/strong><\/p>\n\n\n\n

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

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

The angles that have common vertex and a common arm are known as adjacent angles.<\/p>\n\n\n\n

In fig (i), the adjacent angles are<\/p>\n\n\n\n

\u2220EBA and \u2220ABC<\/p>\n\n\n\n

\u2220ACB and \u2220BCF<\/p>\n\n\n\n

\u2220BAC and \u2220CAD<\/p>\n\n\n\n

In fig (ii), the adjacent angles are<\/p>\n\n\n\n

\u2220BAD and \u2220DAC<\/p>\n\n\n\n

\u2220BDA and \u2220CDA<\/p>\n\n\n\n

3. In fig. 15, write down<\/strong><\/p>\n\n\n\n

(i) Each linear pair<\/strong><\/p>\n\n\n\n

(ii) Each pair of vertically opposite angles.<\/strong><\/p>\n\n\n\n

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

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

(i) The two adjacent angles are said to form a linear pair of angles if their non \u2013 common arms are two opposite rays.<\/p>\n\n\n\n

\u22201 and \u22203<\/p>\n\n\n\n

\u22201 and \u22202<\/p>\n\n\n\n

\u22204 and \u22203<\/p>\n\n\n\n

\u22204 and \u22202<\/p>\n\n\n\n

\u22205 and \u22206<\/p>\n\n\n\n

\u22205 and \u22207<\/p>\n\n\n\n

\u22206 and \u22208<\/p>\n\n\n\n

\u22207 and \u22208<\/p>\n\n\n\n

(ii) The two angles formed by two intersecting lines and have no common arms are called vertically opposite angles.<\/p>\n\n\n\n

\u22201 and \u22204<\/p>\n\n\n\n

\u22202 and \u22203<\/p>\n\n\n\n

\u22205 and \u22208<\/p>\n\n\n\n

\u22206 and \u22207<\/p>\n\n\n\n

4. Are the angles 1 and 2 given in Fig. 16 adjacent angles?<\/strong><\/p>\n\n\n\n

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

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

No, because they don\u2019t have common vertex.<\/p>\n\n\n\n

5. Find the complement of each of the following angles:<\/strong><\/p>\n\n\n\n

(i) 35o<\/sup><\/strong><\/p>\n\n\n\n

(ii) 72o<\/sup><\/strong><\/p>\n\n\n\n

(iii) 45o<\/sup><\/strong><\/p>\n\n\n\n

(iv) 85o<\/sup><\/strong><\/p>\n\n\n\n

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

(i) The two angles are said to be complementary angles if the sum of those angles is 90o<\/sup><\/p>\n\n\n\n

Complementary angle for given angle is<\/p>\n\n\n\n

90o<\/sup> \u2013 35o<\/sup> = 55o<\/sup><\/p>\n\n\n\n

(ii) The two angles are said to be complementary angles if the sum of those angles is 90o<\/sup><\/p>\n\n\n\n

Complementary angle for given angle is<\/p>\n\n\n\n

900<\/sup> \u2013 72o<\/sup> = 18o<\/sup><\/p>\n\n\n\n

(iii) The two angles are said to be complementary angles if the sum of those angles is 90o<\/sup><\/p>\n\n\n\n

Complementary angle for given angle is<\/p>\n\n\n\n

90o <\/sup>\u2013 45o<\/sup> = 45o<\/sup><\/p>\n\n\n\n

(iv) The two angles are said to be complementary angles if the sum of those angles is 90o<\/sup><\/p>\n\n\n\n

Complementary angle for given angle is<\/p>\n\n\n\n

90o<\/sup> \u2013 85o<\/sup> = 5o<\/sup><\/p>\n\n\n\n

6. Find the supplement of each of the following angles:<\/strong><\/p>\n\n\n\n

(i) 70o<\/sup><\/strong><\/p>\n\n\n\n

(ii) 120o<\/sup><\/strong><\/p>\n\n\n\n

(iii) 135o<\/sup><\/strong><\/p>\n\n\n\n

(iv) 90o<\/sup><\/strong><\/p>\n\n\n\n

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

(i) The two angles are said to be supplementary angles if the sum of those angles is 180o<\/sup><\/p>\n\n\n\n

Therefore supplementary angle for the given angle is<\/p>\n\n\n\n

180o<\/sup> \u2013 70o<\/sup> = 110o<\/sup><\/p>\n\n\n\n

(ii) The two angles are said to be supplementary angles if the sum of those angles is 180o<\/sup><\/p>\n\n\n\n

Therefore supplementary angle for the given angle is<\/p>\n\n\n\n

180o<\/sup> \u2013 120o<\/sup> = 60o<\/sup><\/p>\n\n\n\n

(iii) The two angles are said to be supplementary angles if the sum of those angles is 180o<\/sup><\/p>\n\n\n\n

Therefore supplementary angle for the given angle is<\/p>\n\n\n\n

180o<\/sup> \u2013 135o<\/sup> = 45o<\/sup><\/p>\n\n\n\n

(iv) The two angles are said to be supplementary angles if the sum of those angles is 180o<\/sup><\/p>\n\n\n\n

Therefore supplementary angle for the given angle is<\/p>\n\n\n\n

180o<\/sup> \u2013 90o<\/sup> = 90o<\/sup><\/p>\n\n\n\n

7. Identify the complementary and supplementary pairs of angles from the following pairs:<\/strong><\/p>\n\n\n\n

(i) 25o<\/sup>, 65o<\/sup><\/strong><\/p>\n\n\n\n

(ii) 120o<\/sup>, 60o<\/sup><\/strong><\/p>\n\n\n\n

(iii) 63o<\/sup>, 27o<\/sup><\/strong><\/p>\n\n\n\n

(iv) 100o<\/sup>, 80o<\/sup><\/strong><\/p>\n\n\n\n

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

(i) 25o<\/sup> + 65o<\/sup> = 90o<\/sup> so, this is a complementary pair of angle.<\/p>\n\n\n\n

(ii) 120o<\/sup> + 60o<\/sup> = 180o<\/sup> so, this is a supplementary pair of angle.<\/p>\n\n\n\n

(iii) 63o<\/sup> + 27o<\/sup> = 90o<\/sup> so, this is a complementary pair of angle.<\/p>\n\n\n\n

(iv) 100o<\/sup> + 80o<\/sup> = 180o<\/sup> so, this is a supplementary pair of angle.<\/p>\n\n\n\n

8. Can two obtuse angles be supplementary, if both of them be<\/strong><\/p>\n\n\n\n

(i) Obtuse?<\/strong><\/p>\n\n\n\n

(ii) Right?<\/strong><\/p>\n\n\n\n

(iii) Acute?<\/strong><\/p>\n\n\n\n

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

(i) No, two obtuse angles cannot be supplementary<\/p>\n\n\n\n

Because, the sum of two angles is greater than 90o<\/sup> so their sum will be greater than 180o<\/sup><\/p>\n\n\n\n

(ii) Yes, two right angles can be supplementary<\/p>\n\n\n\n

Because, 90o<\/sup> + 90o<\/sup> = 180o<\/sup><\/p>\n\n\n\n

(iii) No, two acute angle cannot be supplementary<\/p>\n\n\n\n

Because, the sum of two angles is less than 90o<\/sup> so their sum will also be less than 90o<\/sup><\/p>\n\n\n\n

9. Name the four pairs of supplementary angles shown in Fig.17.<\/strong><\/p>\n\n\n\n

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

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

The two angles are said to be supplementary angles if the sum of those angles is 180o<\/sup><\/p>\n\n\n\n

The supplementary angles are<\/p>\n\n\n\n

\u2220AOC and \u2220COB<\/p>\n\n\n\n

\u2220BOC and \u2220DOB<\/p>\n\n\n\n

\u2220BOD and \u2220DOA<\/p>\n\n\n\n

\u2220AOC and \u2220DOA<\/p>\n\n\n\n

10. In Fig. 18, A, B, C are collinear points and \u2220DBA = \u2220EBA.<\/strong><\/p>\n\n\n\n

(i) Name two linear pairs.<\/strong><\/p>\n\n\n\n

(ii) Name two pairs of supplementary angles.<\/strong><\/p>\n\n\n\n

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

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

(i) Two adjacent angles are said to be form a linear pair of angles, if their non-common arms are two opposite rays.<\/p>\n\n\n\n

Therefore linear pairs are<\/p>\n\n\n\n

\u2220ABD and \u2220DBC<\/p>\n\n\n\n

\u2220ABE and \u2220EBC<\/p>\n\n\n\n

(ii) We know that every linear pair forms supplementary angles, these angles are<\/p>\n\n\n\n

\u2220ABD and \u2220DBC<\/p>\n\n\n\n

\u2220ABE and \u2220EBC<\/p>\n\n\n\n

11. If two supplementary angles have equal measure, what is the measure of each angle?<\/strong><\/p>\n\n\n\n

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

Let p and q be the two supplementary angles that are equal<\/p>\n\n\n\n

The two angles are said to be supplementary angles if the sum of those angles is 180o<\/sup><\/p>\n\n\n\n

\u2220p = \u2220q<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

\u2220p + \u2220q = 180o<\/sup><\/p>\n\n\n\n

\u2220p + \u2220p = 180o<\/sup><\/p>\n\n\n\n

2\u2220p = 180o<\/sup><\/p>\n\n\n\n

\u2220p = 180o<\/sup>\/2<\/p>\n\n\n\n

\u2220p = 90o<\/sup><\/p>\n\n\n\n

Therefore, \u2220p = \u2220q = 90o<\/sup><\/p>\n\n\n\n

12. If the complement of an angle is 28o<\/sup>, then find the supplement of the angle.<\/strong><\/p>\n\n\n\n

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

Given complement of an angle is 28o<\/sup><\/p>\n\n\n\n

Here, let x be the complement of the given angle 28o<\/sup><\/p>\n\n\n\n

Therefore, \u2220x + 28o<\/sup> = 90o<\/sup><\/p>\n\n\n\n

\u2220x = 90o<\/sup> \u2013 28o<\/sup><\/p>\n\n\n\n

= 62o<\/sup><\/p>\n\n\n\n

So, the supplement of the angle = 180o<\/sup> \u2013 62o<\/sup><\/p>\n\n\n\n

= 118o<\/sup><\/p>\n\n\n\n

13. In Fig. 19, name each linear pair and each pair of vertically opposite angles:<\/strong><\/p>\n\n\n\n

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

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

Two adjacent angles are said to be linear pair of angles, if their non-common arms are two opposite rays.<\/p>\n\n\n\n

Therefore linear pairs are listed below:<\/p>\n\n\n\n

\u22201 and \u22202<\/p>\n\n\n\n

\u22202 and \u22203<\/p>\n\n\n\n

\u22203 and \u22204<\/p>\n\n\n\n

\u22201 and \u22204<\/p>\n\n\n\n

\u22205 and \u22206<\/p>\n\n\n\n

\u22206 and \u22207<\/p>\n\n\n\n

\u22207 and \u22208<\/p>\n\n\n\n

\u22208 and \u22205<\/p>\n\n\n\n

\u22209 and \u222010<\/p>\n\n\n\n

\u222010 and \u222011<\/p>\n\n\n\n

\u222011 and \u222012<\/p>\n\n\n\n

\u222012 and \u22209<\/p>\n\n\n\n

The two angles are said to be vertically opposite angles if the two intersecting lines have no common arms.<\/p>\n\n\n\n

Therefore supplement of the angle are listed below:<\/p>\n\n\n\n

\u22201 and \u22203<\/p>\n\n\n\n

\u22204 and \u22202<\/p>\n\n\n\n

\u22205 and \u22207<\/p>\n\n\n\n

\u22206 and \u22208<\/p>\n\n\n\n

\u22209 and \u222011<\/p>\n\n\n\n

\u222010 and \u222012<\/p>\n\n\n\n

14. In Fig. 20, OE is the bisector of \u2220BOD. If \u22201 = 70o<\/sup>, find the magnitude of \u22202, \u22203 and \u22204.<\/strong><\/p>\n\n\n\n

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

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

Given, \u22201 = 70o<\/sup><\/p>\n\n\n\n

\u22203 = 2(\u22201)<\/p>\n\n\n\n

= 2(70o<\/sup>)<\/p>\n\n\n\n

\u22203 = 140o<\/sup><\/p>\n\n\n\n

\u22203 = \u22204<\/p>\n\n\n\n

As, OE is the angle bisector,<\/p>\n\n\n\n

\u2220DOB = 2(\u22201)<\/p>\n\n\n\n

= 2(70o<\/sup>)<\/p>\n\n\n\n

= 140o<\/sup><\/p>\n\n\n\n

\u2220DOB + \u2220AOC + \u2220COB +\u2220AOD = 360o<\/sup> [sum of the angle of circle = 360o<\/sup>]<\/p>\n\n\n\n

140o<\/sup> + 140o<\/sup> + 2(\u2220COB) = 360o<\/sup><\/p>\n\n\n\n

Since, \u2220COB = \u2220AOD<\/p>\n\n\n\n

2(\u2220COB) = 360o<\/sup> \u2013 280o<\/sup><\/p>\n\n\n\n

2(\u2220COB) = 80o<\/sup><\/p>\n\n\n\n

\u2220COB = 80o<\/sup>\/2<\/p>\n\n\n\n

\u2220COB = 40o<\/sup><\/p>\n\n\n\n

Therefore, \u2220COB = \u2220AOD = 40o<\/sup><\/p>\n\n\n\n

The angles are, \u22201 = 70o<\/sup>, \u22202 = 40o<\/sup>, \u22203 = 140o<\/sup> and \u22204 = 40o<\/sup><\/p>\n\n\n\n

15. One of the angles forming a linear pair is a right angle. What can you say about its other angle?<\/strong><\/p>\n\n\n\n

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

Given one of the angle of a linear pair is the right angle that is 90o<\/sup><\/p>\n\n\n\n

We know that linear pair angle is 180o<\/sup><\/p>\n\n\n\n

Therefore, the other angle is<\/p>\n\n\n\n

180o<\/sup> \u2013 90o<\/sup> = 90o<\/sup><\/p>\n\n\n\n

16. One of the angles forming a linear pair is an obtuse angle. What kind of angle is the other?<\/strong><\/p>\n\n\n\n

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

Given one of the angles of a linear pair is obtuse, then the other angle should be acute, because only then their sum will be 180o<\/sup>.<\/p>\n\n\n\n

17. One of the angles forming a linear pair is an acute angle. What kind of angle is the other?<\/strong><\/p>\n\n\n\n

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

Given one of the Angles of a linear pair is acute, then the other angle should be obtuse, only then their sum will be 180o<\/sup>.<\/p>\n\n\n\n

18. Can two acute angles form a linear pair?<\/strong><\/p>\n\n\n\n

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

No, two acute angles cannot form a linear pair because their sum is always less than 180o<\/sup>.<\/p>\n\n\n\n

19. If the supplement of an angle is 65o<\/sup>, then find its complement.<\/strong><\/p>\n\n\n\n

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

Let x be the required angle<\/p>\n\n\n\n

So, x + 65o<\/sup> = 180o<\/sup><\/p>\n\n\n\n

x = 180o<\/sup> \u2013 65o<\/sup><\/p>\n\n\n\n

x = 115o<\/sup><\/p>\n\n\n\n

The two angles are said to be complementary angles if the sum of those angles is 90o<\/sup> here it is more than 90o<\/sup> therefore the complement of the angle cannot be determined.<\/p>\n\n\n\n

20. Find the value of x in each of the following figures.<\/strong><\/p>\n\n\n\n

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

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

(i)  We know that \u2220BOA + \u2220BOC = 180o<\/sup>[Linear pair: The two adjacent angles are said to form a linear pair of angles if their non\u2013common arms are two opposite rays and sum of the angle is 180o<\/sup>]<\/p>\n\n\n\n

60o<\/sup> + xo<\/sup> = 180o<\/sup><\/p>\n\n\n\n

xo<\/sup> = 180o<\/sup> \u2013 60o<\/sup><\/p>\n\n\n\n

xo<\/sup> = 120o<\/sup><\/p>\n\n\n\n

(ii) We know that \u2220POQ + \u2220QOR = 180o<\/sup>[Linear pair: The two adjacent angles are said to form a linear pair of angles if their non\u2013common arms are two opposite rays and sum of the angle is 180o<\/sup>]<\/p>\n\n\n\n

3xo<\/sup> + 2xo<\/sup> = 180o<\/sup><\/p>\n\n\n\n

5xo<\/sup> = 180o<\/sup><\/p>\n\n\n\n

xo<\/sup> = 180o<\/sup>\/5<\/p>\n\n\n\n

xo<\/sup> = 36o<\/sup><\/p>\n\n\n\n

(iii) We know that \u2220LOP + \u2220PON + \u2220NOM = 180o<\/sup>[Linear pair: The two adjacent angles are said to form a linear pair of angles if their non\u2013common arms are two opposite rays and sum of the angle is 180o<\/sup>]<\/p>\n\n\n\n

Since, 35o<\/sup> + xo<\/sup> + 60o<\/sup> = 180o<\/sup><\/p>\n\n\n\n

xo<\/sup> = 180o<\/sup> \u2013 35o<\/sup> \u2013 60o<\/sup><\/p>\n\n\n\n

xo<\/sup> = 180o<\/sup> \u2013 95o<\/sup><\/p>\n\n\n\n

xo<\/sup> = 85o<\/sup><\/p>\n\n\n\n

(iv) We know that \u2220DOC + \u2220DOE + \u2220EOA + \u2220AOB+ \u2220BOC = 360o<\/sup><\/p>\n\n\n\n

83o<\/sup> + 92o<\/sup> + 47o<\/sup> + 75o<\/sup> + xo<\/sup> = 360o<\/sup><\/p>\n\n\n\n

xo<\/sup> + 297o<\/sup> = 360o<\/sup><\/p>\n\n\n\n

xo<\/sup> = 360o<\/sup> \u2013 297o<\/sup><\/p>\n\n\n\n

xo<\/sup> = 63o<\/sup><\/p>\n\n\n\n

(v) We know that \u2220ROS + \u2220ROQ + \u2220QOP + \u2220POS = 360o<\/sup><\/p>\n\n\n\n

3xo<\/sup> + 2xo<\/sup> + xo<\/sup> + 2xo<\/sup> = 360o<\/sup><\/p>\n\n\n\n

8xo<\/sup> = 360o<\/sup><\/p>\n\n\n\n

xo<\/sup> = 360o<\/sup>\/8<\/p>\n\n\n\n

xo<\/sup> = 45o<\/sup><\/p>\n\n\n\n

(vi)  Linear pair: The two adjacent angles are said to form a linear pair of angles if their non\u2013common arms are two opposite rays and sum of the angle is 180o<\/sup><\/p>\n\n\n\n

Therefore 3xo<\/sup> = 105o<\/sup><\/p>\n\n\n\n

xo<\/sup> = 105o<\/sup>\/3<\/p>\n\n\n\n

xo<\/sup> = 35o<\/sup><\/p>\n\n\n\n

21. In Fig. 22, it being given that \u22201 = 65o<\/sup>, find all other angles.<\/strong><\/p>\n\n\n\n

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

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

Given from the figure 22, \u22201 = \u22203 are the vertically opposite angles<\/p>\n\n\n\n

Therefore, \u22203 = 65o<\/sup><\/p>\n\n\n\n

Here, \u22201 + \u22202 = 180\u00b0 are the linear pair [The two adjacent angles are said to form a linear pair of angles if their non\u2013common arms are two opposite rays and sum of the angle is 180o<\/sup>]<\/p>\n\n\n\n

Therefore, \u22202 = 180o<\/sup> \u2013 65o<\/sup><\/p>\n\n\n\n

= 115o<\/sup><\/p>\n\n\n\n

\u22202 = \u22204 are the vertically opposite angles [from the figure]<\/p>\n\n\n\n

Therefore, \u22202 = \u22204 = 115o<\/sup><\/p>\n\n\n\n

And \u22203 = 65o<\/sup><\/p>\n\n\n\n

22. In Fig. 23, OA and OB are opposite rays:<\/strong><\/p>\n\n\n\n

(i) If x = 25o<\/sup>, what is the value of y?<\/strong><\/p>\n\n\n\n

(ii) If y = 35o<\/sup>, what is the value of x?<\/strong><\/p>\n\n\n\n

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

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

(i) \u2220AOC + \u2220BOC = 180o<\/sup> [The two adjacent angles are said to form a linear pair of angles if their non\u2013common arms are two opposite rays and sum of the angle is 180o<\/sup>]<\/p>\n\n\n\n

2y + 50<\/sup> + 3x = 180o<\/sup><\/p>\n\n\n\n

3x + 2y = 175o<\/sup><\/p>\n\n\n\n

Given If x = 25o<\/sup>, then<\/p>\n\n\n\n

3(25o<\/sup>) + 2y = 175o<\/sup><\/p>\n\n\n\n

75o<\/sup> + 2y = 175o<\/sup><\/p>\n\n\n\n

2y = 175o<\/sup> \u2013 75o<\/sup><\/p>\n\n\n\n

2y = 100o<\/sup><\/p>\n\n\n\n

y = 100o<\/sup>\/2<\/p>\n\n\n\n

y = 50o<\/sup><\/p>\n\n\n\n

(ii) \u2220AOC + \u2220BOC = 180o<\/sup> [The two adjacent angles are said to form a linear pair of angles if their non\u2013common arms are two opposite rays and sum of the angle is 180o<\/sup>]<\/p>\n\n\n\n

2y + 5 + 3x = 180o<\/sup><\/p>\n\n\n\n

3x + 2y = 175o<\/sup><\/p>\n\n\n\n

Given If y = 35o<\/sup>, then<\/p>\n\n\n\n

3x + 2(35o<\/sup>) = 175o<\/sup><\/p>\n\n\n\n

3x + 70o<\/sup> = 175o<\/sup><\/p>\n\n\n\n

3x = 1750<\/sup> \u2013 70o<\/sup><\/p>\n\n\n\n

3x = 105o<\/sup><\/p>\n\n\n\n

x = 105o<\/sup>\/3<\/p>\n\n\n\n

x = 35o<\/sup><\/p>\n\n\n\n

23. In Fig. 24, write all pairs of adjacent angles and all the liner pairs.<\/strong><\/p>\n\n\n\n

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

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

Pairs of adjacent angles are:<\/p>\n\n\n\n

\u2220DOA and \u2220DOC<\/p>\n\n\n\n

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

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

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

Linear pairs: [The two adjacent angles are said to form a linear pair of angles if their non\u2013common arms are two opposite rays and sum of the angle is 180o<\/sup>]<\/p>\n\n\n\n

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

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

24. In Fig. 25, find \u2220x. Further find \u2220BOC, \u2220COD and \u2220AOD.<\/strong><\/p>\n\n\n\n

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

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

(x + 10)o<\/sup> + xo<\/sup> + (x + 20)o<\/sup> = 180o<\/sup>[linear pair]<\/p>\n\n\n\n

On rearranging we get<\/p>\n\n\n\n

3xo<\/sup> + 30o<\/sup> = 180o<\/sup><\/p>\n\n\n\n

3xo<\/sup> = 180o<\/sup> \u2013 30o<\/sup><\/p>\n\n\n\n

3xo<\/sup> = 150o<\/sup><\/p>\n\n\n\n

xo<\/sup> = 150o<\/sup>\/3<\/p>\n\n\n\n

xo<\/sup> = 50o<\/sup><\/p>\n\n\n\n

Also given that<\/p>\n\n\n\n

\u2220BOC = (x + 20)o<\/sup><\/p>\n\n\n\n

= (50 + 20)o<\/sup><\/p>\n\n\n\n

= 70o<\/sup><\/p>\n\n\n\n

\u2220COD = 50o<\/sup><\/p>\n\n\n\n

\u2220AOD = (x + 10)o<\/sup><\/p>\n\n\n\n

= (50 + 10)o<\/sup><\/p>\n\n\n\n

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

25. How many pairs of adjacent angles are formed when two lines intersect in a point?<\/strong><\/p>\n\n\n\n

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

If the two lines intersect at a point, then four adjacent pairs are formed and those are linear.<\/p>\n\n\n\n

26. How many pairs of adjacent angles, in all, can you name in Fig. 26?<\/strong><\/p>\n\n\n\n

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

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

There are 10 adjacent pairs formed in the given figure, they are<\/p>\n\n\n\n

\u2220EOD and \u2220DOC<\/p>\n\n\n\n

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

\u2220COB and \u2220BOA<\/p>\n\n\n\n

\u2220AOB and \u2220BOD<\/p>\n\n\n\n

\u2220BOC and \u2220COE<\/p>\n\n\n\n

\u2220COD and \u2220COA<\/p>\n\n\n\n

\u2220DOE and \u2220DOB<\/p>\n\n\n\n

\u2220EOD and \u2220DOA<\/p>\n\n\n\n

\u2220EOC and \u2220AOC<\/p>\n\n\n\n

\u2220AOB and \u2220BOE<\/p>\n\n\n\n

27. In Fig. 27, determine the value of x.<\/strong><\/p>\n\n\n\n

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

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

From the figure we can write as \u2220COB + \u2220AOB = 180o <\/sup>[linear pair]<\/p>\n\n\n\n

3xo<\/sup> + 3xo<\/sup> = 180o<\/sup><\/p>\n\n\n\n

6xo<\/sup> = 180o<\/sup><\/p>\n\n\n\n

xo<\/sup> = 180o<\/sup>\/6<\/p>\n\n\n\n

xo<\/sup> = 30o<\/sup><\/p>\n\n\n\n

28. In Fig.28, AOC is a line, find x.<\/strong><\/p>\n\n\n\n

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

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

From the figure we can write as<\/p>\n\n\n\n

\u2220AOB + \u2220BOC = 180o<\/sup> [linear pair]<\/p>\n\n\n\n

Linear pair<\/p>\n\n\n\n

2x + 70o<\/sup> = 180o<\/sup><\/p>\n\n\n\n

2x = 180o<\/sup> \u2013 70o<\/sup><\/p>\n\n\n\n

2x = 110o<\/sup><\/p>\n\n\n\n

x = 110o<\/sup>\/2<\/p>\n\n\n\n

x = 55o<\/sup><\/p>\n\n\n\n

29. In Fig. 29, POS is a line, find x.<\/strong><\/p>\n\n\n\n

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

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

From the figure we can write as angles of a straight line,<\/p>\n\n\n\n

\u2220QOP + \u2220QOR + \u2220ROS = 180o<\/sup><\/p>\n\n\n\n

60o<\/sup> + 4x + 40o<\/sup> = 180o<\/sup><\/p>\n\n\n\n

On rearranging we get, 100o<\/sup> + 4x = 180o<\/sup><\/p>\n\n\n\n

4x = 180o<\/sup> \u2013 100o<\/sup><\/p>\n\n\n\n

4x = 80o<\/sup><\/p>\n\n\n\n

x = 80o<\/sup>\/4<\/p>\n\n\n\n

x = 20o<\/sup><\/p>\n\n\n\n

30. In Fig. 30, lines l1 <\/sub>and l2<\/sub> intersect at O, forming angles as shown in the figure. If x = 45o<\/sup>, find the values of y, z and u.<\/strong><\/p>\n\n\n\n

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

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

Given that, \u2220x = 45o<\/sup><\/p>\n\n\n\n

From the figure we can write as<\/p>\n\n\n\n

\u2220x = \u2220z = 45o<\/sup><\/p>\n\n\n\n

Also from the figure, we have<\/p>\n\n\n\n

\u2220y = \u2220u<\/p>\n\n\n\n

From the property of linear pair we can write as<\/p>\n\n\n\n

\u2220x + \u2220y + \u2220z + \u2220u = 360o<\/sup><\/p>\n\n\n\n

45o<\/sup> + 45o<\/sup> + \u2220y + \u2220u = 360o<\/sup><\/p>\n\n\n\n

90o<\/sup> + \u2220y + \u2220u = 360o<\/sup><\/p>\n\n\n\n

\u2220y + \u2220u = 360o<\/sup> \u2013 90o<\/sup><\/p>\n\n\n\n

\u2220y + \u2220u = 270o<\/sup> (vertically opposite angles \u2220y = \u2220u)<\/p>\n\n\n\n

2\u2220y = 270o<\/sup><\/p>\n\n\n\n

\u2220y = 135o<\/sup><\/p>\n\n\n\n

Therefore, \u2220y = \u2220u = 135o<\/sup><\/p>\n\n\n\n

So, \u2220x = 45o<\/sup>, \u2220y = 135o<\/sup>, \u2220z = 45o<\/sup> and \u2220u = 135o<\/sup><\/p>\n\n\n\n

31. In Fig. 31, three coplanar lines intersect at a point O, forming angles as shown in the figure. Find the values of x, y, z and u<\/strong><\/p>\n\n\n\n

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

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

Given that, \u2220x + \u2220y + \u2220z+ \u2220u + 50o<\/sup> + 90o<\/sup> = 360o<\/sup><\/p>\n\n\n\n

Linear pair, \u2220x + 50o<\/sup> + 90o<\/sup> = 180o<\/sup><\/p>\n\n\n\n

\u2220x + 140o<\/sup> = 180o<\/sup><\/p>\n\n\n\n

On rearranging we get<\/p>\n\n\n\n

\u2220x = 180o<\/sup> \u2013 140o<\/sup><\/p>\n\n\n\n

\u2220x = 40o<\/sup><\/p>\n\n\n\n

From the figure we can write as<\/p>\n\n\n\n

\u2220x = \u2220u = 40o<\/sup> are vertically opposite angles<\/p>\n\n\n\n

\u2220z = 90o<\/sup> is a vertically opposite angle<\/p>\n\n\n\n

\u2220y = 50o<\/sup> is a vertically opposite angle<\/p>\n\n\n\n

Therefore, \u2220x = 40o<\/sup>, \u2220y = 50o<\/sup>, \u2220z = 90o<\/sup> and \u2220u = 40o<\/sup><\/p>\n\n\n\n

32. In Fig. 32, find the values of x, y and z.<\/strong><\/p>\n\n\n\n

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

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

\u2220y = 25o<\/sup> vertically opposite angle<\/p>\n\n\n\n

From the figure we can write as<\/p>\n\n\n\n

\u2220x = \u2220z are vertically opposite angles<\/p>\n\n\n\n

\u2220x + \u2220y + \u2220z + 25o<\/sup> = 360o<\/sup><\/p>\n\n\n\n

\u2220x + \u2220z + 25o<\/sup> + 25o<\/sup> = 360o<\/sup><\/p>\n\n\n\n

On rearranging we get,<\/p>\n\n\n\n

\u2220x + \u2220z + 50o<\/sup> = 360o<\/sup><\/p>\n\n\n\n

\u2220x + \u2220z = 360o<\/sup> \u2013 50o<\/sup> [\u2220x = \u2220z]<\/p>\n\n\n\n

2\u2220x = 310o<\/sup><\/p>\n\n\n\n

\u2220x = 155o<\/sup><\/p>\n\n\n\n

And, \u2220x = \u2220z = 155o<\/sup><\/p>\n\n\n\n

Therefore, \u2220x = 155o<\/sup>, \u2220y = 25o<\/sup> and \u2220z = 155o<\/sup><\/p>\n\n\n\n

\n\n\n\n

Exercise 14.2 Page No: 14.20<\/p>\n\n\n\n

1. In Fig. 58, line n is a transversal to line l and m. Identify the following:<\/strong><\/p>\n\n\n\n

(i) Alternate and corresponding angles in Fig. 58 (i)<\/strong><\/p>\n\n\n\n

(ii) Angles alternate to \u2220d and \u2220g and angles corresponding to \u2220f and \u2220h in Fig. 58 (ii)<\/strong><\/p>\n\n\n\n

(iii) Angle alternate to \u2220PQR, angle corresponding to \u2220RQF and angle alternate to \u2220PQE in Fig. 58 (iii)<\/strong><\/p>\n\n\n\n

(iv) Pairs of interior and exterior angles on the same side of the transversal in Fig. 58 (ii)<\/strong><\/p>\n\n\n\n

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

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

(i) A pair of angles in which one arm of both the angles is on the same side of the transversal and their other arms are directed in the same sense is called a pair of corresponding angles.<\/p>\n\n\n\n

In Figure (i) Corresponding angles are<\/p>\n\n\n\n

\u2220EGB and \u2220GHD<\/p>\n\n\n\n

\u2220HGB and \u2220FHD<\/p>\n\n\n\n

\u2220EGA and \u2220GHC<\/p>\n\n\n\n

\u2220AGH and \u2220CHF<\/p>\n\n\n\n

A pair of angles in which one arm of each of the angle is on opposite sides of the transversal and whose other arms include the one segment is called a pair of alternate angles.<\/p>\n\n\n\n

The alternate angles are:<\/p>\n\n\n\n

\u2220EGB and \u2220CHF<\/p>\n\n\n\n

\u2220HGB and \u2220CHG<\/p>\n\n\n\n

\u2220EGA and \u2220FHD<\/p>\n\n\n\n

\u2220AGH and \u2220GHD<\/p>\n\n\n\n

(ii) In Figure (ii)<\/p>\n\n\n\n

The alternate angle to \u2220d is \u2220e.<\/p>\n\n\n\n

The alternate angle to \u2220g is \u2220b.<\/p>\n\n\n\n

The corresponding angle to \u2220f is \u2220c.<\/p>\n\n\n\n

The corresponding angle to \u2220h is \u2220a.<\/p>\n\n\n\n

(iii) In Figure (iii)<\/p>\n\n\n\n

Angle alternate to \u2220PQR is \u2220QRA.<\/p>\n\n\n\n

Angle corresponding to \u2220RQF is \u2220ARB.<\/p>\n\n\n\n

Angle alternate to \u2220POE is \u2220ARB.<\/p>\n\n\n\n

(iv) In Figure (ii)<\/p>\n\n\n\n

Pair of interior angles are<\/p>\n\n\n\n

\u2220a is \u2220e.<\/p>\n\n\n\n

\u2220d is \u2220f.<\/p>\n\n\n\n

Pair of exterior angles are<\/p>\n\n\n\n

\u2220b is \u2220h.<\/p>\n\n\n\n

\u2220c is \u2220g.<\/p>\n\n\n\n

2. In Fig. 59, AB and CD are parallel lines intersected by a transversal PQ at L and M respectively, If \u2220CMQ = 60o<\/sup>, find all other angles in the figure.<\/strong><\/p>\n\n\n\n

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

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

A pair of angles in which one arm of both the angles is on the same side of the transversal and their other arms are directed in the same sense is called a pair of corresponding angles.<\/p>\n\n\n\n

Therefore corresponding angles are<\/p>\n\n\n\n

\u2220ALM = \u2220CMQ = 60o<\/sup> [given]<\/p>\n\n\n\n

Vertically opposite angles are<\/p>\n\n\n\n

\u2220LMD = \u2220CMQ = 60o <\/sup>[given]<\/p>\n\n\n\n

Vertically opposite angles are<\/p>\n\n\n\n

\u2220ALM = \u2220PLB = 60o<\/sup><\/p>\n\n\n\n

Here, \u2220CMQ + \u2220QMD = 180o<\/sup> are the linear pair<\/p>\n\n\n\n

On rearranging we get<\/p>\n\n\n\n

\u2220QMD = 180o<\/sup> \u2013 60o<\/sup><\/p>\n\n\n\n

= 120o<\/sup><\/p>\n\n\n\n

Corresponding angles are<\/p>\n\n\n\n

\u2220QMD = \u2220MLB = 120o<\/sup><\/p>\n\n\n\n

Vertically opposite angles<\/p>\n\n\n\n

\u2220QMD = \u2220CML = 120o<\/sup><\/p>\n\n\n\n

Vertically opposite angles<\/p>\n\n\n\n

\u2220MLB = \u2220ALP = 120o<\/sup><\/p>\n\n\n\n

3. In Fig. 60, AB and CD are parallel lines intersected by a transversal by a transversal PQ at L and M respectively. If \u2220LMD = 35o<\/sup> find \u2220ALM and \u2220PLA.<\/strong><\/p>\n\n\n\n

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

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

Given that, \u2220LMD = 35o<\/sup><\/p>\n\n\n\n

From the figure we can write<\/p>\n\n\n\n

\u2220LMD and \u2220LMC is a linear pair<\/p>\n\n\n\n

\u2220LMD + \u2220LMC = 180o <\/sup>[sum of angles in linear pair = 180o<\/sup>]<\/p>\n\n\n\n

On rearranging, we get<\/p>\n\n\n\n

\u2220LMC = 180o<\/sup> \u2013 35o<\/sup><\/p>\n\n\n\n

= 145o<\/sup><\/p>\n\n\n\n

So, \u2220LMC = \u2220PLA = 145o<\/sup><\/p>\n\n\n\n

And, \u2220LMC = \u2220MLB = 145o<\/sup><\/p>\n\n\n\n

\u2220MLB and \u2220ALM is a linear pair<\/p>\n\n\n\n

\u2220MLB + \u2220ALM = 180o<\/sup> [sum of angles in linear pair = 180o<\/sup>]<\/p>\n\n\n\n

\u2220ALM = 180o<\/sup> \u2013 145o<\/sup><\/p>\n\n\n\n

\u2220ALM = 350<\/sup><\/p>\n\n\n\n

Therefore, \u2220ALM = 35o<\/sup>, \u2220PLA = 145o<\/sup>.<\/p>\n\n\n\n

4. The line n is transversal to line l and m in Fig. 61. Identify the angle alternate to \u222013, angle corresponding to \u222015, and angle alternate to \u222015.<\/strong><\/p>\n\n\n\n

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

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

Given that, l \u2225 m<\/p>\n\n\n\n

From the figure the angle alternate to \u222013 is \u22207<\/p>\n\n\n\n

From the figure the angle corresponding to \u222015 is \u22207 [A pair of angles in which one arm of both the angles is on the same side of the transversal and their other arms are directed in the same sense is called a pair of corresponding angles.]<\/p>\n\n\n\n

Again from the figure angle alternate to \u222015 is \u22205<\/p>\n\n\n\n

5. In Fig. 62, line l \u2225 m and n is transversal. If \u22201 = 40\u00b0, find all the angles and check that all corresponding angles and alternate angles are equal.<\/strong><\/p>\n\n\n\n

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

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

Given that, \u22201 = 40o<\/sup><\/p>\n\n\n\n

\u22201 and \u22202 is a linear pair [from the figure]<\/p>\n\n\n\n

\u22201 + \u22202 = 180o<\/sup><\/p>\n\n\n\n

\u22202 = 180o<\/sup> \u2013 40o<\/sup><\/p>\n\n\n\n

\u22202 = 140o<\/sup><\/p>\n\n\n\n

Again from the figure we can say that<\/p>\n\n\n\n

\u22202 and \u22206 is a corresponding angle pair<\/p>\n\n\n\n

So, \u22206 = 140o<\/sup><\/p>\n\n\n\n

\u22206 and \u22205 is a linear pair [from the figure]<\/p>\n\n\n\n

\u22206 + \u22205 = 180o<\/sup><\/p>\n\n\n\n

\u22205 = 180o<\/sup> \u2013 140o<\/sup><\/p>\n\n\n\n

\u22205 = 40o<\/sup><\/p>\n\n\n\n

From the figure we can write as<\/p>\n\n\n\n

\u22203 and \u22205 are alternate interior angles<\/p>\n\n\n\n

So, \u22205 = \u22203 = 40o<\/sup><\/p>\n\n\n\n

\u22203 and \u22204 is a linear pair<\/p>\n\n\n\n

\u22203 + \u22204 = 180o<\/sup><\/p>\n\n\n\n

\u22204 = 180o<\/sup> \u2013 40o<\/sup><\/p>\n\n\n\n

\u22204 = 140o<\/sup><\/p>\n\n\n\n

Now, \u22204 and \u22206 are a pair of interior angles<\/p>\n\n\n\n

So, \u22204 = \u22206 = 140o<\/sup><\/p>\n\n\n\n

\u22203 and \u22207 are a pair of corresponding angles<\/p>\n\n\n\n

So, \u22203 = \u22207 = 40o<\/sup><\/p>\n\n\n\n

Therefore, \u22207 = 40o<\/sup><\/p>\n\n\n\n

\u22204 and \u22208 are a pair of corresponding angles<\/p>\n\n\n\n

So, \u22204 = \u22208 = 140o<\/sup><\/p>\n\n\n\n

Therefore, \u22208 = 140o<\/sup><\/p>\n\n\n\n

Therefore, \u22201 = 40o<\/sup>, \u22202 = 140o<\/sup>, \u22203 = 40o<\/sup>, \u22204 = 140o<\/sup>, \u22205 = 40o<\/sup>, \u22206 = 140o<\/sup>, \u22207 = 40o<\/sup> and \u22208 = 140o<\/sup><\/p>\n\n\n\n

6. In Fig.63, line l \u2225 m and a transversal n cuts them P and Q respectively. If \u22201 = 75\u00b0, find all other angles.<\/strong><\/p>\n\n\n\n

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

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

Given that, l \u2225 m and \u22201 = 75o<\/sup><\/p>\n\n\n\n

\u22201 = \u22203 are vertically opposite angles<\/p>\n\n\n\n

We know that, from the figure<\/p>\n\n\n\n

\u22201 + \u22202 = 180o<\/sup> is a linear pair<\/p>\n\n\n\n

\u22202 = 180o<\/sup> \u2013 75o<\/sup><\/p>\n\n\n\n

\u22202 = 105o<\/sup><\/p>\n\n\n\n

Here, \u22201 = \u22205 = 75o<\/sup> are corresponding angles<\/p>\n\n\n\n

\u22205 = \u22207 = 75o<\/sup> are vertically opposite angles.<\/p>\n\n\n\n

\u22202 = \u22206 = 105o<\/sup> are corresponding angles<\/p>\n\n\n\n

\u22206 = \u22208 = 105o<\/sup> are vertically opposite angles<\/p>\n\n\n\n

\u22202 = \u22204 = 105o<\/sup> are vertically opposite angles<\/p>\n\n\n\n

So, \u22201 = 75o<\/sup>, \u22202 = 105o<\/sup>, \u22203 = 75o<\/sup>, \u22204 = 105o<\/sup>, \u22205 = 75o<\/sup>, \u22206 = 105o<\/sup>, \u22207 = 75o<\/sup> and \u22208 = 105o<\/sup><\/p>\n\n\n\n

7. In Fig. 64, AB \u2225 CD and a transversal PQ cuts at L and M respectively. If \u2220QMD = 100o<\/sup>, find all the other angles.<\/strong><\/p>\n\n\n\n

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

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

Given that, AB \u2225 CD and \u2220QMD = 100o<\/sup><\/p>\n\n\n\n

We know that, from the figure \u2220QMD + \u2220QMC = 180o<\/sup> is a linear pair,<\/p>\n\n\n\n

\u2220QMC = 180o<\/sup> \u2013 \u2220QMD<\/p>\n\n\n\n

\u2220QMC = 180o<\/sup> \u2013 100\u00b0<\/p>\n\n\n\n

\u2220QMC = 80o<\/sup><\/p>\n\n\n\n

Corresponding angles are<\/p>\n\n\n\n

\u2220DMQ = \u2220BLM = 100o<\/sup><\/p>\n\n\n\n

\u2220CMQ = \u2220ALM = 80o<\/sup><\/p>\n\n\n\n

Vertically Opposite angles are<\/p>\n\n\n\n

\u2220DMQ = \u2220CML = 100o<\/sup><\/p>\n\n\n\n

\u2220BLM = \u2220PLA = 100o<\/sup><\/p>\n\n\n\n

\u2220CMQ = \u2220DML = 80o<\/sup><\/p>\n\n\n\n

\u2220ALM = \u2220PLB = 80o<\/sup><\/p>\n\n\n\n

8. In Fig. 65, l \u2225 m and p \u2225 q. Find the values of x, y, z, t.<\/strong><\/p>\n\n\n\n

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

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

Given that one of the angle is 80o<\/sup><\/p>\n\n\n\n

\u2220z and 80o<\/sup> are vertically opposite angles<\/p>\n\n\n\n

Therefore \u2220z = 80o<\/sup><\/p>\n\n\n\n

\u2220z and \u2220t are corresponding angles<\/p>\n\n\n\n

\u2220z = \u2220t<\/p>\n\n\n\n

Therefore, \u2220t = 80o<\/sup><\/p>\n\n\n\n

\u2220z and \u2220y are corresponding angles<\/p>\n\n\n\n

\u2220z = \u2220y<\/p>\n\n\n\n

Therefore, \u2220y = 80o<\/sup><\/p>\n\n\n\n

\u2220x and \u2220y are corresponding angles<\/p>\n\n\n\n

\u2220y = \u2220x<\/p>\n\n\n\n

Therefore, \u2220x = 80o<\/sup><\/p>\n\n\n\n

9. In Fig. 66, line l \u2225 m, \u22201 = 120o<\/sup> and \u22202 = 100o<\/sup>, find out \u22203 and \u22204.<\/strong><\/p>\n\n\n\n

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

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

Given that, \u22201 = 120o<\/sup> and \u22202 = 100o<\/sup><\/p>\n\n\n\n

From the figure \u22201 and \u22205 is a linear pair<\/p>\n\n\n\n

\u22201 + \u22205 = 180o<\/sup><\/p>\n\n\n\n

\u22205 = 180o<\/sup> \u2013 120o<\/sup><\/p>\n\n\n\n

\u22205 = 60o<\/sup><\/p>\n\n\n\n

Therefore, \u22205 = 60o<\/sup><\/p>\n\n\n\n

\u22202 and \u22206 are corresponding angles<\/p>\n\n\n\n

\u22202 = \u22206 = 100o<\/sup><\/p>\n\n\n\n

Therefore, \u22206 = 100o<\/sup><\/p>\n\n\n\n

\u22206 and \u22203 a linear pair<\/p>\n\n\n\n

\u22206 + \u22203 = 180o<\/sup><\/p>\n\n\n\n

\u22203 = 180o<\/sup> \u2013 100o<\/sup><\/p>\n\n\n\n

\u22203 = 80o<\/sup><\/p>\n\n\n\n

Therefore, \u22203 = 80o<\/sup><\/p>\n\n\n\n

By, angles of sum property<\/p>\n\n\n\n

\u22203 + \u22205 + \u22204 = 180o<\/sup><\/p>\n\n\n\n

\u22204 = 180o<\/sup> \u2013 80o<\/sup> \u2013 60o<\/sup><\/p>\n\n\n\n

\u22204 = 40o<\/sup><\/p>\n\n\n\n

Therefore, \u22204 = 40o<\/sup><\/p>\n\n\n\n

10.  In Fig. 67, l \u2225 m. Find the values of a, b, c, d. Give reasons.<\/strong><\/p>\n\n\n\n

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

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

Given l \u2225 m<\/p>\n\n\n\n

From the figure vertically opposite angles,<\/p>\n\n\n\n

\u2220a = 110o<\/sup><\/p>\n\n\n\n

Corresponding angles, \u2220a = \u2220b<\/p>\n\n\n\n

Therefore, \u2220b = 110o<\/sup><\/p>\n\n\n\n

Vertically opposite angle,<\/p>\n\n\n\n

\u2220d = 85o<\/sup><\/p>\n\n\n\n

Corresponding angles, \u2220d = \u2220c<\/p>\n\n\n\n

Therefore, \u2220c = 85o<\/sup><\/p>\n\n\n\n

Hence, \u2220a = 110o<\/sup>, \u2220b = 110o<\/sup>, \u2220c = 85o<\/sup>, \u2220d = 85o<\/sup><\/p>\n\n\n\n

11. In Fig. 68, AB \u2225 CD and \u22201 and \u22202 are in the ratio of 3: 2. Determine all angles from 1 to 8.<\/strong><\/p>\n\n\n\n

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

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

Given \u22201 and \u22202 are in the ratio 3: 2<\/p>\n\n\n\n

Let us take the angles as 3x, 2x<\/p>\n\n\n\n

\u22201 and \u22202 are linear pair [from the figure]<\/p>\n\n\n\n

3x + 2x = 180o<\/sup><\/p>\n\n\n\n

5x = 180o<\/sup><\/p>\n\n\n\n

x = 180o<\/sup>\/5<\/p>\n\n\n\n

x = 36o<\/sup><\/p>\n\n\n\n

Therefore, \u22201 = 3x = 3(36) = 108o<\/sup><\/p>\n\n\n\n

\u22202 = 2x = 2(36) = 72o<\/sup><\/p>\n\n\n\n

\u22201 and \u22205 are corresponding angles<\/p>\n\n\n\n

Therefore \u22201 = \u22205<\/p>\n\n\n\n

Hence, \u22205 = 108o<\/sup><\/p>\n\n\n\n

\u22202 and \u22206 are corresponding angles<\/p>\n\n\n\n

So \u22202 = \u22206<\/p>\n\n\n\n

Therefore, \u22206 = 72o<\/sup><\/p>\n\n\n\n

\u22204 and \u22206 are alternate pair of angles<\/p>\n\n\n\n

\u22204 = \u22206 = 72o<\/sup><\/p>\n\n\n\n

Therefore, \u22204 = 72o<\/sup><\/p>\n\n\n\n

\u22203 and \u22205 are alternate pair of angles<\/p>\n\n\n\n

\u22203 = \u22205 = 108o<\/sup><\/p>\n\n\n\n

Therefore, \u22203 = 108o<\/sup><\/p>\n\n\n\n

\u22202 and \u22208 are alternate exterior of angles<\/p>\n\n\n\n

\u22202 = \u22208 = 72o<\/sup><\/p>\n\n\n\n

Therefore, \u22208 = 72o<\/sup><\/p>\n\n\n\n

\u22201 and \u22207 are alternate exterior of angles<\/p>\n\n\n\n

\u22201 = \u22207 = 108o<\/sup><\/p>\n\n\n\n

Therefore, \u22207 = 108o<\/sup><\/p>\n\n\n\n

Hence, \u22201 = 108o<\/sup>, \u22202 = 72o<\/sup>, \u22203 = 108o<\/sup>, \u22204 = 72o<\/sup>, \u22205 = 108o<\/sup>, \u22206 = 72o<\/sup>, \u22207 = 108o<\/sup>, \u22208 = 72o<\/sup><\/p>\n\n\n\n

12. In Fig. 69, l, m and n are parallel lines intersected by transversal p at X, Y and Z respectively. Find \u22201, \u22202 and \u22203.<\/strong><\/p>\n\n\n\n

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

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

Given l, m and n are parallel lines intersected by transversal p at X, Y and Z<\/p>\n\n\n\n

Therefore linear pair,<\/p>\n\n\n\n

\u22204 + 60o<\/sup> = 180o<\/sup><\/p>\n\n\n\n

\u22204 = 180o<\/sup> \u2013 60o<\/sup><\/p>\n\n\n\n

\u22204 = 120o<\/sup><\/p>\n\n\n\n

From the figure,<\/p>\n\n\n\n

\u22204 and \u22201 are corresponding angles<\/p>\n\n\n\n

\u22204 = \u22201<\/p>\n\n\n\n

Therefore, \u22201 = 120o<\/sup><\/p>\n\n\n\n

\u22201 and \u22202 are corresponding angles<\/p>\n\n\n\n

\u22202 = \u22201<\/p>\n\n\n\n

Therefore, \u22202 = 120o<\/sup><\/p>\n\n\n\n

\u22202 and \u22203 are vertically opposite angles<\/p>\n\n\n\n

\u22202 = \u22203<\/p>\n\n\n\n

Therefore, \u22203 = 1200<\/sup><\/p>\n\n\n\n

13. In Fig. 70, if l \u2225 m \u2225 n and \u22201 = 60o<\/sup>, find \u22202<\/strong><\/p>\n\n\n\n

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

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

Given that l \u2225 m \u2225 n<\/p>\n\n\n\n

From the figure Corresponding angles are<\/p>\n\n\n\n

\u22201 = \u22203<\/p>\n\n\n\n

\u22201 = 60o<\/sup><\/p>\n\n\n\n

Therefore, \u22203 = 60o<\/sup><\/p>\n\n\n\n

\u22203 and \u22204 are linear pair<\/p>\n\n\n\n

\u22203 + \u22204 = 180o<\/sup><\/p>\n\n\n\n

\u22204 = 180o<\/sup> \u2013 60o<\/sup><\/p>\n\n\n\n

\u22204 = 120o<\/sup><\/p>\n\n\n\n

\u22202 and \u22204 are alternate interior angles<\/p>\n\n\n\n

\u22204 = \u22202<\/p>\n\n\n\n

Therefore, \u22202 = 120o<\/sup><\/p>\n\n\n\n

14. In Fig. 71, if AB \u2225 CD and CD \u2225 EF, find \u2220ACE<\/strong><\/p>\n\n\n\n

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

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

Given that, AB \u2225 CD and CD \u2225 EF<\/p>\n\n\n\n

Sum of the interior angles,<\/p>\n\n\n\n

\u2220CEF + \u2220ECD = 180o<\/sup><\/p>\n\n\n\n

130o<\/sup> + \u2220ECD = 180o<\/sup><\/p>\n\n\n\n

\u2220ECD = 180o<\/sup> \u2013 130o<\/sup><\/p>\n\n\n\n

\u2220ECD = 50o<\/sup><\/p>\n\n\n\n

We know that alternate angles are equal<\/p>\n\n\n\n

\u2220BAC = \u2220ACD<\/p>\n\n\n\n

\u2220BAC = \u2220ECD + \u2220ACE<\/p>\n\n\n\n

\u2220ACE = 70o<\/sup> \u2013 50o<\/sup><\/p>\n\n\n\n

\u2220ACE = 20o<\/sup><\/p>\n\n\n\n

Therefore, \u2220ACE = 20o<\/sup><\/p>\n\n\n\n

15. In Fig. 72, if l \u2225 m, n \u2225 p and \u22201 = 85o<\/sup>, find \u22202.<\/strong><\/p>\n\n\n\n

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

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

Given that, \u22201 = 85o<\/sup><\/p>\n\n\n\n

\u22201 and \u22203 are corresponding angles<\/p>\n\n\n\n

So, \u22201 = \u22203<\/p>\n\n\n\n

\u22203 = 85o<\/sup><\/p>\n\n\n\n

Sum of the interior angles is 180o<\/sup><\/p>\n\n\n\n

\u22203 + \u22202 = 180o<\/sup><\/p>\n\n\n\n

\u22202 = 180o<\/sup> \u2013 85o<\/sup><\/p>\n\n\n\n

\u22202 = 95o<\/sup><\/p>\n\n\n\n

16. In Fig. 73, a transversal n cuts two lines l and m. If \u22201 = 70o<\/sup> and \u22207 = 80o<\/sup>, is l \u2225 m?<\/strong><\/p>\n\n\n\n

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

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

Given \u22201 = 70o<\/sup> and \u22207 = 80o<\/sup><\/p>\n\n\n\n

We know that if the alternate exterior angles of the two lines are equal, then the lines are parallel.<\/p>\n\n\n\n

Here, \u22201 and \u22207 are alternate exterior angles, but they are not equal<\/p>\n\n\n\n

\u22201 \u2260 \u22207<\/p>\n\n\n\n

17. In Fig. 74, a transversal n cuts two lines l and m such that \u22202 = 65o<\/sup> and \u22208 = 65o<\/sup>. Are the lines parallel?<\/strong><\/p>\n\n\n\n

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

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

From the figure \u22202 = \u22204 are vertically opposite angles,<\/p>\n\n\n\n

\u22202 = \u22204 = 65o<\/sup><\/p>\n\n\n\n

\u22208 = \u22206 = 65o<\/sup><\/p>\n\n\n\n

Therefore, \u22204 = \u22206<\/p>\n\n\n\n

Hence, l \u2225 m<\/p>\n\n\n\n

18. In Fig. 75, Show that AB \u2225 EF.<\/strong><\/p>\n\n\n\n

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

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

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

\u2220ACD = \u2220ACE + \u2220ECD<\/p>\n\n\n\n

\u2220ACD = 22o<\/sup> + 35o<\/sup><\/p>\n\n\n\n

\u2220ACD = 57o<\/sup> = \u2220BAC<\/p>\n\n\n\n

Thus, lines BA and CD are intersected by the line AC such that, \u2220ACD = \u2220BAC<\/p>\n\n\n\n

So, the alternate angles are equal<\/p>\n\n\n\n

Therefore, AB \u2225 CD \u2026\u20261<\/p>\n\n\n\n

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

\u2220ECD + \u2220CEF = 35o<\/sup> + 145o<\/sup> = 180o<\/sup><\/p>\n\n\n\n

This, shows that sum of the angles of the interior angles on the same side of the transversal CE is 180o<\/sup><\/p>\n\n\n\n

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

Therefore, EF \u2225 CD \u2026\u2026.2<\/p>\n\n\n\n

From equation 1 and 2<\/p>\n\n\n\n

We conclude that, AB \u2225 EF<\/p>\n\n\n\n

19. In Fig. 76, AB \u2225 CD. Find the values of x, y, z.<\/strong><\/p>\n\n\n\n

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

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

Given that AB \u2225 CD<\/p>\n\n\n\n

Linear pair,<\/p>\n\n\n\n

\u2220x + 125o<\/sup> = 180o<\/sup><\/p>\n\n\n\n

\u2220x = 180o<\/sup> \u2013 125o<\/sup><\/p>\n\n\n\n

\u2220x = 55o<\/sup><\/p>\n\n\n\n

Corresponding angles<\/p>\n\n\n\n

\u2220z = 125o<\/sup><\/p>\n\n\n\n

Adjacent interior angles<\/p>\n\n\n\n

\u2220x + \u2220z = 180o<\/sup><\/p>\n\n\n\n

\u2220x + 125o<\/sup> = 180o<\/sup><\/p>\n\n\n\n

\u2220x = 180o<\/sup> \u2013 125o<\/sup><\/p>\n\n\n\n

\u2220x = 55o<\/sup><\/p>\n\n\n\n

Adjacent interior angles<\/p>\n\n\n\n

\u2220x + \u2220y = 180o<\/sup><\/p>\n\n\n\n

\u2220y + 55o<\/sup> = 180o<\/sup><\/p>\n\n\n\n

\u2220y = 180o<\/sup> \u2013 55o<\/sup><\/p>\n\n\n\n

\u2220y = 125o<\/sup><\/p>\n\n\n\n

20. In Fig. 77, find out \u2220PXR, if PQ \u2225 RS.<\/strong><\/p>\n\n\n\n

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

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

Given PQ \u2225 RS<\/p>\n\n\n\n

We need to find \u2220PXR<\/p>\n\n\n\n

\u2220XRS = 50o<\/sup><\/p>\n\n\n\n

\u2220XPQ = 70o<\/sup><\/p>\n\n\n\n

Given, that PQ \u2225 RS<\/p>\n\n\n\n

\u2220PXR = \u2220XRS + \u2220XPR<\/p>\n\n\n\n

\u2220PXR = 50o<\/sup> + 70o<\/sup><\/p>\n\n\n\n

\u2220PXR = 120o<\/sup><\/p>\n\n\n\n

Therefore, \u2220PXR = 120o<\/sup><\/p>\n\n\n\n

21. In Figure, we have<\/strong><\/p>\n\n\n\n

(i) \u2220MLY = 2\u2220LMQ<\/strong><\/p>\n\n\n\n

(ii) \u2220XLM = (2x \u2013 10)o<\/sup> and \u2220LMQ = (x + 30)o<\/sup>, find x.<\/strong><\/p>\n\n\n\n

(iii) \u2220XLM = \u2220PML, find \u2220ALY<\/strong><\/p>\n\n\n\n

(iv) \u2220ALY = (2x \u2013 15)o<\/sup>, \u2220LMQ = (x + 40)o <\/sup>, find x.<\/strong><\/p>\n\n\n\n

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

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

(i) \u2220MLY and \u2220LMQ are interior angles<\/p>\n\n\n\n

\u2220MLY + \u2220LMQ = 180o<\/sup><\/p>\n\n\n\n

2\u2220LMQ + \u2220LMQ = 180o<\/sup><\/p>\n\n\n\n

3\u2220LMQ = 180o<\/sup><\/p>\n\n\n\n

\u2220LMQ = 180o<\/sup>\/3<\/p>\n\n\n\n

\u2220LMQ = 60o<\/sup><\/p>\n\n\n\n

(ii) \u2220XLM = (2x \u2013 10)o<\/sup> and \u2220LMQ = (x + 30)o<\/sup>, find x.<\/p>\n\n\n\n

\u2220XLM = (2x \u2013 10)o<\/sup> and \u2220LMQ = (x + 30)o<\/sup><\/p>\n\n\n\n

\u2220XLM and \u2220LMQ are alternate interior angles<\/p>\n\n\n\n

\u2220XLM = \u2220LMQ<\/p>\n\n\n\n

(2x \u2013 10)o<\/sup> = (x + 30)o<\/sup><\/p>\n\n\n\n

2x \u2013 x = 30o<\/sup> + 10o<\/sup><\/p>\n\n\n\n

x = 40o<\/sup><\/p>\n\n\n\n

Therefore, x = 40\u00b0<\/p>\n\n\n\n

(iii) \u2220XLM = \u2220PML, find \u2220ALY<\/p>\n\n\n\n

\u2220XLM = \u2220PML<\/p>\n\n\n\n

Sum of interior angles is 180 degrees<\/p>\n\n\n\n

\u2220XLM + \u2220PML = 180o<\/sup><\/p>\n\n\n\n

\u2220XLM + \u2220XLM = 180o<\/sup><\/p>\n\n\n\n

2\u2220XLM = 180o<\/sup><\/p>\n\n\n\n

\u2220XLM = 180o<\/sup>\/2<\/p>\n\n\n\n

\u2220XLM = 90o<\/sup><\/p>\n\n\n\n

\u2220XLM and \u2220ALY are vertically opposite angles<\/p>\n\n\n\n

Therefore, \u2220ALY = 90o<\/sup><\/p>\n\n\n\n

(iv) \u2220ALY = (2x \u2013 15)o<\/sup>, \u2220LMQ = (x + 40)o<\/sup>, find x.<\/p>\n\n\n\n

\u2220ALY and \u2220LMQ are corresponding angles<\/p>\n\n\n\n

\u2220ALY = \u2220LMQ<\/p>\n\n\n\n

(2x \u2013 15)o <\/sup>= (x + 40)o<\/sup><\/p>\n\n\n\n

2x \u2013 x = 40o<\/sup> + 15o<\/sup><\/p>\n\n\n\n

x = 55o<\/sup><\/p>\n\n\n\n

Therefore, x = 55o<\/sup><\/p>\n\n\n\n

22. In Fig. 79, DE \u2225 BC. Find the values of x and y.<\/strong><\/p>\n\n\n\n

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

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

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

ABC, DAB are alternate interior angles<\/p>\n\n\n\n

\u2220ABC = \u2220DAB<\/p>\n\n\n\n

So, x = 40o<\/sup><\/p>\n\n\n\n

And ACB, EAC are alternate interior angles<\/p>\n\n\n\n

\u2220ACB = \u2220EAC<\/p>\n\n\n\n

So, y = 55o<\/sup><\/p>\n\n\n\n

23. In Fig. 80, line AC \u2225 line DE and \u2220ABD = 32o<\/sup>, Find out the angles x and y if \u2220E = 122o<\/sup>.<\/strong><\/p>\n\n\n\n

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

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

Given line AC \u2225 line DE and \u2220ABD = 32o<\/sup><\/p>\n\n\n\n

\u2220BDE = \u2220ABD = 32o<\/sup> \u2013 Alternate interior angles<\/p>\n\n\n\n

\u2220BDE + y = 180o<\/sup>\u2013 linear pair<\/p>\n\n\n\n

32o <\/sup>+ y = 180o<\/sup><\/p>\n\n\n\n

y = 180o<\/sup> \u2013 32o<\/sup><\/p>\n\n\n\n

y = 148o<\/sup><\/p>\n\n\n\n

\u2220ABE = \u2220E = 122o<\/sup> \u2013 Alternate interior angles<\/p>\n\n\n\n

\u2220ABD + \u2220DBE = 122o<\/sup><\/p>\n\n\n\n

32o<\/sup> + x = 122o<\/sup><\/p>\n\n\n\n

x = 122o<\/sup> \u2013 32o<\/sup><\/p>\n\n\n\n

x = 90o<\/sup><\/p>\n\n\n\n

24. In Fig. 81, side BC of \u0394ABC has been produced to D and CE \u2225 BA. If \u2220ABC = 65o<\/sup>, \u2220BAC = 55o<\/sup>, find \u2220ACE, \u2220ECD, \u2220ACD.<\/strong><\/p>\n\n\n\n

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

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

Given \u2220ABC = 65o<\/sup>, \u2220BAC = 55o<\/sup><\/p>\n\n\n\n

Corresponding angles,<\/p>\n\n\n\n

\u2220ABC = \u2220ECD = 65o<\/sup><\/p>\n\n\n\n

Alternate interior angles,<\/p>\n\n\n\n

\u2220BAC = \u2220ACE = 55o<\/sup><\/p>\n\n\n\n

Now, \u2220ACD = \u2220ACE + \u2220ECD<\/p>\n\n\n\n

\u2220ACD = 55o<\/sup> + 65o<\/sup><\/p>\n\n\n\n

= 120o<\/sup><\/p>\n\n\n\n

25. In Fig. 82, line CA \u22a5 AB \u2225 line CR and line PR \u2225 line BD. Find \u2220x, \u2220y, \u2220z.<\/strong><\/p>\n\n\n\n

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

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

Given that, CA \u22a5 AB<\/p>\n\n\n\n

\u2220CAB = 90o<\/sup><\/p>\n\n\n\n

\u2220AQP = 20o<\/sup><\/p>\n\n\n\n

By, angle of sum property<\/p>\n\n\n\n

In \u0394ABC<\/p>\n\n\n\n

\u2220CAB + \u2220AQP + \u2220APQ = 180o<\/sup><\/p>\n\n\n\n

\u2220APQ = 180o<\/sup> \u2013 90o<\/sup> \u2013 20o<\/sup><\/p>\n\n\n\n

\u2220APQ = 70o<\/sup><\/p>\n\n\n\n

y and \u2220APQ are corresponding angles<\/p>\n\n\n\n

y = \u2220APQ = 70o<\/sup><\/p>\n\n\n\n

\u2220APQ and \u2220z are interior angles<\/p>\n\n\n\n

\u2220APQ + \u2220z = 180o<\/sup><\/p>\n\n\n\n

\u2220z = 180o<\/sup> \u2013 70o<\/sup><\/p>\n\n\n\n

\u2220z = 110o<\/sup><\/p>\n\n\n\n

26. In Fig. 83, PQ \u2225 RS. Find the value of x.<\/strong><\/p>\n\n\n\n

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

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

Given, linear pair,<\/p>\n\n\n\n

\u2220RCD + \u2220RCB = 180o<\/sup><\/p>\n\n\n\n

\u2220RCB = 180o<\/sup> \u2013 130o<\/sup><\/p>\n\n\n\n

= 50o<\/sup><\/p>\n\n\n\n

In \u0394ABC,<\/p>\n\n\n\n

\u2220BAC + \u2220ABC + \u2220BCA = 180o<\/sup><\/p>\n\n\n\n

By, angle sum property<\/p>\n\n\n\n

\u2220BAC = 180o<\/sup> \u2013 55o<\/sup> \u2013 50o<\/sup><\/p>\n\n\n\n

\u2220BAC = 75o<\/sup><\/p>\n\n\n\n

27. In Fig. 84, AB \u2225 CD and AE \u2225 CF, \u2220FCG = 90o<\/sup> and \u2220BAC = 120o<\/sup>. Find the value of x, y and z.<\/strong><\/p>\n\n\n\n

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

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

Alternate interior angle<\/p>\n\n\n\n

\u2220BAC = \u2220ACG = 120o<\/sup><\/p>\n\n\n\n

\u2220ACF + \u2220FCG = 120o<\/sup><\/p>\n\n\n\n

So, \u2220ACF = 120o<\/sup> \u2013 90o<\/sup><\/p>\n\n\n\n

= 30o<\/sup><\/p>\n\n\n\n

Linear pair,<\/p>\n\n\n\n

\u2220DCA + \u2220ACG = 180o<\/sup><\/p>\n\n\n\n

\u2220x = 180o<\/sup> \u2013 120o<\/sup><\/p>\n\n\n\n

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

\u2220BAC + \u2220BAE + \u2220EAC = 360o<\/sup><\/p>\n\n\n\n

\u2220CAE = 360o<\/sup> \u2013 120o<\/sup> \u2013 (60o<\/sup> + 30o<\/sup>)<\/p>\n\n\n\n

= 150o<\/sup><\/p>\n\n\n\n

28. In Fig. 85, AB \u2225 CD and AC \u2225 BD. Find the values of x, y, z.<\/strong><\/p>\n\n\n\n

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

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

(i)  Since, AC \u2225 BD and CD \u2225 AB, ABCD is a parallelogram<\/p>\n\n\n\n

Adjacent angles of parallelogram,<\/p>\n\n\n\n

\u2220CAB + \u2220ACD = 180o<\/sup><\/p>\n\n\n\n

\u2220ACD = 180o<\/sup> \u2013 65o<\/sup><\/p>\n\n\n\n

= 115o<\/sup><\/p>\n\n\n\n

Opposite angles of parallelogram,<\/p>\n\n\n\n

\u2220CAB = \u2220CDB = 65o<\/sup><\/p>\n\n\n\n

\u2220ACD = \u2220DBA = 115o<\/sup><\/p>\n\n\n\n

(ii)  Here,<\/p>\n\n\n\n

AC \u2225 BD and CD \u2225 AB<\/p>\n\n\n\n

Alternate interior angles,<\/p>\n\n\n\n

\u2220CAD = x = 40o<\/sup><\/p>\n\n\n\n

\u2220DAB = y = 35o<\/sup><\/p>\n\n\n\n

29. In Fig. 86, state which lines are parallel and why?<\/strong><\/p>\n\n\n\n

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

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

Let, F be the point of intersection of the line CD and the line passing through point E.<\/p>\n\n\n\n

Here, \u2220ACD and \u2220CDE are alternate and equal angles.<\/p>\n\n\n\n

So, \u2220ACD = \u2220CDE = 100o<\/sup><\/p>\n\n\n\n

Therefore, AC \u2225 EF<\/p>\n\n\n\n

30. In Fig. 87, the corresponding arms of \u2220ABC and \u2220DEF are parallel. If \u2220ABC = 75o<\/sup>, find \u2220DEF.<\/strong><\/p>\n\n\n\n

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

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

Let, G be the point of intersection of the lines BC and DE<\/p>\n\n\n\n

Since, AB \u2225 DE and BC \u2225 EF<\/p>\n\n\n\n

The corresponding angles are,<\/p>\n\n\n\n

\u2220ABC = \u2220DGC = \u2220DEF = 75o<\/sup><\/p>\n\n\n\n

RD Sharma Solutions for Class 7 Maths Chapter 14: Download PDF<\/strong><\/h2>\n\n\n\n

RD Sharma Solutions for Class 7 Maths Chapter 14\u2013Lines And Angles<\/p>\n\n\n\n

Download PDF<\/strong>: RD Sharma Solutions for Class 7 Maths Chapter 14\u2013Lines And Angles PDF<\/a><\/p>\n\n\n\n

Chapterwise RD Sharma Solutions for Class 7 Maths :<\/strong><\/h2>\n\n\n\n