{"id":545471,"date":"2021-10-04T10:52:18","date_gmt":"2021-10-04T10:52:18","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=545471"},"modified":"2021-10-05T09:23:25","modified_gmt":"2021-10-05T09:23:25","slug":"rd-sharma-solutions-for-class-9-maths-chapter-8-lines-and-angles","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-9-maths-chapter-8-lines-and-angles\/","title":{"rendered":"RD Sharma Solutions for Class 9 Maths Chapter 8\u2013Lines and Angles"},"content":{"rendered":"\n

Class 9: Maths Chapter 8 solutions. Complete Class 9 Maths Chapter 8 Notes.<\/p>\n\n\n\n

RD Sharma Solutions for Class 9 Maths Chapter 8\u2013Lines and Angles<\/h2>\n\n\n\n

RD Sharma 9th Maths Chapter 8, Class 9 Maths Chapter 8 solutions<\/p>\n\n\n\n

Exercise 8.1<\/u><\/strong><\/p>\n\n\n\n

Question 1: Write the complement of each of the following angles:<\/strong><\/p>\n\n\n\n

(i)200<\/sup><\/strong><\/p>\n\n\n\n

(ii)350<\/sup><\/strong><\/p>\n\n\n\n

(iii)900<\/sup><\/strong><\/p>\n\n\n\n

(iv) 770<\/sup><\/strong><\/p>\n\n\n\n

(v)300<\/sup><\/strong><\/p>\n\n\n\n

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

(i)<\/strong> The sum of an angle and its complement = 900<\/sup><\/strong><\/p>\n\n\n\n

Therefore, the complement of 200 <\/sup><\/strong>= 900<\/sup><\/strong> \u2013 200<\/sup><\/strong> = 700<\/sup><\/strong><\/p>\n\n\n\n

(ii)<\/strong> The sum of an angle and its complement = 900<\/sup><\/strong><\/p>\n\n\n\n

Therefore, the complement of 35\u00b0 = 90\u00b0 \u2013 35\u00b0 = 55<\/p>\n\n\n\n

(iii)<\/strong> The sum of an angle and its complement = 900<\/sup><\/strong><\/p>\n\n\n\n

Therefore, the complement of 900<\/sup><\/strong> = 900<\/sup><\/strong> \u2013 900<\/sup><\/strong> = 00<\/sup><\/strong><\/p>\n\n\n\n

(iv)<\/strong> The sum of an angle and its complement = 900<\/sup><\/strong><\/p>\n\n\n\n

Therefore, the complement of 770<\/sup><\/strong> = 90\u00b0 \u2013 770<\/sup><\/strong> = 130<\/sup><\/strong><\/p>\n\n\n\n

(v)<\/strong> The sum of an angle and its complement = 900<\/sup><\/strong><\/p>\n\n\n\n

Therefore, the complement of 300<\/sup><\/strong> = 900<\/sup><\/strong> \u2013 300<\/sup><\/strong> = 600<\/sup><\/strong><\/p>\n\n\n\n

Question 2 : Write the supplement of each of the following angles:<\/strong><\/p>\n\n\n\n

(i) 540<\/sup><\/strong><\/p>\n\n\n\n

(ii) 1320<\/sup><\/strong><\/p>\n\n\n\n

(iii) 1380<\/sup><\/strong><\/p>\n\n\n\n

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

(i)<\/strong> The sum of an angle and its supplement = 1800<\/sup><\/strong>.<\/p>\n\n\n\n

Therefore supplement of angle 540<\/sup><\/strong> = 1800<\/sup><\/strong> \u2013 540<\/sup><\/strong> = 1260<\/sup><\/strong><\/p>\n\n\n\n

(ii)<\/strong> The sum of an angle and its supplement = 1800<\/sup><\/strong>.<\/p>\n\n\n\n

Therefore supplement of angle 1320<\/sup><\/strong> = 1800<\/sup><\/strong> \u2013 1320<\/sup><\/strong> = 480<\/sup><\/strong><\/p>\n\n\n\n

(iii)<\/strong> The sum of an angle and its supplement = 1800<\/sup><\/strong>.<\/p>\n\n\n\n

Therefore supplement of angle 1380 <\/sup><\/strong>= 1800 <\/sup><\/strong>\u2013 1380<\/sup><\/strong> = 420<\/sup><\/strong><\/p>\n\n\n\n

Question 3: If an angle is 280<\/sup> less than its complement, find its measure?<\/strong><\/p>\n\n\n\n

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

Let the measure of any angle is \u2018 a \u2018 degrees<\/p>\n\n\n\n

Thus, its complement will be (90 \u2013 a) 0<\/sup><\/strong><\/p>\n\n\n\n

So, the required angle = Complement of a \u2013 28<\/p>\n\n\n\n

a = ( 90 \u2013 a ) \u2013 28<\/p>\n\n\n\n

2a = 62<\/p>\n\n\n\n

a = 31<\/p>\n\n\n\n

Hence, the angle measured is 310<\/sup><\/strong>.<\/p>\n\n\n\n

Question 4 : If an angle is 30\u00b0 more than one half of its complement, find the measure of the angle?<\/strong><\/p>\n\n\n\n

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

Let an angle measured by \u2018 a \u2018 in degrees<\/p>\n\n\n\n

Thus, its complement will be (90 \u2013 a) 0<\/sup><\/strong><\/p>\n\n\n\n

Required Angle = 300<\/sup><\/strong> + complement\/2<\/p>\n\n\n\n

a = 300<\/sup><\/strong> + ( 90 \u2013 a ) 0<\/sup><\/strong> \/ 2<\/p>\n\n\n\n

a + a\/2 = 300<\/sup><\/strong> + 450<\/sup><\/strong><\/p>\n\n\n\n

3a\/2 = 750<\/sup><\/strong><\/p>\n\n\n\n

a = 500<\/sup><\/strong><\/p>\n\n\n\n

Therefore, the measure of required angle is 500<\/sup><\/strong>.<\/p>\n\n\n\n

Question 5 : Two supplementary angles are in the ratio 4:5. Find the angles?<\/strong><\/p>\n\n\n\n

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

Two supplementary angles are in the ratio 4:5.<\/p>\n\n\n\n

Let us say, the angles are 4a and 5a (in degrees)<\/p>\n\n\n\n

Since angle are supplementary angles;<\/p>\n\n\n\n

Which implies, 4a + 5a = 1800<\/sup><\/strong><\/p>\n\n\n\n

9a = 1800<\/sup><\/strong><\/p>\n\n\n\n

a = 200<\/sup><\/strong><\/p>\n\n\n\n

Therefore, 4a = 4 (20) = 800 <\/sup><\/strong>and<\/p>\n\n\n\n

5(a) = 5 (20) = 1000<\/sup><\/strong><\/p>\n\n\n\n

Hence, required angles are 80\u00b0 and 1000<\/sup><\/strong>.<\/p>\n\n\n\n

Question 6 : Two supplementary angles differ by 480<\/sup>. Find the angles?<\/strong><\/p>\n\n\n\n

Solution<\/strong>: Given: Two supplementary angles differ by 480<\/sup><\/strong>.<\/p>\n\n\n\n

Consider a0<\/sup><\/strong> be one angle then its supplementary angle will be equal to (180 \u2013 a) 0<\/sup><\/strong><\/p>\n\n\n\n

According to the question;<\/p>\n\n\n\n

(180 \u2013 a ) \u2013 x = 48<\/p>\n\n\n\n

(180 \u2013 48 ) = 2a<\/p>\n\n\n\n

132 = 2a<\/p>\n\n\n\n

132\/2 = a<\/p>\n\n\n\n

Or a = 660<\/sup><\/strong><\/p>\n\n\n\n

Therefore, 180 \u2013 a = 1140<\/sup><\/strong><\/p>\n\n\n\n

Hence, the two angles are 660<\/sup><\/strong> and 1140<\/sup><\/strong>.<\/p>\n\n\n\n

Question 7: An angle is equal to 8 times its complement. Determine its measure?<\/strong><\/p>\n\n\n\n

Solution:<\/strong> Given: Required angle = 8 times of its complement<\/p>\n\n\n\n

Consider a0<\/sup><\/strong> be one angle then its complementary angle will be equal to (90 \u2013 a) 0<\/sup><\/strong><\/p>\n\n\n\n

According to the question;<\/p>\n\n\n\n

a = 8 times of its complement<\/p>\n\n\n\n

a = 8 ( 90 \u2013 a )<\/p>\n\n\n\n

a = 720 \u2013 8a<\/p>\n\n\n\n

a + 8a = 720<\/p>\n\n\n\n

9a = 720<\/p>\n\n\n\n

a = 80<\/p>\n\n\n\n

Therefore, the required angle is 800<\/sup><\/strong>.<\/p>\n\n\n\n

Exercise 8.2<\/u><\/strong><\/p>\n\n\n\n

Question 1: In the below Fig. OA and OB are opposite rays:<\/strong><\/p>\n\n\n\n

(i) If x = 250<\/sup>, what is the value of y?<\/strong><\/p>\n\n\n\n

(ii) If y = 350<\/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)<\/strong> Given: x = 25<\/p>\n\n\n\n

From figure: \u2220AOC and \u2220BOC form a linear pair<\/p>\n\n\n\n

Which implies, \u2220AOC + \u2220BOC = 1800<\/sup><\/p>\n\n\n\n

From the figure, \u2220AOC = 2y + 5 and \u2220BOC = 3x<\/p>\n\n\n\n

\u2220AOC + \u2220BOC = 1800<\/sup><\/p>\n\n\n\n

(2y + 5) + 3x = 180<\/p>\n\n\n\n

(2y + 5) + 3 (25) = 180<\/p>\n\n\n\n

2y + 5 + 75 = 180<\/p>\n\n\n\n

2y + 80 = 180<\/p>\n\n\n\n

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

y = 100\/2 = 50<\/p>\n\n\n\n

Therefore, y = 500<\/sup> <\/sub><\/p>\n\n\n\n

(ii)<\/strong> Given: y = 350<\/sup><\/p>\n\n\n\n

From figure: \u2220AOC + \u2220BOC = 180\u00b0 (Linear pair angles)<\/p>\n\n\n\n

(2y + 5) + 3x = 180<\/p>\n\n\n\n

(2(35) + 5) + 3x = 180<\/p>\n\n\n\n

75 + 3x = 180<\/p>\n\n\n\n

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

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

Therefore, x = 350<\/sup><\/p>\n\n\n\n

Question 2: In the below figure, write all pairs of adjacent angles and all the linear pairs.<\/strong><\/p>\n\n\n\n

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

Solution<\/strong>: From figure, pairs of adjacent angles are :<\/p>\n\n\n\n

(\u2220AOC, \u2220COB) ; (\u2220AOD, \u2220BOD) ; (\u2220AOD, \u2220COD) ; (\u2220BOC, \u2220COD)<\/p>\n\n\n\n

And Linear pair of angles are (\u2220AOD, \u2220BOD) and (\u2220AOC, \u2220BOC).[As \u2220AOD + \u2220BOD = 1800<\/sup> and \u2220AOC+ \u2220BOC = 1800<\/sup>.]<\/p>\n\n\n\n

Question 3 : In the given figure, find x. Further find \u2220BOC , \u2220COD and \u2220AOD.<\/strong><\/p>\n\n\n\n

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

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

From figure, \u2220AOD and \u2220BOD form a linear pair,<\/p>\n\n\n\n

Therefore, \u2220AOD+ \u2220BOD = 1800<\/sup><\/p>\n\n\n\n

Also, \u2220AOD + \u2220BOC + \u2220COD = 1800<\/sup><\/p>\n\n\n\n

Given: \u2220AOD = (x+10) 0<\/sup> , \u2220COD = x0<\/sup> and \u2220BOC = (x + 20) 0<\/sup><\/p>\n\n\n\n

( x + 10 ) + x + ( x + 20 ) = 180<\/p>\n\n\n\n

3x + 30 = 180<\/p>\n\n\n\n

3x = 180 \u2013 30<\/p>\n\n\n\n

x = 150\/3<\/p>\n\n\n\n

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

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

\u2220AOD=(x+10) =50 + 10 = 60<\/p>\n\n\n\n

\u2220COD = x = 50<\/p>\n\n\n\n

\u2220BOC = (x+20) = 50 + 20 = 70<\/p>\n\n\n\n

Hence, \u2220AOD=600<\/sup>, \u2220COD=500<\/sup> and \u2220BOC=700<\/sup><\/p>\n\n\n\n

Question 4: In figure, rays OA, OB, OC, OD and OE have the common end point 0. Show that \u2220AOB+\u2220BOC+\u2220COD+\u2220DOE+\u2220EOA=360\u00b0.<\/strong><\/p>\n\n\n\n

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

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

Given: Rays OA, OB, OC, OD and OE have the common endpoint O.<\/p>\n\n\n\n

Draw an opposite ray OX to ray OA, which make a straight line AX.<\/p>\n\n\n\n

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

From figure:<\/p>\n\n\n\n

\u2220AOB and \u2220BOX are linear pair angles, therefore,<\/p>\n\n\n\n

\u2220AOB +\u2220BOX = 1800<\/sup><\/p>\n\n\n\n

Or, \u2220AOB + \u2220BOC + \u2220COX = 1800<\/sup> \u2014\u2013\u2014\u2013(1)<\/p>\n\n\n\n

Also,<\/p>\n\n\n\n

\u2220AOE and \u2220EOX are linear pair angles, therefore,<\/p>\n\n\n\n

\u2220AOE+\u2220EOX =180\u00b0<\/p>\n\n\n\n

Or, \u2220AOE + \u2220DOE + \u2220DOX = 1800<\/sup> \u2014\u2013(2)<\/p>\n\n\n\n

By adding equations, (1) and (2), we get;<\/p>\n\n\n\n

\u2220AOB + \u2220BOC + \u2220COF + \u2220AOE + \u2220DOE + \u2220DOX = 1800<\/sup> + 1800<\/sup><\/p>\n\n\n\n

\u2220AOB + \u2220BOC + \u2220COD + \u2220DOE + \u2220EOA = 3600<\/sup><\/p>\n\n\n\n

Hence Proved.<\/p>\n\n\n\n

Question 5 : In figure, \u2220AOC and \u2220BOC form a linear pair. If a \u2013 2b = 30\u00b0, find a and b?<\/strong><\/p>\n\n\n\n

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

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

Given : \u2220AOC and \u2220BOC form a linear pair.<\/p>\n\n\n\n

=> a + b = 1800 <\/sup><\/strong>\u2026..(1)<\/p>\n\n\n\n

a \u2013 2b = 300<\/sup> \u2026(2) (given)<\/p>\n\n\n\n

On subtracting equation (2) from (1), we get<\/p>\n\n\n\n

a + b \u2013 a + 2b = 180 \u2013 30<\/p>\n\n\n\n

3b = 150<\/p>\n\n\n\n

b = 150\/3<\/p>\n\n\n\n

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

Since, a \u2013 2b = 300<\/sup><\/p>\n\n\n\n

a \u2013 2(50) = 30<\/p>\n\n\n\n

a = 30 + 100<\/p>\n\n\n\n

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

Therefore, the values of a and b are 130\u00b0 and 50\u00b0 respectively.<\/p>\n\n\n\n

Question 6: How many pairs of adjacent angles are formed when two lines intersect at a point?<\/strong><\/p>\n\n\n\n

Solution<\/strong>: Four pairs of adjacent angles are formed when two lines intersect each other at a single point.<\/p>\n\n\n\n

For example, Let two lines AB and CD intersect at point O.<\/p>\n\n\n\n

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

The 4 pair of adjacent angles are :<\/p>\n\n\n\n

(\u2220AOD,\u2220DOB),(\u2220DOB,\u2220BOC),(\u2220COA, \u2220AOD) and (\u2220BOC,\u2220COA).<\/p>\n\n\n\n

Question 7: How many pairs of adjacent angles, in all, can you name in figure given?<\/strong><\/p>\n\n\n\n

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

Solution<\/strong>: Number of Pairs of adjacent angles, from the figure, are :<\/p>\n\n\n\n

\u2220EOC and \u2220DOC<\/p>\n\n\n\n

\u2220EOD and \u2220DOB<\/p>\n\n\n\n

\u2220DOC and \u2220COB<\/p>\n\n\n\n

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

\u2220DOC and \u2220COA<\/p>\n\n\n\n

\u2220BOC and \u2220BOA<\/p>\n\n\n\n

\u2220BOA and \u2220BOD<\/p>\n\n\n\n

\u2220BOA and \u2220BOE<\/p>\n\n\n\n

\u2220EOC and \u2220COA<\/p>\n\n\n\n

\u2220EOC and \u2220COB<\/p>\n\n\n\n

Hence, there are 10 pairs of adjacent angles.<\/p>\n\n\n\n

Question 8: In figure, determine the value of x.<\/strong><\/p>\n\n\n\n

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

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

The sum of all the angles around a point O is equal to 360\u00b0.<\/p>\n\n\n\n

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

3x + 3x + 150 + x = 3600<\/sup><\/p>\n\n\n\n

7x = 3600<\/sup> \u2013 1500<\/sup><\/p>\n\n\n\n

7x = 2100<\/sup><\/p>\n\n\n\n

x = 210\/7<\/p>\n\n\n\n

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

Hence, the value of x is 30\u00b0.<\/p>\n\n\n\n

Question 9: In figure, 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, \u2220AOB and \u2220BOC are linear pairs,<\/p>\n\n\n\n

\u2220AOB +\u2220BOC =180\u00b0<\/p>\n\n\n\n

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

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

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

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

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

Therefore, the value of x is 550<\/sup>.<\/p>\n\n\n\n

Question 10: In figure, 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 figure, \u2220POQ and \u2220QOS are linear pairs.<\/p>\n\n\n\n

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

\u2220POQ + \u2220QOS=1800<\/sup><\/p>\n\n\n\n

\u2220POQ + \u2220QOR+\u2220SOR=1800<\/sup><\/p>\n\n\n\n

600<\/sup> + 4x +400<\/sup> = 1800<\/sup><\/p>\n\n\n\n

4x = 1800<\/sup> -1000<\/sup><\/p>\n\n\n\n

4x = 800<\/sup><\/p>\n\n\n\n

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

Hence, the value of x is 200<\/sup>.<\/p>\n\n\n\n

Exercise 8.3<\/u><\/strong><\/p>\n\n\n\n

Question 1: In figure, lines l1<\/sub><\/em>, and l<\/em>2<\/sub> intersect at O, forming angles as shown in the figure. If x = 45. 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: x = 450<\/sup><\/p>\n\n\n\n

Since vertically opposite angles are equal, therefore z = x = 450<\/sup><\/p>\n\n\n\n

z and u are angles that are a linear pair, therefore, z + u = 1800<\/sup><\/p>\n\n\n\n

Solve, z + u = 1800<\/sup> , for u<\/p>\n\n\n\n

u = 1800<\/sup> \u2013 z<\/p>\n\n\n\n

u = 1800<\/sup> \u2013 45<\/p>\n\n\n\n

u = 1350<\/sup><\/p>\n\n\n\n

Again, x and y angles are a linear pair.<\/p>\n\n\n\n

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

y = 1800<\/sup> \u2013 x<\/p>\n\n\n\n

y =1800<\/sup> \u2013 450<\/sup><\/p>\n\n\n\n

y = 1350<\/sup><\/p>\n\n\n\n

Hence, remaining angles are y = 1350<\/sup>, u = 1350<\/sup> and z = 450<\/sup>.<\/p>\n\n\n\n

Question 2 : In figure, 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

(\u2220BOD, z); (\u2220DOF, y ) are pair of vertically opposite angles.<\/p>\n\n\n\n

So, \u2220BOD = z = 900<\/sup><\/p>\n\n\n\n

\u2220DOF = y = 500<\/sup>[Vertically opposite angles are equal.]<\/p>\n\n\n\n

Now, x + y + z = 180 [Linear pair] [AB is a straight line]<\/p>\n\n\n\n

x + y + z = 180<\/p>\n\n\n\n

x + 50 + 90 = 180<\/p>\n\n\n\n

x = 180 \u2013 140<\/p>\n\n\n\n

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

Hence values of x, y, z and u are 400<\/sup>, 500<\/sup>, 900<\/sup> and 400<\/sup> respectively.<\/p>\n\n\n\n

Question 3 : In figure, 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

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

y = 250<\/sup> [Vertically opposite angles are equal]<\/p>\n\n\n\n

Now \u2220x + \u2220y = 1800<\/sup> [Linear pair of angles]<\/p>\n\n\n\n

x = 180 \u2013 25<\/p>\n\n\n\n

x = 155<\/p>\n\n\n\n

Also, z = x = 155 [Vertically opposite angles]<\/p>\n\n\n\n

Answer: y = 250<\/sup> and z = 1550<\/sup><\/p>\n\n\n\n

Question 4 : In figure, find the value of x.<\/strong><\/p>\n\n\n\n

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

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

\u2220AOE = \u2220BOF = 5x [Vertically opposite angles]<\/p>\n\n\n\n

\u2220COA+\u2220AOE+\u2220EOD = 1800<\/sup> [Linear pair]<\/p>\n\n\n\n

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

10x = 180<\/p>\n\n\n\n

x = 180\/10<\/p>\n\n\n\n

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

The value of x = 180<\/sup><\/p>\n\n\n\n

Question 5 : Prove that bisectors of a pair of vertically opposite angles are in the same straight line.<\/strong><\/p>\n\n\n\n

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

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

Lines AB and CD intersect at point O, such that<\/p>\n\n\n\n

\u2220AOC = \u2220BOD (vertically angles) \u2026(1)<\/p>\n\n\n\n

Also OP is the bisector of AOC and OQ is the bisector of BOD<\/p>\n\n\n\n

To Prove: POQ is a straight line.<\/p>\n\n\n\n

OP is the bisector of \u2220AOC:<\/p>\n\n\n\n

\u2220AOP = \u2220COP \u2026(2)<\/p>\n\n\n\n

OQ is the bisector of \u2220BOD:<\/p>\n\n\n\n

\u2220BOQ = \u2220QOD \u2026(3)<\/p>\n\n\n\n

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

Sum of the angles around a point is 360o<\/sup>.<\/p>\n\n\n\n

\u2220AOC + \u2220BOD + \u2220AOP + \u2220COP + \u2220BOQ + \u2220QOD = 3600<\/sup><\/p>\n\n\n\n

\u2220BOQ + \u2220QOD + \u2220DOA + \u2220AOP + \u2220POC + \u2220COB = 3600<\/sup><\/p>\n\n\n\n

2\u2220QOD + 2\u2220DOA + 2\u2220AOP = 3600<\/sup> (Using (1), (2) and (3))<\/p>\n\n\n\n

\u2220QOD + \u2220DOA + \u2220AOP = 1800<\/sup><\/p>\n\n\n\n

POQ = 1800<\/sup><\/p>\n\n\n\n

Which shows that, the bisectors of pair of vertically opposite angles are on the same straight line.<\/p>\n\n\n\n

Hence Proved.<\/p>\n\n\n\n

Question 6 : If two straight lines intersect each other, prove that the ray opposite to the bisector of one of the angles thus formed bisects the vertically opposite angle.<\/strong><\/p>\n\n\n\n

Solution:<\/strong> Given AB and CD are straight lines which intersect at O.<\/p>\n\n\n\n

OP is the bisector of \u2220 AOC.<\/p>\n\n\n\n

To Prove : OQ is the bisector of \u2220BOD<\/p>\n\n\n\n

Proof :<\/p>\n\n\n\n

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

AB, CD and PQ are straight lines which intersect in O.<\/p>\n\n\n\n

Vertically opposite angles: \u2220 AOP = \u2220 BOQ<\/p>\n\n\n\n

Vertically opposite angles: \u2220 COP = \u2220 DOQ<\/p>\n\n\n\n

OP is the bisector of \u2220 AOC : \u2220 AOP = \u2220 COP<\/p>\n\n\n\n

Therefore,  \u2220BOQ = \u2220 DOQ<\/p>\n\n\n\n

Hence, OQ is the bisector of \u2220BOD.<\/p>\n\n\n\n

Exercise 8.4<\/u><\/strong><\/p>\n\n\n\n

Question 1: In figure, AB, CD and \u22201 and \u22202 are in the ratio 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

Let \u22201 = 3x and \u22202 = 2x<\/p>\n\n\n\n

From figure: \u22201 and \u22202 are linear pair of angles<\/p>\n\n\n\n

Therefore, \u22201 + \u22202 = 180<\/p>\n\n\n\n

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

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

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

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

So, \u22201 = 3x = 1080<\/sup> and \u22202 = 2x = 720<\/sup><\/p>\n\n\n\n

As we know, vertically opposite angles are equal.<\/p>\n\n\n\n

Pairs of vertically opposite angles are:<\/p>\n\n\n\n

(\u22201 = \u22203); (\u22202 = \u22204) ; (\u22205, \u22207) and (\u22206 , \u22208)<\/p>\n\n\n\n

\u22201 = \u22203 = 108\u00b0<\/p>\n\n\n\n

\u22202 = \u22204 = 72\u00b0<\/p>\n\n\n\n

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

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

We also know, if a transversal intersects any parallel lines, then the corresponding angles are equal<\/p>\n\n\n\n

\u22201 = \u22205 = \u22207 = 108\u00b0<\/p>\n\n\n\n

\u22202 = \u22206 = \u22208 = 72\u00b0<\/p>\n\n\n\n

Answer: \u22201 = 108\u00b0, \u22202 = 72\u00b0, \u22203 = 108\u00b0, \u22204 = 72\u00b0, \u22205 = 108\u00b0, \u22206 = 72\u00b0, \u22207 = 108\u00b0 and \u22208 = 72\u00b0<\/p>\n\n\n\n

Question 2: In figure, I, 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> From figure:<\/p>\n\n\n\n

\u2220Y = 120\u00b0 [Vertical opposite angles]<\/p>\n\n\n\n

\u22203 + \u2220Y = 180\u00b0 [Linear pair angles theorem]<\/p>\n\n\n\n

=> \u22203= 180 \u2013 120<\/p>\n\n\n\n

=> \u22203= 60\u00b0<\/p>\n\n\n\n

Line l is parallel to line m,<\/p>\n\n\n\n

\u22201 = \u22203 [ Corresponding angles]<\/p>\n\n\n\n

\u22201 = 60\u00b0<\/p>\n\n\n\n

Also, line m is parallel to line n,<\/p>\n\n\n\n

\u22202 = \u2220Y [Alternate interior angles are equal]<\/p>\n\n\n\n

\u22202 = 120\u00b0<\/p>\n\n\n\n

Answer: \u22201 = 60\u00b0, \u22202 = 120\u00b0 and \u22203 = 60\u00b0.<\/p>\n\n\n\n

Question 3: In figure, AB || CD || EF and GH || KL. Find \u2220HKL.<\/strong><\/p>\n\n\n\n

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

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

Extend LK to meet line GF at point P.<\/p>\n\n\n\n

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

From figure, CD || GF, so, alternate angles are equal.<\/p>\n\n\n\n

\u2220CHG =\u2220HGP = 60\u00b0<\/p>\n\n\n\n

\u2220HGP =\u2220KPF = 60\u00b0 [Corresponding angles of parallel lines are equal]<\/p>\n\n\n\n

Hence, \u2220KPG =180 \u2013 60 = 120\u00b0<\/p>\n\n\n\n

=> \u2220GPK = \u2220AKL= 120\u00b0 [Corresponding angles of parallel lines are equal]<\/p>\n\n\n\n

\u2220AKH = \u2220KHD = 25\u00b0 [alternate angles of parallel lines]<\/p>\n\n\n\n

Therefore, \u2220HKL = \u2220AKH + \u2220AKL = 25 + 120 = 145\u00b0<\/p>\n\n\n\n

Question 4: In figure, show that AB || EF.<\/strong><\/p>\n\n\n\n

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

Solution: <\/strong>Produce EF to intersect AC at point N.<\/p>\n\n\n\n

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

From figure, \u2220BAC = 57\u00b0 and<\/p>\n\n\n\n

\u2220ACD = 22\u00b0+35\u00b0 = 57\u00b0<\/p>\n\n\n\n

Alternative angles of parallel lines are equal<\/p>\n\n\n\n

=> BA || EF \u2026..(1)<\/p>\n\n\n\n

Sum of Co-interior angles of parallel lines is 180\u00b0<\/p>\n\n\n\n

EF || CD<\/p>\n\n\n\n

\u2220DCE + \u2220CEF = 35 + 145 = 180\u00b0 \u2026(2)<\/p>\n\n\n\n

From (1) and (2)<\/p>\n\n\n\n

AB || EF[Since, Lines parallel to the same line are parallel to each other]<\/p>\n\n\n\n

Hence Proved.<\/p>\n\n\n\n

Question 5 : In figure, if AB || CD and CD || EF, find \u2220ACE.<\/strong><\/p>\n\n\n\n

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

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

Given: CD || EF<\/p>\n\n\n\n

\u2220 FEC + \u2220ECD = 180\u00b0[Sum of co-interior angles is supplementary to each other]<\/p>\n\n\n\n

=> \u2220ECD = 180\u00b0 \u2013 130\u00b0 = 50\u00b0<\/p>\n\n\n\n

Also, BA || CD<\/p>\n\n\n\n

=> \u2220BAC = \u2220ACD = 70\u00b0[Alternative angles of parallel lines are equal]<\/p>\n\n\n\n

But, \u2220ACE + \u2220ECD =70\u00b0<\/p>\n\n\n\n

=> \u2220ACE = 70\u00b0 \u2014 50\u00b0 = 20\u00b0<\/p>\n\n\n\n

Question 6: In figure, PQ || AB and PR || BC. If \u2220QPR = 102\u00b0, determine \u2220ABC. Give reasons.<\/strong><\/p>\n\n\n\n

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

Solution:<\/strong> Extend line AB to meet line PR at point G.<\/p>\n\n\n\n

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

Given: PQ || AB,<\/p>\n\n\n\n

\u2220QPR = \u2220BGR =102\u00b0[Corresponding angles of parallel lines are equal]<\/p>\n\n\n\n

And PR || BC,<\/p>\n\n\n\n

\u2220RGB+ \u2220CBG =180\u00b0[Corresponding angles are supplementary]<\/p>\n\n\n\n

\u2220CBG = 180\u00b0 \u2013 102\u00b0 = 78\u00b0<\/p>\n\n\n\n

Since, \u2220CBG = \u2220ABC<\/p>\n\n\n\n

=>\u2220ABC = 78\u00b0<\/p>\n\n\n\n

Question 7 : In figure, 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

We know, If a transversal intersects two lines such that a pair of alternate interior angles are equal, then the two lines are parallel<\/p>\n\n\n\n

From figure:<\/p>\n\n\n\n

=> \u2220EDC = \u2220DCA = 100\u00b0<\/p>\n\n\n\n

Lines DE and AC are intersected by a transversal DC such that the pair of alternate angles are equal.<\/p>\n\n\n\n

So, DE || AC<\/p>\n\n\n\n

Question 8: In figure, if l||m, n || p and \u22201 = 85\u00b0, find \u22202.<\/strong><\/p>\n\n\n\n

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

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

Given: \u22201 = 85\u00b0<\/p>\n\n\n\n

As we know, when a line cuts the parallel lines, the pair of alternate interior angles are equal.<\/p>\n\n\n\n

=> \u22201 = \u22203 = 85\u00b0<\/p>\n\n\n\n

Again, co-interior angles are supplementary, so<\/p>\n\n\n\n

\u22202 + \u22203 = 180\u00b0<\/p>\n\n\n\n

\u22202 + 55\u00b0 =180\u00b0<\/p>\n\n\n\n

\u22202 = 180\u00b0 \u2013 85\u00b0<\/p>\n\n\n\n

\u22202 = 95\u00b0<\/p>\n\n\n\n

Question 9 : If two straight lines are perpendicular to the same line, prove that they are parallel to each other.<\/strong><\/p>\n\n\n\n

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

Let lines l and m are perpendicular to n, then<\/p>\n\n\n\n

\u22201= \u22202=90\u00b0<\/p>\n\n\n\n

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

Since, lines l and m cut by a transversal line n and the corresponding angles are equal, which shows that, line l is parallel to line m.<\/p>\n\n\n\n

Question 10: Prove that if the two arms of an angle are perpendicular to the two arms of another angle, then the angles are either equal or supplementary.<\/strong><\/p>\n\n\n\n

Solution:<\/strong> Let the angles be \u2220ACB and \u2220ABD<\/p>\n\n\n\n

Let AC perpendicular to AB, and CD is perpendicular to BD.<\/p>\n\n\n\n

To Prove : \u2220ACD = \u2220ABD OR \u2220ACD + \u2220ABD =180\u00b0<\/p>\n\n\n\n

Proof :<\/p>\n\n\n\n

In a quadrilateral,<\/p>\n\n\n\n

\u2220A+ \u2220C+ \u2220D+ \u2220B = 360\u00b0[ Sum of angles of quadrilateral is 360\u00b0 ]<\/p>\n\n\n\n

=> 180\u00b0 + \u2220C + \u2220B = 360\u00b0<\/p>\n\n\n\n

=> \u2220C + \u2220B = 360\u00b0 \u2013180\u00b0<\/p>\n\n\n\n

Therefore, \u2220ACD + \u2220ABD = 180\u00b0<\/p>\n\n\n\n

And \u2220ABD = \u2220ACD = 90\u00b0<\/p>\n\n\n\n

Hence, angles are equal as well as supplementary.<\/p>\n\n\n\n

Exercise VSAQs<\/u><\/strong><\/p>\n\n\n\n

Question 1: Define complementary angles.<\/strong><\/p>\n\n\n\n

Solution:<\/strong> When the sum of two angles is 90 degrees, then the angles are known as complementary angles.<\/p>\n\n\n\n

Question 2: Define supplementary angles.<\/strong><\/p>\n\n\n\n

Solution:<\/strong> When the sum of two angles is 180\u00b0, then the angles are known as supplementary angles.<\/p>\n\n\n\n

Question 3: Define adjacent angles.<\/strong><\/p>\n\n\n\n

Solution:<\/strong> Two angles are Adjacent when they have a common side and a common vertex.<\/p>\n\n\n\n

Question 4: The complement of an acute angle is _____.<\/strong><\/p>\n\n\n\n

Solution<\/strong>: An acute angle<\/p>\n\n\n\n

Question 5: The supplement of an acute angle is _____.<\/strong><\/p>\n\n\n\n

Solution:<\/strong> An obtuse angle<\/p>\n\n\n\n

Question 6: The supplement of a right angle is _____.<\/strong><\/p>\n\n\n\n

Solution:<\/strong> A right angle<\/p>\n\n\n\n

RD Sharma Solutions for Class 9 Maths Chapter 8: Download PDF<\/strong><\/h2>\n\n\n\n

RD Sharma Solutions for Class 9 Maths Chapter 8\u2013Lines and Angles<\/p>\n\n\n\n

Download PDF<\/strong>: RD Sharma Solutions for Class 9 Maths Chapter 8\u2013Lines and Angles PDF<\/a><\/p>\n\n\n\n

Chapterwise RD Sharma Solutions for Class 9 Maths :<\/strong><\/h2>\n\n\n\n