RS Aggarwal Class 9 Solutions Lines And Triangles

RS Aggarwal Class 9 Solutions Chapter 4

Q1: Define the following terms

(i)angle

(ii) Interior of an angle.

(iii) Obtuse angle

(iv) Reflex angle

(v) Complementary angles

(vi) Supplementary angles

Answer :

(i) Two rays OA and OB, with a common end-point 0, form an angle AOB that is represented as AOB .

(ii) The interior of an angle is the set of all points in its plane, which lie on the same side of OA as B and also on the same side of OB as A.

– ► B

(iii) An angle greater than 90° but less than 180° is called an obtuse angle.

(iv) An angle greater than 180° but less than 360° is called a reflex angle.

Reflex angle

(v) Two angles are said to be complementary if the sum of their measures is 90°.

(vi) Two angles are said to be supplementary if the sum of their measures is 180°.

Q2: If A=362746 and B=284339, find A+B.

Ans:

angle A + angle B = [(36o27’46”) + (28o43’39”)]

We know that    1o = 60 mins or 60’

1’ = 60 sec or 60”

 

Deg Min Sec
36 27 46
28 43 39
65 11 25

Hence, the measure of the required angle = 65o 11’ 25”

Q3: Find the difference between two angles measuring 36and242830.

Ans:

We know that 36o = 35o 59’ 60”

Deg Min Sec
35 59 60
24 28 30
11 31 30

Hence, the measure of the required angle = 11o 31’ 60”

Q4: Find the complement of each of the following angles.

(i) 58o

(ii) 16o

(iii) 23 of a right angle

(iv) 46o 30’

(v) 52o 43’ 20”

(vi) 68o 35’ 45”

Ans:

(i) Complement of 58o = (90 -58)o = 32o

(ii) Complement if 16o = (90 – 16)o = 74o

(iii) 23 of a right angle = (90×23)o=60o

Complement of 23 of a right angle = (90 – 60)o = 30o

(iv) We know that 90o = 89o 60’

Complement of 46o 30’ = (89o 60’ – 46o 30’)o = 43o 30’

(v) We know that 90o = 89o 60’

Complement of 52o 43’ 20” = [(89o 59’ 60”) – (52o 43’ 20”)]

Deg Min Sec
89 59 60
52 43 20
37 16 40

Hence, the measure of angle = 37o 16’ 40”

(vi) We know that 90o = 89o 59’ 60”

Complement of 68o 35’ 45” – [(89o 59’ 60”) – (68o 35’ 45”)]

Deg Min Sec
89 59 60
68 35 45
21 24 15

Hence, the measure of the required angle = 21o 24’ 15”

Q5: Find the supplement of each of the following angles.

(i) 63°

(ii) 138°

(iii) 35 of a right angle

(iv) 75° 36′

(v) 124° 20′ 40″

Ans:

(i) Supplement of 63° = (180 – 63)o =117o

(ii) Supplement of 138° = (180 – 138)° =42o

(iii) 35 of a right angle = (35×90) = 54°

Supplement of 35 of a right angle = (180 – 54)o = 126°

(iv) We know that 180° = 179° 60′

Supplement of 75° 36′ = (179° 60′ – 75° 36′) = 104° 24′

(v) We know that 180° = 179° 59′ 60″

Supplement of 124° 20′ 40″ = (179′ 59′ 60″ —124° 20′ 40″)

Deg Min Sec
179 59 60
124 20 40
55 39 20

Hence, the measure of the required angle = 55o 39’ 20”

Q6: Find the measure of an angle which is

(i) Equal to its complement.

(ii) Equal to its supplement.

Ans:

(i) Let the measure of the required angle be x°.

Then, in case of complementary angles:

x + x = 90°

= 2x = 90°

= x = 45°

Hence, measure of the angle that is equal to its complement is 45°.

(ii) Let the measure of the required angle be x°.

Then, in case of supplementary angles:

x—x=180°

= 2x = 180°

= x = 90°

Hence, measure of the angle that is equal to its supplement is 90°.

Q7: Find the measure of an angle which is 36 more than its complement.

Ans:

Let the measure of the required angle be x°.

Then, measure of its complement = (90 – x)o .

Therefore,

X – (90° – x) = 36°

2x = 126°

X = 63°

Hence, the measure of the required angle is 63°.

Q8: Find the measure of an angle which is 25 less than its supplement.

Ans:

Let the measure of the required angle be x.

Then, measure of its supplement = (180° – x).

Therefore, (180° – x) – x = 25°

2x = 155°

x = 77.5°

Hence, the measure of the required angle is 77.5°.

Q9: Find the angle which is four times its complement.

Ans:

Let the measure of the required angle be x.

Then, measure of its complement = (90° – x).

Therefore,

x = (90° – x) 4

x = 360° – 4x

5x = 360°

x = 72°

Hence, the measure of the required angle is 72°.

Q10: Find the angle which is five times its supplement.

Ans:

Let the measure of the required angle be x.

Then, measure of its supplement = 180° – x.

Therefore, x = (180° – x) 5

X = 900° – 5x

6x = 900°

x = 150°

Hence, the measure of the required angle is 150°.

Q11: Find the angle whose supplement is four times its complement.

Sol:

Let the measure of the required angle be x°.

Then, measure of its complement = (90 – x)o

And, measure of its supplement= (180 – x)o

Therefore,

(180 – x) = 4(90 – x)

180 – x = 360 – 4x

3x = 180

x = 60

Hence, the measure of the required angle is 60°.

Q12: Find the angle whose complement is one-third of its supplement.

Answer :

Let the measure of the required angle be x°.

Then, the measure of its complement = (90 – x).

And the measure of its supplement = 180 – x°.

Therefore, 90 – x =13(180 – x)

3(90 – x) = (180 – x)

270 – 3x = 180 – x

2x = 90

x = 45

Hence, the measure of the required angle is 45°.

Q13: Two supplementary angles are in the ratio 3:2. Find the angles.

Answer :

Let the two angles be 3x and 2x, respectively.

Then,

3x + 2x = 180

5x = 180

X = 36°

Therefore, the two angles are

3x = 3 x 36° = 108° and 2x = 2 x 36° = 72°.

Q14: Two complementary angles are in the ratio 4:5. Find the angles.

Answer :

Let the two angles be 4x and 5x, respectively.

Then, 4x + 5x = 90

9x = 90

X = 10°

Hence, the two angles are

4x = 4 x 10° = 40° and 5x = 5 x 10° = 50°.

Q15: Find the measure of an angle, if 7 times its complement is 10 less than three times its supplement.

Answer :

Let the measure of the required angle be x°.

Then, the measure of its complement = (90 – x)°.

And the measure of its supplement = (180 – x)°.

Therefore, 7(90 – x) = 3(180-x) – 10

= 630 – 7x = 540 – 3x-10

= 4x = 100

= x = 25

Hence, the measure of the required angle is 25°.

Question 16: In the adjoining figure, AOB is a straight line. Find the value of x.

                              

Ans: We know that the sum of angles in a linear pair is 180o.

Therefore,

AOC+BOC = 180o

62o+xo=180o xo=(180o62o) x=118o

Hence, the value of x is 118o

Question 17: In the adjoining figure, AOB is a straight line. Find the value of x. Hence find AOC and BOD.

                          

Ans:  As AOB is a straight line, the sum of angles on the same side of AOB, at a point O on it, is 180o

Therefore,

AOC+COD+BOD=180o (3x6)o+55o+(x+20)o=180o 4x=111o x=27.5o

Hence,

AOC=3x6

=3 x 27.5-6

=76.5o

And BOD=x+20

= 27.5+20

= 47.5o

Question 18: In the adjoining figure, AOB is a straight line. Find the value of x. Hence find AOC, COD and BOD.

                        

Ans: AOB Is a straight line. Therfore,

AOC+COD+BOD (3x+7)o+(2x19)o+xo=180o 6x=192o x=32o

Therefore,

AOC = 3×32o+7=103o

COD =2×32o – 19 = 45o and

BOD = 32o

Question 19: In the adjoining figure, x:y:z = 5:4:6 . If XOY is a straight line, find the value of x,y,z .

                                    

Ans: Let x = 5a; y = 4a and z = 6a

XOY is a straight line. Therefore

XOP+POQ+YOQ=180o

5a + 4a + 6a = 180o

15a = 180o

a = 12o

Therefore,

x = 5 x 12o = 60o

y = 4 x 12o = 48o

and z = 6 x 12o =72o

Question 20: In the adjoining figure, what value of x will make AOB, a straight line?                                      

Ans:

AOB will be a straight line, if

3x + 20 + 4x – 36 = 180o

7x = 196o

X = 28o

Hence, x = 28 will make AOB a straight line.

Question 21: Two line AB and CD intersect at O. If AOC=50, find AOC,BODandBOC.                                

Ans: We know that if two line intersect then the vertically opposite angles are equal. Therefore, AOC=BOD=50o

Let AODBOC=xo

Also, we know that the sum of all angles around a point is 360o

Therefore,

AOC+AOD+BOD+BOC=360o

50 + x + 50 + x = 360o

2x = 260o

X = 130o

Hence, AOD=BOC=130o

Therefore,

AOD=130o,BOD=50oandBOC=130o

Question 22: In the adjoining figure, three coplanar lines AB,CD and EF intersect at a point O, forming angles as shown. Find the value of x,y,z and t.                                

Ans: We know that if two lines intersect, then the vertically opposite angles are equal.

therefore, BOD=AOC=90o

Hence, t=90o

Also,

DOF=COE=50o

Since, AOB is a straight line, we have

AOC+COE+BOE=180o

90 + 50 + y = 180o

140 + y =180o

Y = 40o

Also,

BOE=AOF=40o

Hence, x = 40o

Therefore, x = 40o, y = 40o, z = 50o and t = 90o

Question 23: In the adjoining figure, three coplanar lines AB, CD and EF intersect at a point O. Find the value of x. Hence find AOD,COEandAOE.                 

                             

Ans: We know that if two line intersect, then the vertically opposite angles are equal.

therefore, DOF=COE=5xo

AOD=BOC=2xo and

AOE=BOF=3xo

Since, AOB is a straight line, we have

AOE+COE+BOC=180o

3x + 5x + 2x =180o

10x = 180o

X = 18o

Therefore,

AOD=2×18o=36o COE=5×18o=90o AOE=3×18o=54o

Question 24: Two adjacent angles on a straight line are in the ration 5:4. Find the measure of each one of these angles.

Ans:  Let the two adjacent angles be 5x and 4x, respectively

Then,

5x + 4x = 180o

9x = 180o

x = 20o

Hence, the two angles are 5 x20o = 100o and 4 x 20o = 80o

Question 25: If two straight lines intersect each other in such a way that one of the angles formed measures 90, show that each of the remaining angles measures 90.

Ans:

We know that if two lines intersect, then the vertically opposite angles are equal.

AOC=90o. Then AOC=BOD=90o

And let BOC=AOD=x

Also, we know that the sum of all angles around a point is 360o.

AOC+BOD+AOD+BOC=360o

90o +90o + x + x = 360o

2x = 180o

x =90o

Hence, BOC=AOD=90o

Therefore, angle AOC = angle BOD = angle BOC = angle AOD = 90^{o}

Hence, the measure of each of the remaining angles are 90o

Question 26: Two lines AB and CD intersect at a point O such that angleBOC+angleAOD=280, as shown in the figure. Find all the four angles.                           

                               

Ans:  We know that if two lines intersect, then the vertically-opposite angles are equal.

Let BOC=AOD

Then,

x + x = 280

2x = 280

x = 140o

therefore, BOC=AOD=140o

Also, let AOC=BOD=yo

We know that the sum of all angles around a point is 360o.

therefore, AOC+BOC+BOD+AOD=360o

y + 140 + y + 140 = 360o

2y = 80o

Y = 40o

Hence, AOC=BOD=40o

Therefore, BOC=AOD=140o and AOC=BOD=40o

Question 27: In the given figure, ray OC is the bisector of AOB and OD is the ray opposite to OC. Show that AOD=BOD.

                          

Ans:

Since DOC is a straight line, we have

BOD+COB=180o …(1)

AOC+AOD=180o …(2)

From (1) and (2), we get:

BOD+COB=AOC+AOD

Also, COB=AOC [Since OC is a bisector of AOB ]

Therefore, AOD=BOD

Related Links
NCERT Books NCERT Solutions RS Aggarwal
Lakhmir Singh RD Sharma Solutions NCERT Solutions Class 6 to 12
More RS Aggarwal Solutions
RS Aggarwal Solutions Class 9 Solutions Real NumbeRSRS Aggarwal Solutions Class 10 Solutions Volume And Surface Areas Of Solids
RS Aggarwal Solutions Class 10 Solutions Chapter 10 Quadratic Equations Exercise 10 1RS Aggarwal Solutions Class 10 Solutions Chapter 10 Quadratic Equations Exercise 10 5
RS Aggarwal Solutions Class 10 Solutions Chapter 11 Arithmetic Progressions Exercise 11 1RS Aggarwal Solutions Class 10 Solutions Chapter 11 Arithmetic Progressions Exercise 11 5
RS Aggarwal Solutions Class 10 Solutions Chapter 13 Constructions Exercise 13 1RS Aggarwal Solutions Class 10 Solutions Chapter 15 Probability Exercise 15 3