A line which is also known as a straight line, was initially conceptualized by ancient mathematicians which does not have width nor depth. The different types of lines are:
- Parallel lines
- Perpendicular lines
- Vertical lines
- Horizontal lines
- Skew lines
A triangle is a type of polygon which contains three edges and vertices. Triangles are made up of three lines and there are unique internal angles that are formed by the lines. The different types of triangles are:
- Isosceles Triangle
- Equilateral Triangle
- Scalene Triangle
- Right Triangle
- Obtuse Triangle
- Acute Triangle
Learn about the RS Aggarwal Class 9 Solutions Chapter 4 Lines and Triangles below:
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=36∘27′46” and ∠B=28∘43′39”, 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 36∘and24∘28′30”.
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 the case of complementary angles:
x + x = 90°
= 2x = 90°
= x = 45°
Hence, the measure of the angle that is equal to its complement is 45°.
(ii) Let the measure of the required angle be x°.
Then, in the case of supplementary angles:
x—x=180°
= 2x = 180°
= x = 90°
Hence, the 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, the 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, the 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=(180o–62o)
⇒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
⇒(3x−6)o+55o+(x+20)o=180o
⇒4x=111o
⇒x=27.5o
Hence,
∠AOC=3x−6
=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+(2x−19)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 lines AB and CD intersect at O. If ∠AOC=50∘, find ∠AOC,∠BODand∠BOC.
Ans: We know that if two line intersects then the vertically opposite angles are equal. Therefore, ∠AOC=∠BOD=50o
Let ∠AOD–∠BOC=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=50oand∠BOC=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,∠COEand∠AOE.
Ans: We know that if two lines 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