RS Aggarwal Class 9 Solutions Chapter 4 - Lines And Triangles

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:

1. Parallel lines
2. Perpendicular lines
3. Vertical lines
4. Horizontal lines
5. 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:

1. Isosceles Triangle
2. Equilateral Triangle
3. Scalene Triangle
4. Right Triangle
5. Obtuse Triangle
6. 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 $\angle AOB$ .

(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 $\angle A=36^{\circ}{27}'{46}”$ and $\angle B=28^{\circ}{43}'{39}”$, find $\angle A+\angle B$.

Ans:

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

We know that    1^o = 60 mins or 60’

1’ = 60 sec or 60”

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

Q3: Find the difference between two angles measuring $36^{\circ}\; and \; 24^{\circ}{28}'{30}”$.

Ans:

We know that 36^o = 35^o 59’ 60”

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

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

(i) 58^o

(ii) 16^o

(iii) $\frac{2}{3}$ of a right angle

(iv) 46^o 30’

(v) 52^o 43’ 20”

(vi) 68^o 35’ 45”

Ans:

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

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

(iii) $\frac{2}{3}$ of a right angle = $(90 \times \frac{2}{3})^{o} = 60^{o}$

Complement of $\frac{2}{3}$ of a right angle = (90 – 60)^o = 30^o

(iv) We know that 90^o = 89^o 60’

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

(v) We know that 90^o = 89^o 60’

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

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

(vi) We know that 90^o = 89^o 59’ 60”

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

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

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

(i) 63°

(ii) 138°

(iii) $\frac{3}{5}$ of a right angle

(iv) 75° 36′

(v) 124° 20′ 40″

Ans:

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

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

(iii) $\frac{3}{5}$ of a right angle = $(\frac{3}{5} \times 90)$ = 54°

Supplement of $\frac{3}{5}$ 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″)

Hence, the measure of the required angle = 55^o 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 $36^{\circ}$ 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 $25^{\circ}$ 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 =$\frac{1}{3}$(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 $10^{\circ}$ 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 180^o.

Therefore,

$\angle AOC + \angle BOC$ = 180^o

$\Rightarrow 62^{o} + x^{o} = 180^{o}$
$\Rightarrow x^{o} = (180^{o} – 62^{o})$
$\Rightarrow x = 118^{o}$

Hence, the value of x is 118^o

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

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

Therefore,

$\angle AOC + \angle COD + \angle BOD = 180^{o}$
$\Rightarrow (3x-6)^{o} +55^{o} +(x+20)^{o} = 180^{o}$
$\Rightarrow 4x=111^{o}$
$\Rightarrow x = 27.5^{o}$

Hence,

$\angle AOC = 3x-6$

=3 x 27.5-6

=76.5^o

And $\angle BOD = x + 20$

= 27.5+20

= 47.5^o

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

Ans: AOB Is a straight line. Therfore,

$\angle AOC + \angle COD + \angle BOD$
$\Rightarrow (3x+7)^{o} + (2x -19)^{o} +x^{o} =180^{o}$
$\Rightarrow 6x=192^{o}$
$\Rightarrow x = 32^{o}$

Therefore,

$\angle AOC$ = 3×32^o+7=103^o

$\angle COD$ =2×32^o – 19 = 45^o and

$\angle BOD$ = 32^o

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

$\angle XOP + \angle POQ + \angle YOQ = 180^{o}$

5a + 4a + 6a = 180^o

15a = 180^o

a = 12^o

Therefore,

x = 5 x 12^o = 60^o

y = 4 x 12^o = 48^o

and z = 6 x 12^o =72^o

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 = 180^o

7x = 196^o

X = 28^o

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

Question 21: Two lines AB and CD intersect at O. If $\angle AOC=50^{\circ}$, find $\angle AOC,\angle BOD\; and \; \angle BOC$.                                

Ans: We know that if two line intersects then the vertically opposite angles are equal. Therefore, $\angle AOC = \angle BOD = 50^{o}$

Let $\angle AOD – \angle BOC = x^{o}$

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

Therefore,

$\angle AOC + \angle AOD + \angle BOD + \angle BOC = 360^{o}$

50 + x + 50 + x = 360^o

2x = 260^o

X = 130^o

Hence, $\angle AOD = \angle BOC = 130^{o}$

Therefore,

$\angle AOD = 130^{o} , \angle BOD = 50^{o} and \angle BOC =130^{o}$

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, $\angle BOD = \angle AOC = 90^{o}$

Hence, t=90^o

Also,

$\angle DOF = \angle COE = 50^{o}$

Since, AOB is a straight line, we have

$\angle AOC + \angle COE + \angle BOE = 180^{o}$

90 + 50 + y = 180^o

140 + y =180^o

Y = 40^o

Also,

$\angle BOE = \angle AOF = 40^{o}$

Hence, x = 40^o

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

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 $\angle AOD,\angle COE\; and \; \angle AOE$.                 

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

therefore, $\angle DOF = \angle COE = 5x^{o}$

$\angle AOD = \angle BOC = 2x^{o}$ and

$\angle AOE = \angle BOF = 3x^{o}$

Since, AOB is a straight line, we have

$\angle AOE + \angle COE + \angle BOC = 180^{o}$

3x + 5x + 2x =180^o

10x = 180^o

X = 18^o

Therefore,

$\angle AOD = 2 \times 18^{o} = 36^{o}$
$\angle COE = 5 \times 18^{o} = 90^{o}$
$\angle AOE = 3 \times 18^{o} = 54^{o}$

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 = 180^o

9x = 180^o

x = 20^o

Hence, the two angles are 5 x20^o = 100^o and 4 x 20^o = 80^o

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

Ans:

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

$\angle AOC = 90^{o}.\ Then\ \angle AOC = \angle BOD = 90^{o}$

And let $\angle BOC = \angle AOD = x$

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

$\angle AOC + \angle BOD + \angle AOD + \angle BOC = 360o$

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

2x = 180^o

x =90^o

Hence, $\angle BOC = \angle AOD = 90^{o}$

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

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

Question 26: Two lines AB and CD intersect at a point O such that $angle BOC+angle AOD =280^{\circ}$, 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 $\angle BOC = \angle AOD$

Then,

x + x = 280

2x = 280

x = 140^o

therefore, $\angle BOC = \angle AOD = 140^{o}$

Also, let $\angle AOC = \angle BOD = y^{o}$

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

therefore, $\angle AOC + \angle BOC + \angle BOD + \angle AOD = 360^{o}$

y + 140 + y + 140 = 360^o

2y = 80^o

Y = 40^o

Hence, $\angle AOC = \angle BOD = 40^{o}$

Therefore, $\angle BOC = \angle AOD = 140^{o}\ and\ \angle AOC = \angle BOD = 40^{o}$

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

Ans:

Since DOC is a straight line, we have

$\angle BOD + \angle COB = 180^{o}$ …(1)

$\angle AOC + \angle AOD = 180^{o}$ …(2)

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

$\angle BOD + \angle COB = \angle AOC + \angle AOD$

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

Therefore, $\angle AOD = \angle BOD$