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 $\angle AOB$ .

(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 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 $36^{\circ}$ 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 $25^{\circ}$ 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, 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 line 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 intersect 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 line 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, $\angle AOD = \angle BOD$