Maths RS Aggarwal Solutions Class 9

RS Aggarwal Class 9 Solutions Chapter 4 Lines and Triangles Ex 4D

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 4 Lines and Triangles Ex 4D.

Other Exercises

RS Aggarwal Solutions Class 9 Chapter 4 Lines and Triangles Ex 4A
RS Aggarwal Solutions Class 9 Chapter 4 Lines and Triangles Ex 4B
RS Aggarwal Solutions Class 9 Chapter 4 Lines and Triangles Ex 4C
RS Aggarwal Solutions Class 9 Chapter 4 Lines and Triangles Ex 4D

Question 1.
Solution:
In ∆ABC,
∠B = 76° and ∠C = 48°
But ∠A + ∠B + ∠C = 180°
(Sum of angles of a triangle)
=> ∠A + 76° + 48° = 180°
=> ∠ A + 124° = 180°
=> ∠A= 180° – 124° = 56°

Question 2.
Solution:
Angles of a triangle are in the ratio = 2:3:4
Let first angle = 2x
then second angle = 3x
and third angle = 4x
2x + 3x + 4x = 180°
(Sum of angles of a triangle)
=> 9x = 180°
=> x = \(\frac { { 180 }^{ o } }{ 9 } \) = 20°
First angle = 2x = 2 x 20° = 40°
Second angle = 3x = 3 x 20° = 60°
and third angle = 4x = 4 x 20° = 80° Ans.

Question 3.
Solution:
In ∆ABC,
3∠A = 4∠B = 6∠C = x (Suppose)

Question 4.
Solution:
In ∆ABC,
∠ A + ∠B = 108° …(i)
∠B + ∠C – 130° …(ii)
But ∠A + ∠B + ∠C = 180° …(iii)
(sum of angles of a triangle)
Subtracting (i) from (iii),
∠C = 180° – 108° = 72°
Subtracting (ii) from (iii),
∠A = 180°- 130° = 50°
But ∠ A + ∠B = 108° (from i)
50° + ∠B = 108°
=> ∠B = 108° – 50° = 58°
Hence ∠A = 50°, ∠B = 58° and ∠C = 72° Ans.

Question 5.
Solution:
In ∆ABC,
∠A+∠B = 125° …(i)
∠A + ∠C = 113° …(ii)
But ∠A + ∠B + ∠C = 180° …(iii)
(sum of angles of a triangles) Subtracting, (i), from (iii),
∠C = 180°- 125° = 55°
Subtracting (ii) from (iii),
∠B = 180°- 113° – 67°
∠A + ∠B = 125°
∠ A + 67° = 125°
=> ∠ A = 125° – 67°
∠A = 58°
Hence ∠A = 58°, ∠B = 67° and ∠ C = 55° Ans.

Question 6.
Solution:
In ∆ PQR,
∠ P – ∠ Q = 42°
=> ∠P = 42°+∠Q …(i)
∠Q – ∠R = 21°
∠Q – 21°=∠R …(ii)
But ∠P + ∠Q + ∠R = 180°
(Sum of angles of a triangles)
42° + ∠Q + ∠Q + ∠Q – 21°= 180°
=> 21° + 3∠Q = 180°
=> 3∠Q = 180°- 21° = 159°
from ∠Q = \(\frac { { 159 }^{ o } }{ 3 } \) = 53°
(i)∠P = 42° + ∠Q = 42° + 53° = 95°
and from (ii) ∠R = ∠Q – 21°
= 53° – 25° = 32°
Hence ∠P = 95°, ∠Q = 53° and ∠R = 32° Ans.

Question 7.
Solution:
Let ∠ A, ∠ B and ∠ C are the three angles of A ABC.
and ∠A + ∠B = 116° …(i)

Question 8.
Solution:
Let ∠ A, ∠ B and ∠ C are the three angles of the ∆ ABC
Let ∠ A = ∠ B = x
then ∠C = x + 48°
But ∠A + ∠B + ∠C = 180°
(Sum of angles of a triangle)
x + x + x + 18° = 180°
=> 3x + 18° = 180°
=> 3x = 180° – 18° = 162°
x = \(\frac { { 162 }^{ o } }{ 3 } \) = 54°
∠A = 54°, ∠B = 54° and ∠C = 54° + 18° = 72°
Hence angles are 54°, 54 and 72° Ans.

Question 9.
Solution:
Let the smallest angle of a triangle = x°
their second angle = 2x°
and third angle = 3x°
But sum of angle of a triangle = 180°
x + 2x + 3x = 180°
=> 6x = 180°
=> x – \(\frac { { 180 }^{ o } }{ 6 } \) = 30°
Hence smallest angle = 30°
Second angle = 2 x 30° = 60°
and third angle = 3 x 30° = 90° Ans.

Question 10.
Solution:
In a right angled triangle.
one angle is = 90°
Sum of other two acute angles = 90°
But one acute angle = 53°
Second acute angle = 90° – 53° = 37°
Hence angle of the triangle with be 90°, 53°, 37° Ans.

Question 11.
Solution:

Given : In ∆ ABC,
∠ A = ∠B + ∠C
To Prove : ∆ABC is a right-angled
Proof : We know that in ∆ABC,
∠A + ∠B + ∠C = 180°
(angles of a triangle)
But ∠ A = ∠ B + C given
∠A + (∠B + ∠C) = 180°
=> ∠A + ∠A = 180°
=> 2∠A = 180°
=> ∠ A = \(\frac { { 180 }^{ o } }{ 2 } \) = 90°
∠ A = 90°
Hence ∆ ABC is a right-angled Hence proved.

Question 12.
Solution:
Given. In ∆ ABC, ∠A = 90°
AL ⊥ BC.

To Prove : ∠BAL = ∠ACB
Proof : In ∆ ABC, AL ⊥ BC
In right angled ∆ALC,
∠ ACB + ∠ CAL = 90° …(i)
( ∴∠L = 90°)
But ∠ A = 90° ‘
=> ∠ BAL + ∠ CAL = 90° …(ii)
From (i) and (ii),
∠BAL + ∠ CAL= ∠ ACB+ ∠CAL
=> ∠ BAL = ∠ ACB Hence proved.

Question 13.
Solution:
Given. In ∆ABC,
Each angle is less than the sum of the other two angles
∠A< ∠B + ∠C
∠B < ∠C + ∠A
and ∠C< ∠A + ∠C
Proof : ∠ A < ∠B + ∠C
Adding ∠ A both sides,
∠A + ∠A < ∠A + ∠B + ∠C => 2 ∠ A < 180°
(∴ ∠A+∠B+∠C=180°)
∠A < \(\frac { { 180 }^{ o } }{ 2 } \) => ∠A< 90
Similarly, we can prove that,
∠B < 90° and ∠C < 90°
∴ each angle is less than 90°
Hence, triangle is an acute angled triangle. Hence proved.

Question 14.
Solution:
Given. In ∆ABC,
∠B > ∠A + ∠C

Question 15.
Solution:
In ∆ABC
∠ ABC = 43° and Ext. ∠ ACD = 128°

Question 16.
Solution:
∠ ABC + ∠ ABD = 180°
(Linear pair)

Question 17.
Solution:
(i)In the figure, ∠BAE =110° and ∠ACD = 120°.

Question 18.
Solution:
In the figure,
∠A = 55°, ∠B = 45°, ∠C = 30° Join AD and produce it to E

Question 19.
Solution:
In the figure,
∠EAC = 108°,
AD divides ∠ BAC in the ratio 1 : 3
and AD = DB
∠EAC + ∠ BAC = 180°
(Linear pair)

Question 20.

Solution:
Sides BC, CA and AB
are produced in order forming exterior
angles ∠ ACD,

Question 21.
Solution:
Given : Two ∆ s DFB and ACF intersect each other as shown in the figure.
To Prove : ∠A + ∠B + ∠C + ∠D + ∠E + ∠F = 360°
Proof : In ∆ DFB,
∠D + ∠F + ∠B = 180°
(sum of angles of a triangle)
Similarly,in ∆ ACE
∠A + ∠C + ∠E = 180° …(ii)
Adding (i) and (ii), we get :
∠D + ∠F + ∠B+ ∠A+ ∠C + ∠E = 180° + 180°
=> ∠A+∠B+∠C+∠D+∠E + ∠ F = 360°
Hence proved.

Question 22.
Solution:
In the figure,
ABC is a triangle
and OB and OC are the angle
bisectors of ∠ B and ∠ C meeting each other at O.
∠ A = 70°
In ∆ ABC,
∠A + ∠B + ∠C = 180°
(sum of angles of a triangle)

Question 23.
Solution:
In ∆ABC, ∠ A = 40°
Sides AB and AC are produced forming exterior angles ∠ CBD and ∠ BCE

Question 24.
Solution:
In the figure, ∆ABC is triangle and ∠A : ∠B : ∠C = 3 : 2 : 1
AC ⊥ CD.
∠ A + ∠B + ∠C = 180°
(sum of angles of a triangle)
But ∠A : ∠B : ∠C = 3 : 2 : 1

Question 25.
Solution:
In ∆ ABC
AN is the bisector of ∠ A
∠NAB =\(\frac { 1 }{ 2 } \) ∠A.
Now in right angled ∆ AMB,
∠B + ∠MAB = 90° (∠M = 90°)

Question 26.
Solution:
(i) False: As a triangle has only one right angle
(ii) True: If two angles will be obtuse, then the third angle will not exist.
(iii) False: As an acute-angled triangle all the three angles are acute.
(iv) False: As if each angle will be less than 60°, then their sum will be less than 60° x 3 = 180°, which is not true.
(v) True: As the sum of three angles will be 60° x 3 = 180°, which is true.
(vi) True: A triangle can be possible if the sum of its angles is 180°
But the given triangle having angles 10° + 80° + 100° = 190° is not possible.

Hope given RS Aggarwal Solutions Class 9 Chapter 4 Lines and Triangles Ex 4D are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 9 Solutions Chapter 4 Lines and Triangles Ex 4C

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 4 Lines and Triangles Ex 4C.

Other Exercises

RS Aggarwal Solutions Class 9 Chapter 4 Lines and Triangles Ex 4A
RS Aggarwal Solutions Class 9 Chapter 4 Lines and Triangles Ex 4B
RS Aggarwal Solutions Class 9 Chapter 4 Lines and Triangles Ex 4C
RS Aggarwal Solutions Class 9 Chapter 4 Lines and Triangles Ex 4D

Question 1.
Solution:
AB || CD and a line t intersects them at E and F forming angles ∠1, ∠ 2, ∠3, ∠4, ∠5, ∠6, ∠7 and ∠8.

Question 2.
Solution:
AB || CD and a transversal t intersects them at E and F respectively forming angles ∠l, ∠2, ∠3, ∠4, ∠5, ∠ 6, ∠ 7 and ∠ 8

Question 3.
Solution:

Given. In quadrilateral ABCD, AB || DC and AD || BC
To Prove : ∠ADC = ∠ABC
Proof : AB || DC and AD intersects their
∠DAB + ∠ADC = 180°
(sum of co-interior angles)
Similarly ∴ AD || BC
∠DAB + ∠ABC = 180° …(ii)
from (i) and (ii),
∠ DAB + ∠ ADC = ∠DAB + ∠ABC
∴∠ADC = ∠ABC. Hence proved.

Question 4.
Solution:
(i) In the figure, AB || CD
∠ABE = 35° and ∠EDC = 65°
Draw FEG || AB or CD
∴ AS || FG (const.)

Question 5.
Solution:
In the figure, AB || CD || EF,
∠ ABC = 70° and ∠ CEF = 130°
∴EF || CD (given)

∴∠ CEF +∠1 – 180°
(sum of Co-interior angles)
=> 130° + ∠ 1 = 180°
=> ∠ 1 = 180° – 130° = 50°
Again, AB || CD (given)
∴ ∠ ABC = ∠ BCD (Alternate angles)
=> 70° = ∠ BCD = x + ∠ 1
=> x + 50° – 70°
=> x = 70° – 50° = 20°
Hence x = 20° Ans.

Question 6.
Solution:
In the figure, AB || CD.
∠DCE = 130°
and ∠ CEA – 20°
From E, draw EF || AB or CD.

Question 7.
Solution:
Given. In the given figure, AB || CD.

Question 8.
Solution:
In the figure, AB || CD

Question 9.
Solution:
In the figure, AB || CD, ∠AEF = p

Question 10.
Solution:
In the figure, AB || PQ.
A transversal LM cuts them at E and F
∠ LEB = 75°
∠BEG = 20°
∠ EFG = 25°
∠ EGF = x° and ∠ GFD = y°
∴∠ LEB + ∠ BEF = 180° (Linear pair)

Question 11.
Solution:
In the figure

Question 12.
Solution:
In the figure, AB || CD
∠PEF = 85°, ∠QHC = 115°
∴ ∠ GHF = ∠ QHC
(Vertically opposite angles)
∠ GHF = 115°
∴AB || CD
∴∠ PEF = ∠ EGH
(Corresponding angles)
∴∠EGH = 85°
But ∠ QGH + ∠ EGH = 180°, (Linear pair)
=> ∠QGH + 85° = 180°
=> QGH = 180° – 85° = 95°
In ∆ GHQ,
Ext. ∠GHF = ∠QGH + ∠GQH
=> 115° = 95° + x
=> x = 115° – 95°
Hence, x = 20° Ans.

Question 13.
Solution:
In the figure, AB || CD
∠BAD = 75°, ∠ BCD = 35°
∴AB || CD
∴∠ABC = ∠BCD (Alternate angles)
=> x = 35°
and ∠ BAD = ∠ ADC (Alternate angles)
=> 75° = z

Question 14.
Solution:
In the figure, AB || CD
∠APQ = 75°, ∠ PRD = 125°
∴AB || CD.
∠ APQ = ∠ PQR (Alternate angles)
∴75° = y°
=> y° = 75°

Question 15.
Solution:
In the figure, AB || CD and EF || GH
∠APR = 110°, ∠LRF = 60°
∴∠PRQ = ∠LRF
(vertically opposite angles)

Question 16.
Solution:
(i) l is parallel to m
if 3x – 20° = 2x +10°
(Alternate angles are equal)

Question 17.
Solution:
Given. Two lines AB and CD are perpendiculars on EF
To Prove : AB ⊥ CD.

Hope given RS Aggarwal Solutions Class 9 Chapter 4 Lines and Triangles Ex 4C are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 9 Solutions Chapter 4 Lines and Triangles Ex 4B

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 4 Lines and Triangles Ex 4B.

Other Exercises

RS Aggarwal Solutions Class 9 Chapter 4 Lines and Triangles Ex 4A
RS Aggarwal Solutions Class 9 Chapter 4 Lines and Triangles Ex 4B
RS Aggarwal Solutions Class 9 Chapter 4 Lines and Triangles Ex 4C
RS Aggarwal Solutions Class 9 Chapter 4 Lines and Triangles Ex 4D

Question 1.
Solution:
AOB is a straight line
∠AOC + ∠BOC = 180° (Linear pair)
=> 62° + x= 180°
=> x = 180° – 62°
=> x = 118°
Hence, x = 118° Ans.

Question 2.
Solution:
AOB is straight line
∠AOC + ∠COD + ∠DOB – 180°
=> (3x – 5)° + 55° + (x + 20)° = 180°
=> 3x – 5° + 55° + x + 20° = 180°
=> 4x – 5° + 75° = 180°
=> 4x + 70° = 180°
=> 4x = 180° – 70°
=> 4x = 110°
=> \(x=\frac { { 110 }^{ o } }{ 4 } =27.{ 5 }^{ o }\)
Hence x = 27.5°
and ∠AOC = 3x – 5° = 3 x 27.5° – 5°
= 82.5° – 5° = 77.5°
∠BOD = x + 20° = 27.5° + 20°
= 47.5° Ans.

Question 3.
Solution:
AOB is a straight line
∠AOC + ∠COD + ∠DOB = 180°
{angles on the same side of line AB}
=> (3x + 7)° + (2x – 19)° + x = 180°
=> 3x + 7° + 2x – 19° + x = 180°
=> 6x – 12° – 180°
=> 6x = 180° + 12° = 192°
=> \(x=\frac { { 192 }^{ o } }{ 6 } =32^{ o }\)
Here x = 32°
∠AOC = 3x + 7° = 3 x 32° + 7°
= 96° + 7°= 103°
∠COD = 2x – 19° = 2 x 32° – 19°
= 64 – 19° = 45°
and ∠BOD = x = 32° Ans.

Question 4.
Solution:
In the given figure,
x + y + z = 180°
But x : y : z = 5:4:6
Let ∠XOP = x° – 5a
∠POQ =y° = 4a
and ∠QOY = z = 6a
then 5a + 4a + 6a = x + y + z = 180°
=> 15a = 180°
=> a = \(\frac { { 180 }^{ o } }{ 15 } ={ 12 }^{ o }\)
=> x = 5a = 5 x 12° = 60°
y = 4a = 4 x 12° = 48°
and z = 6x = 6 x 12° = 72° Ans.

Question 5.
Solution:
AOB will be a straight line
If ∠AOC + ∠COB = 180°
If (3x + 20)° + (4x – 36)° = 180°
If 3x + 20 + 4x – 36 = 180°
If 7x – 16 = 180°
If 7x = 180° + 16 = 196°
If \(x=\frac { { 196 }^{ o } }{ 7 } ={ 28 }^{ o }\)
Hence, if x = 28°, then AOB will be a straight line.

Question 6.
Solution:
AB and CD intersect each other at O
AOC = ∠BOD and ∠BOC = ∠AOD (vertically opposite angles)
But ∠AOC = 50°
∠BOD = ∠AOC = 50°
But ∠AOC + ∠BOC = 180° (Linear pair)
=> 50° + ∠BOC = 180°
=> ∠BOC = 180° – 50° = 130°
∠AOD = ∠BOC = 130°
Hence,∠AOD = 30°,∠BOD = 50° and ∠BOC = 130° Ans.

Question 7.
Solution:
In the figure,
AB, CD and EF are coplanar lines intersecting at O.
∠AOF = ∠BOE
∠DOF = ∠COE and ∠BOD = ∠AOC (Vertically opposite angles)
x = y,
z = 50°
t = 90°
But AOF + ∠DOF + ∠BOD = 180° (Angles on the same side of a st. line)
=> x + 50° + 90° = 180°
=> x° + 140° + 180°
=> x = 180° – 140° = 40°
Hence, x = 40°, y = x = 40°, z = 50° and t = 90° Ans.

Question 8.
Solution:
Three coplanar lines AB, CD and EF intersect at a point O
∠AOD = ∠BOC
∠DOF = ∠COE
and ∠AOE = ∠BOF
(Vertically opposite angles)
But ∠AOD = 2x
∠BOC = 2x
and ∠BOF = 3x
∠AOE = 3x
and ∠COE = 5x
∠DOF = 5x
But ∠AOD + ∠DOF + ∠BOF + ∠BOC + ∠COE + ∠AOE = 360° (Angles at a point)
=> 2x + 5x + 3x + 2x + 5x + 3x = 360°
=> 20x = 360° => x = \(\frac { { 360 }^{ o } }{ 20 } \) = 18°
Hence x = 18°
∠AOD = 2x = 2 x 18° = 36°
∠COE = 5x = 5 x 18° = 90°
and ∠AOE = 3x = 3 x 18° = 54° Ans.

Question 9.
Solution:
AOB is a line and CO stands on it forming ∠AOC and ∠BOC
But ∠AOC : ∠BOC = 5:4
Let ∠AOC = 5x and ∠BOC = 4x
But ∠AOC + ∠BOC = 180° (Linear pair)
=> 5x + 4x = 180° => 9x = 180°
=> x = \(\frac { { 180 }^{ o } }{ 9 } \) = 20°
∠AOC = 5x = 5 x 20° = 100°
and ∠BOC = 4x = 4 x 20° = 80° Ans.

Question 10.
Solution:
Two lines AB and CD intersect each other at O and
∠AOC = 90°
∠AOC = ∠BOD
(Vertically opposite angles)
∠BOD = 90°
But ∠AOC + ∠BOC = 180° (Linear pair)
=> 90° + ∠BOC – 180°
=> ∠BOC = 180° – 90° = 90°
But ∠AOD = ∠BOC
(Vertically opposite angles)
∠AOD = 90°
Hence each of the remaining angle is 90°.

Question 11.
Solution:
Two lines AB and CD intersect each other at O and

∠BOC + ∠AOD = 280°
∠AOD = ∠BOC
(vertically opposite angles)
∠BOC + ∠BOC = 280°
(∠AOD = ∠BOC)
=> 2 ∠BOC = 280°
=> ∠BOC = \(\frac { { 280 }^{ o } }{ 2 } \) = 140°
But ∠BOC + ∠AOC = 180° (Linear pair)
=> 140° + ∠AOC = 180°
=> ∠AOC = 180° – 140° = 40°
But ∠BOD = ∠AOC
(vertically opposite angles)
∠BOD = 40°
Hence ∠AOC = 40°, ∠BOC = 140°,
∠BOD = 40°
and ∠AOD = 140° Ans.

Question 12.
Solution:
OC is the bisector of ∠AOB. and OD is the ray opposite to OC.
Now ∠AOC = ∠BOC (OC is bisector of ∠AOB)
But ∠BOC + ∠BOD = 180° (Linear pair)
Similarly, ∠AOD + ∠AOC = 180°
=> ∠BOC + ∠BOD = ∠AOD + ∠AOC
But ∠AOC = ∠BOC (Given)
∠BOD = ∠AOD
=> ∠AOD = ∠BOD
Hence proved.

Question 13.
Solution:
AB is the mirror.
PQ is the incident ray, QR is its reflected ray.
=> ∠BQR = ∠PQA
But ∠BQR + ∠PQR + ∠PQA = 180° (Angles on one side of a straight line)
=> ∠PQA + ∠PQA + 112° = 180°
=> 2∠PQA + 112° = 180°
=> 2∠PQA = 180° – 112° = 68°
∠PQA = \(\frac { { 68 }^{ o } }{ 2 } \) = 34° Ans.

Question 14.
Solution:
Given. Two lines AB and CD intersect each other at O.
OE is the bisector of ∠BOD and EO is produced to F.
To Prove : OF bisects ∠AOC.
Proof : AB and CD intersect each other at O
∠AOC = ∠BOD
(Vertically opposite angles)
OE is the bisector of ∠BOD
∠1 = ∠2
But ∠1 = ∠3
and ∠2 = ∠4 (Vertically opposite angles)
and ∠1 = ∠2 (proved)
∠3 = ∠4
Hence, OF is the bisector of ∠AOC.
Hence proved.

Question 15.
Solution:
Given ∠AOC and ∠BOC are supplementary angles
OE is the bisector of ∠BOC and OF is the bisector of ∠AOC
To Prove : ∠EOF = 90°
Proof : ∠1 = ∠2
∠3 = ∠4
{OE and OF are the bisectors of ∠BOC and ∠AOC respectively}
But ∠AOC + ∠BOC = 180°
(Linear pair)
=> ∠1 + ∠2 + ∠3 + ∠4 = 180°
=> ∠1 + ∠1 + ∠3 + ∠3 = 180°
=> 2∠1 + 2∠3 = 180°
=> 2(∠1 + ∠3) = 180°
=> ∠1 + ∠3 = \(\frac { { 180 }^{ o } }{ 2 } \) 90°
=> ∠EOF = 90°
Hence proved.

Hope given RS Aggarwal Solutions Class 9 Chapter 4 Lines and Triangles Ex 4B are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 9 Solutions Chapter 4 Lines and Triangles Ex 4A

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 4 Lines and Triangles Ex 4A.

Other Exercises

RS Aggarwal Solutions Class 9 Chapter 4 Lines and Triangles Ex 4A
RS Aggarwal Solutions Class 9 Chapter 4 Lines and Triangles Ex 4B
RS Aggarwal Solutions Class 9 Chapter 4 Lines and Triangles Ex 4C
RS Aggarwal Solutions Class 9 Chapter 4 Lines and Triangles Ex 4D

Question 1.
Solution:
(i)Angle : When two rays OA and OB meet at a point o, then ∠AOB is called an angle.

(ii) Interior of angle : The interior an angle is a 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.
(iii) Obtuse angle : An angle greater than 90° but less than 180° is called an obtuse angle.
(iv) Reflex angle : An angle more than 180° but less than 360° is called a reflex angle.
(v) Complementary angles : Two angles are said to be complementary angles if their sum is 90°.
(vi) Supplementary angles : Two angles are said to be supplementary angles if their sum is 180°.

Question 2.
Solution:
∠A = 36°27’46”
∠B = 28° 43’39”
Adding, ∠A + ∠B = 64° 70′ 85″
We know that 60″ = 1′ and 60′ = 1°
∠A+ ∠B = 65° 11′ 25″ Ans.

Question 3.
Solution:
36° – 24° 28′ 30″
= 35° 59’60” – 25° 28’30”

= 10° 31′ 30″ Ans.

Question 4.
Solution:
We know that two angles are complementary of their sum is 90°. Each of these two angles is complement to the other, therefore.
(i) Complement of 58° = 90° – 58° = 32°
(ii) Complement of 16° = 90° – 16° = 74°
(iii) Complement of \(\frac { 2 }{ 3 } \) of a right angle i.e.
of \(\frac { 2 }{ 3 } \) x 90° or 60° = 90° – 60° = 30°
= \(\frac { 2 }{ 3 } \) of right angle,
(iv) Complement of 46° 30′
= 90° – 46° 30′
= 43° 30′
(v) Complement of 52° 43′ 20°= 90° – 52° 43′ 20″
= 37° 16′ 40″
(vi) Complement of 68° 35′ 45″
= 90° – 68° 35′ 45″
= 21° 24′ 15″ Ans.

Question 5.
Solution:
We know that two angles are said to be supplement to each other of their sum is 180° therefore
(i) Supplement of 68° = 180° – 68° =112°
(ii) Supplement of 138° = 180° – 138° = 42°
(iii) Supplement of \(\frac { 3 }{ 5 } \) of a right angle or \(\frac { 3 }{ 5 } \) x 90° or 54°
= 180° – 54° = 126°
(iv) Supplement of 75° 36′ = 180° – 75° 36′ = 104° 24′
(v) Supplement of 124° 20′ 40″
= 180° – 124° 20′ 40″
= 55° 39′ 20″
(vi) Supplement of 108° 48′ 32″ = 180° – 108″ 48′ 32″ = 71° 11′ 28″ Ans.

Question 6.
Solution:
(i) Let the measure of required angle = x ,
their its complement = 90° – x
According to the condition,
x = 90° – x => 2x = 90°
=>x = \(\frac { { 90 }^{ o } }{ 2 } \) = 45°
Required angle = 45°
(ii) Let the measure of required angle = x then its supplement = 180° – x
According to the condition,
x = 180° – x => 2x = 180° = 90°
=>x = \(\frac {{ 180 }^{ o }}{ 2 }\) = 90°
Hence required angle = 90° Ans.

Question 7.
Solution:
Let required angle = x
then its complement = 90° – x
According to the condition,
x – (90° – x) = 36°
=> x – 90° + x = 36°
=> 2x = 36° + 90° = 126°
= \(\frac { { 126 }^{ o } }{ 2 } \) = 63°
Required angle = 63° Ans.

Question 8.
Solution:
Let the required angle = x
then its supplement = 180° – x
According to the condition,
(180° – x) – x = 25°
=> 180° – x – x = 25°
=> – 2x = 25° – 180°
=> – 2x = – 155°
=> x = \(\frac { { – 155 }^{ o } }{ – 2 } \)
= 77.5°
Hence required angle = 77.5° Ans.

Question 9.
Solution:
Let required angle = x
Then its complement = 90° – x
According to the condition,
x = 4 (90° – x) => x = 360° – 4x
=> x + 4x = 360° => 5x = 360°
x = \(\frac { { 360 }^{ o } }{ 5 } \) = 72°
Required angle = 72° Ans.

Question 10.
Solution:
Let required angle = x
Then its supplement = 180° – x
According to the condition,
x = 5 (180° – x)
=> x = 900° – 5x
=> x + 5x = 900°
=> 6x = 900°
=> x = \(\frac { { 900 }^{ o } }{ 6 } \) = 150°
Hence, required angle = 150° Ans

Question 11.
Solution:
Let required angle = x
then its supplement = 180° – x
and complement = 90° – x
According to the condition,
180° – x = 4 (90°- x)
=> 180° – x = 360° – 4x
=> – x + 4x — 360° – 180°
=> 3x= 180°
=> x = \(\frac { { 180 }^{ o } }{ 3 } \) = 60°
Required angle = 60° Ans.

Question 12.
Solution:
Let required angle = x
Then, its complement = 90° – x
and its supplement = 180° – x
According to the condition,
90° – x = \(\frac { 1 }{ 3 } \) (180° – x)
=> 90° – x = 60° – \(\frac { 1 }{ 3 } \) x
=> 90° – 60° = x – \(\frac { 1 }{ 3 } \) x
=> \(\frac { 2 }{ 3 } \) x = 30° =>x = \(\frac { { 30 }^{ o }X3\quad \quad }{ 2 } \) => x = 45° Ans.

Question 13.
Solution:
Let one angle = x
Then, its supplement = 180° – x
According to the condition,
x : (180° – x) = 3:2
=> \(\frac { x }{ { 180 }^{ o }-x } =\frac { 3 }{ 2 } \)
=>2x = 3(180°- x)
=> 2x = 540° – 3x
=> 2x + 3x = 540°
=> 5x = 540° => x = \(\frac { { 540 }^{ o } }{ 5 } \) = 108°
Angle = 108° and its supplement = 180° – 108° = 72°
Hence, angles are 108° and 72° Ans.

Question 14.
Solution:
Let angle = x
Then, its complementary angle = 90° – x
According to the condition,
x : (90° – x) = 4 : 5
=> \(\frac { x }{ { 90 }^{ o }-x } =\frac { 4 }{ 5 } \)
=> 5x = 4 (90° – x)
=> 5x = 360° – 4x
=> 5x + 4x = 360°
=> 9x = 360°
=> x = \(\frac { { 360 }^{ o } }{ 9 } ={ 40 }^{ o } \)
and its complement = 90° – 40° = 50°
Hence, angles are 40° and 50° Ans.

Question 15.
Solution:
Let the required angle = x
.’. its complement = 90° – x
and supplement = 180° – x
According to the condition,
7(90° – x) = 3(180° – x) – 10°
=> 630° – 7x = 3 (180° – x) – 10°
=> 630° – 7x = 540° – 3x – 10°
=> – 7x + 3x = 540° – 10° – 630°
– 4x = – 100°
x = \(\frac { { -100 }^{ o } }{ -4 } ={ 25 }^{ o }\)
Hence required angle = 25° Ans.

Hope given RS Aggarwal Solutions Class 9 Chapter 4 Lines and Triangles Ex 4A are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 9 Solutions Chapter 3 Introduction to Euclid’s Geometry Ex 3A

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Class 9 Solutions Chapter 3 Introduction to Euclid’s Geometry Ex 3A.

Question 1.
Solution:
A theorem is a statement that requires a proof while an axiom is the basic fact which is taken for granted without proof.

Question 2.
Solution:
(i) Line segment: The straight line between two points A and B is a called a line segment \(\overline { AB } \)

(ii) Ray : A line segment \(\overline { AB } \) when extended indefinitely is one direction is called a ray \(\overrightarrow { AB } \) It has no definitely length.

(iii) Intersecting lines : Two lines having one common point are called intersecting lines and the common point is called the point of intersection.

(iv) Parallel Lines : If two lines lying in the same plane do not intersect each other when produced on either side, then these two lines are called parallel lines. The distance between two parallel hues always remains the same.

(v) Half line : If we take a point P on a line \(\overleftrightarrow { AB } \), its divides the line into two parts. Each part is called half line or two ray i.e. \(\overrightarrow { PA } \) and \(\overrightarrow { PB } \) .

(vi) Concurrent lines : Three or more lines intersecting at the same point are called concurrent lines.

(vii) Collinear points : Three or more points lying on the same line are called collinear points.

(viii) Plane : A plane is a surface such that every point of the line joining any two points on it, lies on it.

Question 3.
Solution:
(i) Six points are : A, B, C, D, E and F
(ii) Five line segments are : \(\overline { EG }\), \(\overline { FH }\), \(\overline { EF }\), \(\overline { GH }\) and \(\overline { MN }\)
(iii) Four rays are : \(\overrightarrow { EP } \) , \(\overrightarrow { GR } \),\(\overrightarrow { GB } \) and \(\overrightarrow { HD } \)
(iv) Four lines are : \(\overrightarrow { AB } \),\(\overrightarrow { CD } \),\(\overrightarrow { PQ } \) and \(\overrightarrow { RS } \)
(v) Four collinear points are M, E, G, B. Ans

Question 4.
Solution:
(i) \(\overleftrightarrow { EF } \) and \(\overleftrightarrow { GH } \) is a pair of intersecting line whose point of intersection is R
and second pair of intersecting lines is \(\overleftrightarrow { AB } \) and \(\overleftrightarrow { CD } \) and point of intersection is P.
(ii) Three concurrent lines are \(\overleftrightarrow { AB } \), \(\overleftrightarrow { EF } \) and \(\overleftrightarrow { GH } \) and the point of intersection is R.
(iii) Three rays are \(\overleftrightarrow { RB } \),\(\overleftrightarrow { RH } \) and \(\overleftrightarrow { RG } \)
(iv) Two line segments are \(\overleftrightarrow { RQ } \) and \(\overleftrightarrow { RP } \)

Question 5.
Solution:
(i) Through a given point, infinitely many lines can be drawn.
(ii) Only one line can be drawn to pass through two given points.
(iii) Two lines can intersect each other at the most one point
(iv) A, B and C are three collinear points. Then the line segments will be \(\overline { AB } \), \(\overline { BC } \) and \(\overline { AC } \).

Question 6.
Solution:
(iv), (vi), (vii), (viii) and (ix) are true and others are not true.

Hope given RS Aggarwal Class 9 Solutions Chapter 3 Introduction to Euclid’s Geometry Ex 3A are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.