RS Aggarwal Class 9 Solutions Chapter 4 - Lines And Triangles Ex 4B

RS Aggarwal Class 9 Chapter 4 - Lines And Triangles Ex 4B Solutions Free PDF

The RS Aggarwal Class 9 Solutions Chapter 4 Angles, Lines and Triangles Ex 4B have been formulated in accordance with the latest syllabus of the CBSE, which makes these solutions highly effective from exam point of view. It helps students to improve their performance level and giving the ability to solve difficult questions with ease. It is the best resource to prepare for the exam as it provides ample number of questions to practice from each and every topic.

Students are advised to solve these solutions so that they feel confident while attempting the final question paper. The RS Aggarwal Class 9 Solutions are simpler and can be practiced easily and are highly recommended for students as reference material. It will help you clear all your doubts quite well and attempt a great number of questions during the exam.

Download PDF of RS Aggarwal Class 9 Solutions Chapter 4– Angles, Lines and Triangles Ex 4B

Question 1: In \(\bigtriangleup ABC\), if \(\angle B=76 ^{0}\), find \(\angle A\).

Ans:

\(\angle A + \angle B + \angle C = 180 ^{0}\) [Sum of the angles of a triangle] \(\Rightarrow + 76^{0} + 48^{0} = 180^{0}\) \(\Rightarrow + 124^{0} = 180^{0}\) \(\Rightarrow = 56^{0}\)

Question 2: The angles of a triangle are in the ratio 2:3:4 . Find the angles.

Ans:

Let the angles of the given triangle measure (2x)0, (3x)0 and (4x)0, respectively.

Then,

2x + 3x + 4x = 1800 [Sum of the angles of a triangle]

9x = 1800

x =200

Hence, the measures of the angles are 2 x 200 = 400, 3 x 200 = 600 and 4 x 200=800.

Question 3: In \(\bigtriangleup ABC\), if \(3\angle A = 4\angle B=6\angle C\), calculate \(\angle A , \angle B\; and \;\angle C\).

Ans:

Let \(3 \angle A = 4 \angle B = 6 \angle C = x^{0}\)

Then,

\(\angle A = (\frac{x}{3})^{0} , \angle B = (\frac{x}{4})^{0}\ and\ \angle C = (\frac{x}{6})^{0}\)

Therefore, \( \frac{x}{3} + \frac{x}{4} + \frac{x}{6} = 180^{0}\) [Sum of the angles of a triangle]

4x + 3x + 2x = 21600

9x = 21600

x = 2400

Therefore,

\(\angle A = (\frac{240}{3})^{0} = 80^{0}\) \(\angle B = (\frac{240}{4})^{0} = 60^{0}\) \(\angle c = (\frac{240}{6})^{0} = 40^{0}\)

Question 4: In \(\bigtriangleup ABC,\angle A +\angle B=108^{0}\; and \;\angle B+\angle C =130^{0}\), find \(\angle A,\angle B\; and \angle C\) .

Ans:

Let \(\angle A + \angle B = 108^{0}\ and\ \angle B + \angle C = 130^{0}\) \(\Rightarrow \angle A + \angle B + \angle B + \angle C = (108+130)^{0} \) \(\Rightarrow (\angle A + \angle B + \angle C) + \angle B = 238^{0}\ [\angle A + \angle B + \angle C = 180^{0}]\) \(\Rightarrow 180^{0} + \angle B = 238^{0}\) \(\Rightarrow \angle B = 58^{0}\)

therefre, \( \angle C = 130^{0} – \angle B\)

= (130 – 58)0

= 720

Therefore, \( \angle A = 108^{0} – \angle B\)

= (108 – 58)0

= 500

Question 5: In \(\bigtriangleup ABC,\angle A +\angle B=125^{0}\; and \;\angle A+\angle C =113^{0}\), find \(\angle A,\angle B\; and \angle C\) .

Ans:

Let \(\angle A + \angle B = 125^{0}\; and\; \angle A + \angle C = 113^{0}\)

Then,

\(\angle A + \angle B + \angle A + \angle C = (125+113)^{0}\) \(\Rightarrow (\angle A + \angle B + \angle C ) + \angle A = 238^{0}\) \(\Rightarrow 180^{0} + \angle A = 238^{0}\) \(\Rightarrow \angle A = 58^{0}\)

therefre, \( \angle B = 125^{0} – \angle A\)

= (125 – 58)0

=670

Therefore, \( \angle C = 113^{0} – \angle A\)

= (113-58)0

=550

Question 6: In \(\bigtriangleup PQR, \angle P -\angle Q = 42 ^{0}\; and \;\angle Q – \angle R = 21^{0}\),  find \(\angle P ,\angle Q \; and \;\angle R \) .

Ans:

Given :\(\angle P – \angle Q = 42^{0}\; and\; \angle Q – \angle R = 21^{0}\)

Then,

\(\angle P = 42^{0} + \angle Q\; and\; \angle R = \angle Q – 21^{0}\)

Therefore, \( 42^{0} + \angle Q + \angle Q + \angle Q – 21^{0} = 180^{0}\) [Sum of the angles of a triangle] \(\Rightarrow 3 \angle Q = 159^{0}\) \(\Rightarrow \angle Q = 53^{0}\)

Therefore, \( \angle P = 42^{0} + \angle Q\)

= (42 + 53)0

= 950

Therefore, \( \angle R = \angle Q – 21^{0}\)

= (53 – 21)0

= 320

Question 7: The sum of two angles of a triangle is \(116^{0}\) and their difference is \(24^{0}\). Find the measure of each angle of the triangle.

Ans:

Let \(\angle A + \angle B = 116^{0}\ and \ \angle A – \angle B = 24^{0} \)

Then,

Therefore, \( \angle A + \angle B + \angle A – \angle B = (116+24)^{0}\) \(\Rightarrow 2 \angle A = 140^{0}\) \(\Rightarrow \angle A = 70^{0}\)

Therefore, \( \angle B = 116^{0} – \angle A\)

= (116 – 70)0

= 460

Also, in \(\triangle ABC\) \(\angle A + \angle B + \angle C = 180^{0}\) [Sum of the angles of a triangle]

Therefore, \( 70^{0} + 46^{0} + \angle C = 180^{0}\)

Therefore, \( \angle C = 64^{0}\)

Question 8: Two angles of a triangle are equal and the third angles is greater than each one of them by \(18^{0}\). Find the angles.

Ans:

Let \(\angle A = \angle B\ and\ \angle C = \angle A + 18^{0}\)

Then,

\(\angle A + \angle B + \angle C = 180^{0}\) [Sum of the angles of a triangle] \(\angle A + \angle A + \angle A + 18^{0} = 180^{0}\) \(\Rightarrow 3 \angle A = 162^{0}\) \(\Rightarrow \angle A = 54^{0}\)

Since,

\(\angle A = \angle B\) \(\Rightarrow \angle B = 54^{0}\)

Therefore, \( \angle C = \angle A + 18^{0}\)

= (54 + 18)0

= 720

Question 9: Of the three angles of a triangle, one is twice the smallest and another one is thrice the smallest. Find the angles.

Ans:

Let the smallest angle of the triangle be \(\angle C\ and\ let\ \angle A = 2 \angle C and \angle B = 3 \angle C\)

Then,

\(\angle A + \angle B + \angle C = 180^{0}\) [Sum of the angles of a triangle] \(\Rightarrow 2 \angle C + 3 \angle C + \angle C = 180^{0}\) \(\Rightarrow 6 \angle C = 180^{0}\) \(\Rightarrow \angle C = 30^{0}\)

Therefore, \( \angle A = 2 \angle C \)

= 2(30)0

= 600

Also,

\(\angle B = 3 \angle C \)

= 3(30)0

=900

Question 10: In a right-angled triangle, one of the acute angles measure \(53^{0}\). Find the measure of each angle of the triangle.

Ans:

Let ABC be a triangle right-angled at B.

Then, \(\angle B = 90^{0}\ and\ let\ \angle A=53^{0}\)

theref0re, \( \angle A + \angle B + \angle C = 180^{0}\) [Sum of the angles of a triangle] \(\Rightarrow 53^{0} + 90^{0} + \angle C = 180^{0}\) \(\Rightarrow \angle C = 37^{0}\)

Hence, \(\angle A = 53^{0}, \angle B = 90^{0}, \angle C = 37^{0}\)

Question 11: If one angle of a triangle is equal to the sum of the other two, show that the triangle is right angled.

Ans:

Let ABC be triangle

Then \(\angle A = \angle B + \angle C\)

Therefore, \( \angle A + \angle B + \angle C = 180^{0}\) [Sum of the angles of a triangle] \(\Rightarrow \angle B + \angle C + \angle B + \angle C = 180^{0}\) \(\Rightarrow 2 \angle B + \angle C = 180^{0}\) \(\Rightarrow \angle B + \angle C = 90^{0}\) \(\Rightarrow\angle A = 90^{0}\) [\( \angle A = \angle B + \angle C\)]

This implies that the triangle is right-angled at A.

Question 12: A \( \bigtriangleup ABC\) is right angled at A. If \(AL\perp BC\), prove that \(\angle BAL=\angle ACB\).

Ans: We know that the sum of two acute angles of a right angled triangle is 900.

From the right \(\triangle ABL\), we have:

Therefore, \( \angle BAL + \angle ABL = 90^{0}\) \(\Rightarrow \angle BAL = 90^{0} – \angle ABL\) \(\Rightarrow \angle BAL = 90^{0} – \angle ABC\) …(1)

Also, from the right \(\triangle ABL \) , we have:

\( \angle ABC+ \angle ACB = 90^{0}\) \( \Rightarrow \angle ACB = 90^{0} – \angle ABC\) …(2)

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

\( \angle ACB = \angle BAL [\angle BAL = 90^{0} – \angle ABC]\)

Therefore, \( \angle BAL = \angle ACB\)

Question 13: If each of a triangle is less than the sum of the other two, show that the triangle is acute angled.

Ans;

Let ABC be the triangle

Let \( \angle A < \angle B + \angle C\)

Then,

\( 2\angle A < \angle A + \angle B + \angle  C \) [Adding Angle A to both sides ] \( \Rightarrow 2 \angle A < 180^{0}\) \( \Rightarrow \angle A < 90^{0}\)

Also, let \( \angle B < \angle A + \angle C\)

Then,

\( 2\angle B < \angle A + \angle B + \angle C\) [Adding Angle B to both sides ] \( \Rightarrow 2 \angle B < 180^{0}\) \( \Rightarrow \angle B < 90^{0}\)

And let \( \angle C < \angle A + \angle B\)

Then,

\( 2 \angle C < \angle A + \angle B + \angle C\) [Adding Angle C to both sides ] \(\Rightarrow 2 \angle C < 180^{0}  [\angle A + \angle B + \angle C = 180^{0}] \) \( \Rightarrow C < 90^{0}\)

Hence, each angle of the triangle is less than 900

Therefore, the triangle is acute-angled.

Question 14: If one of a triangle is greater than the sum of the other two, show that the triangle is obtuse-angled.

Ans:

Let ABC be a triangle and let \(\angle C > \angle A + \angle B \)

Then, we have

\( 2 \angle C > \angle A + \angle B + \angle C\)  [Adding Angle C to both sides ] \( \Rightarrow 2 \angle C > 180^{0}  [\angle A + \angle B + \angle C = 180^{0}]\) \( \Rightarrow \angle C > 90^{0}\)

Since one of the angles of the triangle is greater than 900. the triangle is obtuse-angled.

Question 15: In the given figure, side BC of \(\bigtriangleup ABC\) is produced to D. If \(\angle ACD= 128^{0}\) and \(\angle ABC= 43^{0}\), find \(\angle BAC\; and \; \angle ACB\).

Ans:

Side BC of triangle ABC is produced to D.

Therefore, \( \angle ACD = \angle A + \angle B \) [Exterior angle property] \( \Rightarrow 128^{0} = \angle A + 43^{0}\) \(\Rightarrow \angle A = (128 – 43)^{0} \) \( \Rightarrow \angle A = 85^{0} \) \( \Rightarrow \angle BAC = 85^{0}\)

Also, in triangle ABC,

\( \angle BAC + \angle ABC + \angle ACB = 180^{0}\) [Sum of the angles of a triangle] \( \Rightarrow 85^{0} + 43^{0} + \angle ACB = 180^{0}\) \( \Rightarrow 128^{0} + \angle ACB = 180^{0}\) \( \Rightarrow \angle ACB = 52^{0} \)

Question 16: In the given figure, the side BC of \(\bigtriangleup ABC\) has been produced on both sides on the left to D and on the right to E. If \(\angle ABC= 106^{0}\; and \; \angle ACE = 118^{0}\), find the measure of each angle of the triangle.

Ans:

Side BC of triangle ABC is produced to D.

Therefore, \( ABC = \angle A + \angle C\) \( \Rightarrow 106^{0} = \angle A + \angle C \) …(1)

Also, side BC of triangle ABC is produced to E.

\( \angle ACE = \angle A + \angle B \) \( \Rightarrow 118^{0} = \angle A + \angle B \) …(ii)

Adding (i) and (ii), we get:

\( \angle A + \angle A + \angle B + \angle C = (106 + 118)^{0} \) \( \Rightarrow ( \angle A + \angle B + \angle C ) + \angle A = 224^{0} \;[\angle A + \angle B + \angle C = 180^{0}] \) \(\Rightarrow 180^{0} + \angle A = 224^{0} \) \(\Rightarrow \angle A = 44^{0}\)

therefore, \( B = 118^{0} – \angle A\ [Using (ii)] \) \( \Rightarrow \angle B = (118 – 44)^{0} \) \( \Rightarrow \angle B = 74^{0} \)

And,

\( \angle C = 106^{0} – \angle A\ [Using (i)] \) \(\Rightarrow \angle C = (106 – 44)^{0} \) \( \Rightarrow \angle C = 62^{0} \)

Question 17: Calculate the value of x in each of the following figures.

Ans:

(i) Side AC of a triangle ABC is produced to E.

Therefore, \( \angle EAB = \angle B + \angle C \) \( \Rightarrow 110^{0} = x + \angle C\) ….(i)

Also,

\( \angle ACD + \angle ACB = 180^{0}\) [Linear Pair] \( \Rightarrow 120^{0} + \angle ACB = 180^{0} \) \( \Rightarrow \angle ACB = 60^{0} \) \( \angle C = 60^{0} \)

Substituting the value of \( \angle C \) in (i), we get x = 50

(ii)

From \( \triangle ABC \) we have:

\( \angle A + \angle B + \angle C = 180^{0} \) [Sum of the angles of a triangle] \( \Rightarrow 30^{0} + 40^{0} +\angle C = 180^{0} \) \( \Rightarrow \angle C = 110^{0} \) \( \Rightarrow \angle ACB = 110^{0} \)

Also,

\( \angle ECB + \angle ECD = 180^{0} \) \( \Rightarrow 110^{0} + \angle ECD = 180^{0} \) \( \Rightarrow \angle ECD = 70^{0} \)

Now, in \( \triangle ECD, \)

Therefore, \( \angle AED = \angle ECD + \angle EDC \) [exterior angle property] \( \Rightarrow x = 70^{0} + 50^{0} \) \( \Rightarrow x = 120^{0} \)

(iii)

\( \angle ACB + \angle ACD = 180^{0} \) [Linear Pair] \( \Rightarrow \angle ACB + 115^{0} = 180^{0} \) \( \Rightarrow \angle ACB = 65^{0} \)

Also,

\( \angle EAF = \angle BAC\) [Vertically opposite angles] \( \Rightarrow \angle BAC = 60^{0} \)

Therefore, \( \angle BAC + \angle ABC + \angle ACB = 180^{0} \) [Sum of the angles of a triangle] \( \Rightarrow 60^{0} + x + 65^{0} = 180^{0} \) \( \Rightarrow x = 55^{0} \)

(iv)

\( \angle BAE = \angle CDE\) [Alternate angles] \( \Rightarrow \angle CDE = 60^{0} \)

Therefore, \( \angle ECD + \angle CDE + \angle CED = 180^{0} \) [Sum of the angles of a triangle] \( \Rightarrow 45^{0} + 60^{0} + x = 180^{0} \) \( \Rightarrow x = 75^{0} \)

(v)

From \( \triangle ABC \), we have:

\( \angle BAC + \angle ABC + \angle ACB = 180^{0} \) \( \Rightarrow 40^{0} + \angle ABC + 90^{0} = 180^{0} \) \( \Rightarrow \angle ABC = 50^{0} \)

Also, form \( \triangle EBD \) , we have:

\( \angle BED + \angle EBD + \angle BDE = 180^{0} \) [Sum of the angles of a triangle] \( \Rightarrow 100^{0} + 50^{0} + x = 180^{0} [\angle ABC = \angle EBD] \) \( \Rightarrow x = 30^{0} \)

(vi)

From \( \triangle ABE, \) we have:

\( \angle BAE + \angle ABE + \angle AEB = 180^{0} \) [Sum of the angles of a triangle] \( \Rightarrow 75^{0} + 65^{0} + \angle AEB = 180^{0} \) \( \Rightarrow \angle AEB = 40^{0} \)

Therefore, \( \angle AEB = \angle CED \) [Vertically opposite angles]

Therefore, \( \angle CED = 40^{0} \)

Also, from \( \triangle CDE,\) we have:

\( \angle ECD + \angle CDE + \angle CED = 180^{0} \) [Sum of the angles of a triangle] \( \Rightarrow 110^{0} + x + 40^{0} = 180^{0} \) \( \Rightarrow x = 30^{0} \)

Question 18: Calculate the value of x in the given figure.

Ans:

Join A and D to produce AD to E.

Then,

\( \angle CAD + \angle DAB = 55^{0} \) \( \angle CDE + \angle EDB = x^{0} \)

Side AD of triangle ACD is produced to E.

Therefore, \( \angle CDE = \angle CAD + \angle ACD \) …(i) (Exterior angle property)

Side AD of triangle ABD is produced to E.

Therefore, \( EDB = \angle DAB + \angle ABD \) …(ii) (Exterior angle property)

Adding (i) and (ii) we get.

\( \angle CDE + \angle EDB = \angle CAD + \angle ACD + \angle DAB + \angle ABD \) \( \Rightarrow x^{0} = ( \angle CAD + \angle DAB) + 30^{0} + 45^{0} \) \(\Rightarrow x^{0} = 55^{0} + 30^{0} + 45^{0} \) \(\Rightarrow x^{0} = 130^{0} \)

Key Features of RS Aggarwal Class 9 Solutions Chapter 4– Angles, Lines And Triangles Ex 4B

  • The solutions can be referred by the students to clear their doubts.
  • All the solutions are solved accurately and in a simple language.
  • It is the best study material for students if they want to score good marks in their exam.
  • It provides easy methods to solve tricky and difficult questions.

Practise This Question

The cost of painting the curved surface area of a cylindrical pillar of height 10 m and radius 3.5m at Rs.10 per sq. meter is Rs.2200.
(Take pi=227)

Leave a Comment

Your email address will not be published. Required fields are marked *