RS Aggarwal Solutions Class 9 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.

\( \angle CDE + \angle EDB = \angle CAD + \angle ACD + \angle DAB + \angle ABD \)

\( \Rightarrow x^{0} = ( \angle CAD + \angle DAB) + 30^{0} + 45^{0} \)

\(\Rightarrow x^{0} = 55^{0} + 30^{0} + 45^{0} \)

\(\Rightarrow x^{0} = 130^{0} \)


Practise This Question

The surface area of a box which is in the form of a cuboid whose dimensions are l×b×h is__________.