# RS Aggarwal Solutions Class 9 Ex 4B

## RS Aggarwal Class 9 Ex 4B Chapter 4

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

A crop needs to be protected from rodents like rats, fungi and bacteria. They affect the stored grains and are called the  _________factors.