RD Sharma Solutions Class 9 Triangle And Its Angles Exercise 9.1

RD Sharma Solutions Class 9 Chapter 9 Exercise 9.1

RD Sharma Class 9 Solutions Chapter 9 Ex 9.1 Free Download

Q1) \(In\;a\;\Delta ABC,\;if\;\angle A=55^{0},\;\angle B=40^{0},\;Find\;\angle C.\)

Solution:

Given Data:

\(\;\angle B=55^{0},\;\angle B=40^{0},\;then\;\angle C=?\)

We know that

\(In\;a\;\Delta ABC\) sum of all angles of a triangle is 1800

i.e., \(\;\angle A+\angle B+\angle C=180^{0}\)

\(\Rightarrow 55^{0}+40^{0}+\angle C=180^{0}\)

\(\Rightarrow 95^{0}+\angle C=180^{0}\)

\(\Rightarrow \angle C=180^{0}-95^{0}\)

\(\Rightarrow \angle C=85^{0}\)

Q2) If the angles of a triangle are in the ratio 1:2:3, determine three angles.

Solution:

Given that,

Angles of a triangle are in the ratio 1:2:3

Let the angles be x, 2x, 3x

\(∴\) We know that,

Sum of all angles of triangles is \(180^{0}\)

x+2x+3x = \(180^{0}\)

=>6x = \(180^{0}\)

=>\( x=\frac{180^{0}}{6}\)

=>x=\(30^{0}\)

Since x=\(30^{0}\)

2x = 2\((30)^{0}\) = \(60^{0}\)

3x = 3\((30)^{0}\) = \(90^{0}\)

Therefore, angles are \(30^{0}\), \(60^{0}\), \(90^{0}\)

Q3) The angles of a triangle are \((x-40^{0}), (x-20^{0})\;and\;(\frac{1}{2}x-10^{0})\). Find the value of x.

Solution:

Given that,

The angles of a triangle are

\((x-40^{0}), (x-20^{0})\;and\;(\frac{1}{2}x-10^{0})\)

We know that,

Sum of all angles of triangle is \(180^{0}\)

\(∴ (x-40^{0})+(x-20^{0})+(\frac{1}{2}x-10^{0})=180^{0}\)

\(2x+\frac{1}{2}x-70^{0}=180^{0}\)

\(\frac{5}{2}x=180^{0}+70^{0}\)

\({5}x=2(250)^{0}\)

\(x=\frac{500^{0}}{5}\)

\(∴ x=100^{0}\)

 Q4) The angles of a triangle are arranged in ascending order of magnitude. If the difference between two consecutive angles is \(10^{0}\), find the three angles.

Solution:

Given that,

The difference between two consecutive angles is \(10^{0}\)

Let x, x+\(10^{0}\), x+\(20^{0}\) be the consecutive angles that differ by \(10^{0}\)

We know that,

Sum of all angles in a triangle is \(180^{0}\)

x+x+\(10^{0}\)+x+\(20^{0}\) = \(180^{0}\)

3x+\(30^{0}\) = \(180^{0}\)

=>3x = \(180^{0}\)\(30^{0}\)

=>3x = \(150^{0}\)

=>x = \(50^{0}\)

Therefore, the required angles are

x = \(50^{0}\)

x+\(10^{0}\) = \(50^{0}\) + \(10^{0}\) = \(60^{0}\)

x+\(20^{0}\) = \(50^{0}\) + \(20^{0}\) = \(70^{0}\)

As the difference between two consecutive angles is \(10^{0}\), the three angles are \(50^{0},\;60^{0},\;70^{0}.\)

Q5) Two angles of a triangle are equal and the third angle is greater than each of those angles by \(30^{0}\). Determine all the angles of the triangle.

Solution:

Given that,

Two angles of a triangle are equal and the third angle is greater than each of those angles by \(30^{0}\).

Let x, x, x+\(30^{0}\) be the angles of a triangle

We know that,

Sum of all angles in a triangle is \(180^{0}\)

x + x + x + \(30^{0}\) = \(180^{0}\)

3x + \(30^{0}\) = \(180^{0}\)

3x = \(180^{0}-30^{0}\)

3x = \(150^{0}\)

x = \(50^{0}\)

Therefore, the three angles are \(50^{0},\;50^{0},\;80^{0}.\)

Q6) If one angle of a triangle is equal to the sum of the other two, show that the triangle is a right angle triangle.

Solution:

If one angle of a triangle is equal to the sum of the other two angles

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

In \(\Delta ABC\),

Sum of all angles of a triangle is \(180^{0}\)

=>\(\angle A+\angle B+\angle C=180^{0}\)

=>\(\angle B+\angle B=180^{0}\)[∠ B=∠ A+∠C]

=>\(2\angle B=180^{0}\)

=>\(\angle B=\frac{180^{0}}{2}\)

=>\(\angle B=90^{0}\)

Therefore, ABC is a right angled triangle.

Q7) ABC is a triangle in which \(\angle A=72^{0}\), the internal bisectors of angles B and C meet in O. Find the magnitude of \(\angle BOC\).

1

Solution:

Given,

ABC is a triangle where \(\angle A=72^{0}\) and the internal bisector of angles B and C meeting O.

In \(\Delta ABC\),

\(\angle A+\angle B+\angle C=180^{0}\)

=>\( 72^{0}+\angle B+\angle C=180^{0}\)

=>\( \angle B+\angle C=180^{0}-72^{0}\)

Dividing both sides by ‘2’

=>\(\frac{\angle B}{2}+\frac{\angle C}{2}=\frac{108^{0}}{2}\)

=>\(\angle OBC+\angle OCB=54^{0}\)

Now, \(In\;\Delta BOC\Rightarrow \angle OBC+\angle OCB+\angle BOC=180^{0}\)

=>\( 54^{0}+\angle BOC=180^{0}\)

=>\(\angle BOC=180^{0}- 54^{0}= 126^{0}\)

\(∴ \angle BOC=126^{0}\)

Q8) The bisectors of base angles of a triangle cannot enclose a right angle in any case.

Solution:

In \(\Delta XYZ\),

Sum of all angles of a triangle is \(180^{0}\)

i.e., \(\angle X+\angle Y+\angle Z=180^{0}\)

Dividing both sides by ‘2’

=>\(\frac{1}{2}\angle X+\frac{1}{2}\angle Y+\frac{1}{2}\angle Z=180^{0}\)

=>\(\frac{1}{2}\angle X+\angle OYZ+\angle OYZ=90^{0}\)  [∵ OY, OZ,∠ Y and ∠ Z]

2

\(\Rightarrow \angle OYZ+\angle OZY=90^{0}-\frac{1}{2}\angle X\)

Now in \(\Delta YOZ\)

\(∴ \angle YOZ+\angle OYZ+\angle OZY=180^{0}\)

\(\Rightarrow \angle YOZ+90^{0}-\frac{1}{2}\angle X=180^{0}\)

\(\Rightarrow \angle YOZ=90^{0}-\frac{1}{2}\angle X\)

Therefore, the bisectors of a base angle cannot enclosure right angle.

Q9) If the bisectors of the base angles of a triangle enclose an angle of \(135^{0}\), prove that the triangle is a right angle.

Solution:

3

Given the bisectors of the base angles of a triangle enclose an angle of \(135^{0}\)

i.e., \(\angle BOC=135^{0}\)

But, We know that

\(\angle BOC=90^{0}+\frac{1}{2}\angle A\)

=>\(135^{0}=90^{0}+\frac{1}{2}\angle A\)

=>\(\frac{1}{2}\angle A=135^{0}-90^{0}\)

=>\(\angle A=45^{0}(2)\)

=>\(\angle A=90^{0}\)

Therefore, \(\Delta ABC\) is a right angle triangle that is right angled at A.

Q10) In a \(\Delta ABC\), \(\angle ABC=\angle ACB\) and the bisectors of \(\angle ABC\;and\;\angle ACB\) intersect at O such that \(\angle BOC=120^{0}\). Show that \(\angle A=\angle B=\angle C=60^{0}\).

Solution:

4

Given,

In \(\Delta ABC\),

\(\angle ABC=\angle ACB\)

Dividing both sides by ‘2’

\(\frac{1}{2}\angle ABC=\frac{1}{2}\angle ACB\)

=>\(\angle OBC=\angle OCB\;\;\;\;\;\;\;\;\;\;[∴ OB,\;OC\;bisects\;\angle B\;and\;\angle C]\)

Now,

\(\angle BOC=90^{0}+\frac{1}{2}\angle A\)

=>\(120^{0}-90^{0}=\frac{1}{2}\angle A\)

=>\(30^{0}*(2)=\angle A\)

=>\( \angle A=60^{0} \)

Now in \(\Delta ABC\)

\(\angle A+\angle ABC+\angle ACB=180^{0}\;\;\;\;\;\;(Sum\;of\;all\;angles\;of\;a\;triangle)\)

=>\(60^{0}+2\angle ABC=180^{0}\;\;\;\;\;\;\;[∴ \angle ABC=\angle ACB]\)

=>\( 2\angle ABC=180^{0}-60^{0}\)

=>\(\angle ABC=\frac{120^{0}}{2}=60^{0}\)

=>\(\angle ABC=\angle ACB\)

\(∴ \angle ACB=60^{0}\)

Hence Proved.

 Q11)  Can a triangle have:

 (i) Two right angles?

 (ii) Two obtuse angles?

(iii) Two acute angles?

(iv) All angles more than 60°?

(v) All angles less than 60°?

(vi) All angles equal to 60″?

Justify your answer in each case.

Sol:

(i) No,

Two right angles would up to 180°. So the third angle becomes zero. This is not possible, so a triangle cannot have two right angles. [Since sum of angles in a triangle is 1800]

(ii) No,

A triangle can’t have 2 obtuse angles. Obtuse angle means more than 90° So that the sum of the two sides will exceed 180° which is not possible. As the sum of all three angles of a triangle is 180°.

(iii) Yes

A triangle can have 2 acute angles. Acute angle means less the 90″ angle.

(iv) No

Having angles more than 600 make that sum more than 1800. This is not possible.  [Since the sum of all the internal angles of a triangle is 1800]

(v) No

Having all angles less than 600 will make that sum less than 1800 which is not possible.[Therefore, the sum of all the internal angles of a triangle is 1800]

(vi) Yes

A triangle can have three angles equal to 600 . Then the sum of three angles equal to the 1800.  Such triangles are called as equilateral triangle. [Since, the sum of all the internal angles of a triangle is1800]

Q12) If each angle of a triangle is less than the sum of the other two, show that the triangle is acute angled.

Solution

Given each angle of a triangle less than the sum of the other two

\(∴ \angle X+\angle Y+\angle Z\)

=>\(\angle X+\angle X<\angle X+\angle Y+\angle Z\)

=>\(2\angle X<180^{0}\;\;\;\;\;\;[Sum\;of\;all\;the\;angles\;of\;a\;triangle]\)

=>\(\angle X<90^{0}\)

Similarly \(\angle Y<90^{0}\;and\;\angle Z<90^{0}\)

Hence, the triangles are acute angled.