The document RD Sharma Solutions -Ex-9.1, Triangle And Its Angles, Class 9, Maths Class 9 Notes | EduRev is a part of the Class 9 Course RD Sharma Solutions for Class 9 Mathematics.

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**Q1) InaÎ”ABC,ifâˆ A = 55 ^{0},âˆ B = 40^{0},Findâˆ C.**

**Solution:**

Given Data:

âˆ B = 55^{0},âˆ B = 40^{0},thenâˆ C = ?

We know that

InaÎ”ABC sum of all angles of a triangle is 180^{0}

i.e., âˆ A+âˆ B+âˆ C = 180^{0}

â‡’ 55^{0}+40^{0}+âˆ C = 180^{0}

â‡’ 95^{0}+âˆ C = 180^{0}

â‡’ âˆ C = 180^{0}âˆ’95^{0}

â‡’ âˆ 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

âˆ´ We know that,

Sum of all angles of triangles is 180^{0}

x+2x+3x = 180^{0}

â‡’ 6x = 180^{0}

â‡’ x =

â‡’ 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 (xâˆ’40 ^{0}),(xâˆ’20^{0})and(**

**Solution:**

Given that,

The angles of a triangle are

(xâˆ’40^{0}),(xâˆ’20^{0})and(xâˆ’10^{0})

We know that,

Sum of all angles of triangle is 180^{0}

âˆ´(xâˆ’40^{0})+(xâˆ’20^{0})+(xâˆ’10^{0}) = 180^{0}

2x+xâˆ’70^{0 }= 180^{0}

x = 180^{0}+70^{0}

5x = 2(250)^{0}

x =

âˆ´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}

â‡’ 3x = 180^{0}â€“ 30^{0}

â‡’ 3x = 150^{0}

â‡’ 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}âˆ’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

â‡’ âˆ B = âˆ A+âˆ C

In Î”ABC,

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

â‡’ âˆ A+âˆ B+âˆ C = 180^{0}

â‡’ âˆ B+âˆ B = 180^{0}[âˆ B = âˆ A+âˆ C]

â‡’ 2âˆ B = 180^{0}

â‡’ âˆ B =

â‡’ âˆ B = 90^{0}

Therefore, ABC is a right angled triangle.

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

**Solution:**

Given,

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

In Î”ABC,

âˆ A+âˆ B+âˆ C = 180^{0}

â‡’ 720+âˆ B+âˆ C = 180^{0}

â‡’ âˆ B+âˆ C = 180^{0}âˆ’72^{0}

Dividing both sides by â€˜2â€™

â‡’ âˆ OBC+âˆ OCB = 54^{0}

Now, InÎ”BOCâ‡’ âˆ OBC+âˆ OCB+âˆ BOC = 180^{0}

â‡’ 54^{0}+âˆ BOC = 180^{0}

â‡’ âˆ BOC = 180^{0}âˆ’54^{0} = 126^{0}

âˆ´âˆ BOC = 126^{0}

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

**Solution:**

In Î”XYZ,

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

i.e., âˆ X+âˆ Y+âˆ Z = 180^{0}

Dividing both sides by â€˜2â€™

â‡’ âˆ X+âˆ Y+âˆ Z = 180^{0}

â‡’ âˆ X+âˆ OYZ+âˆ OYZ = 90^{0} [âˆµ OY, OZ,âˆ Y and âˆ Z]

â‡’ âˆ OYZ+âˆ OZY = 90^{0}âˆ’âˆ X

Now in Î”YOZ

âˆ´âˆ YOZ+âˆ OYZ+âˆ OZY = 180^{0}

â‡’ âˆ YOZ+90^{0}âˆ’âˆ X = 180^{0}

â‡’ âˆ YOZ = 90^{0}âˆ’âˆ 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:**

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

i.e., âˆ BOC = 135^{0}

But, We know that

âˆ BOC = 90^{0}+âˆ A

â‡’ 135^{0} = 90^{0}+âˆ A

â‡’ âˆ A = 135^{0}âˆ’90^{0}

â‡’ âˆ A = 45^{0}(2)

â‡’ âˆ A = 90^{0}

Therefore, Î”ABC is a right angle triangle that is right angled at A.

**Q10) In a Î”ABC, âˆ ABC = âˆ ACB and the bisectors of âˆ ABCandâˆ ACB intersect at O such that âˆ BOC = 120 ^{0}. Show that âˆ A = âˆ B = âˆ C = 60^{0}.**

**Solution:**

Given,

In Î”ABC,

âˆ ABC = âˆ ACB

Dividing both sides by â€˜2â€™

âˆ ABC = âˆ ACB

â‡’ âˆ OBC = âˆ OCB [âˆ´OB,OC bisects âˆ B and âˆ C]

Now,

âˆ BOC = 90^{0}+âˆ A

â‡’ 120^{0}âˆ’90^{0 }= âˆ A

â‡’ 30^{0}âˆ—(2) = âˆ A

â‡’ âˆ A = 60^{0}

Now in Î”ABC

âˆ A+âˆ ABC+âˆ ACB = 1800(Sumofallanglesofatriangle)

â‡’ 60^{0}+2âˆ ABC = 180^{0 }[âˆ´âˆ ABC = âˆ ACB]

â‡’ 2âˆ ABC = 180^{0}âˆ’60^{0}

â‡’ âˆ ABC = = 60^{0}

â‡’ âˆ ABC = âˆ ACB

âˆ´âˆ 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 60Â°?**

**(v) All angles less than 60Â°?**

**(vi) All angles equal to 60â€³?**

**Justify your answer in each case.**

**Sol:**

(i) No,

Two right angles would up to 180Â°. 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 180^{0}]

(ii) No,

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

(iii) Yes

A triangle can have 2 acute angles. Acute angle means less the 90â€³ angle.

(iv) No

Having angles more than 60^{0} make that sum more than 180^{0}. This is not possible. [Since the sum of all the internal angles of a triangle is 180^{0}]

(v) No

Having all angles less than 60^{0} will make that sum less than 180^{0} which is not possible.[Therefore, the sum of all the internal angles of a triangle is 180^{0}]

(vi) Yes

A triangle can have three angles equal to 60^{0} . Then the sum of three angles equal to the 180^{0}. Such triangles are called as equilateral triangle. [Since, the sum of all the internal angles of a triangle is180^{0}]

**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

âˆ´âˆ X+âˆ Y+âˆ Z

â‡’ âˆ X+âˆ X<âˆ X+âˆ Y+âˆ Z

â‡’ 2âˆ X<180^{0 }[Sumofalltheanglesofatriangle]

â‡’ âˆ X<90^{0}

Similarly âˆ Y<90^{0}andâˆ Z<90^{0}

Hence, the triangles are acute angled.