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.

All you need of Class 9 at this link: Class 9

**Q1) InaΔABC,if∠A = 55 ^{0},∠B = 40^{0},Find∠C.**

**Solution:**

Given Data:

∠B = 55^{0},∠B = 40^{0},then∠C = ?

We know that

InaΔABC sum of all angles of a triangle is 180^{0}

i.e., ∠A+∠B+∠C = 180^{0}

⇒ 55^{0}+40^{0}+∠C = 180^{0}

⇒ 95^{0}+∠C = 180^{0}

⇒ ∠C = 180^{0}−95^{0}

⇒ ∠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 =

⇒ 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(**

**Solution:**

Given that,

The angles of a triangle are

(x−40^{0}),(x−20^{0})and(x−10^{0})

We know that,

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

∴(x−40^{0})+(x−20^{0})+(x−10^{0}) = 180^{0}

2x+x−70^{0 }= 180^{0}

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

5x = 2(250)^{0}

x =

∴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

⇒ ∠B = ∠A+∠C

In ΔABC,

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

⇒ ∠A+∠B+∠C = 180^{0}

⇒ ∠B+∠B = 180^{0}[∠ B = ∠ A+∠C]

⇒ 2∠B = 180^{0}

⇒ ∠B =

⇒ ∠B = 90^{0}

Therefore, ABC is a right angled triangle.

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

**Solution:**

Given,

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

In ΔABC,

∠A+∠B+∠C = 180^{0}

⇒ 720+∠B+∠C = 180^{0}

⇒ ∠B+∠C = 180^{0}−72^{0}

Dividing both sides by ‘2’

⇒ ∠OBC+∠OCB = 54^{0}

Now, InΔBOC⇒ ∠OBC+∠OCB+∠BOC = 180^{0}

⇒ 54^{0}+∠BOC = 180^{0}

⇒ ∠BOC = 180^{0}−54^{0} = 126^{0}

∴∠BOC = 126^{0}

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

**Solution:**

In ΔXYZ,

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

i.e., ∠X+∠Y+∠Z = 180^{0}

Dividing both sides by ‘2’

⇒ ∠X+∠Y+∠Z = 180^{0}

⇒ ∠X+∠OYZ+∠OYZ = 90^{0} [∵ OY, OZ,∠ Y and ∠ Z]

⇒ ∠OYZ+∠OZY = 90^{0}−∠X

Now in ΔYOZ

∴∠YOZ+∠OYZ+∠OZY = 180^{0}

⇒ ∠YOZ+90^{0}−∠X = 180^{0}

⇒ ∠YOZ = 90^{0}−∠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., ∠BOC = 135^{0}

But, We know that

∠BOC = 90^{0}+∠A

⇒ 135^{0} = 90^{0}+∠A

⇒ ∠A = 135^{0}−90^{0}

⇒ ∠A = 45^{0}(2)

⇒ ∠A = 90^{0}

Therefore, ΔABC is a right angle triangle that is right angled at A.

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

**Solution:**

Given,

In ΔABC,

∠ABC = ∠ACB

Dividing both sides by ‘2’

∠ABC = ∠ACB

⇒ ∠OBC = ∠OCB [∴OB,OC bisects ∠B and ∠C]

Now,

∠BOC = 90^{0}+∠A

⇒ 120^{0}−90^{0 }= ∠A

⇒ 30^{0}∗(2) = ∠A

⇒ ∠A = 60^{0}

Now in ΔABC

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

⇒ 60^{0}+2∠ABC = 180^{0 }[∴∠ABC = ∠ACB]

⇒ 2∠ABC = 180^{0}−60^{0}

⇒ ∠ABC = = 60^{0}

⇒ ∠ABC = ∠ACB

∴∠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 180^{0}]

(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 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

∴∠X+∠Y+∠Z

⇒ ∠X+∠X<∠X+∠Y+∠Z

⇒ 2∠X<180^{0 }[Sumofalltheanglesofatriangle]

⇒ ∠X<90^{0}

Similarly ∠Y<90^{0}and∠Z<90^{0}

Hence, the triangles are acute angled.

Offer running on EduRev: __Apply code STAYHOME200__ to get INR 200 off on our premium plan EduRev Infinity!