Ex-9.1, Triangle And Its Angles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics PDF Download

Q1) InaΔABC,if∠A = 55⁰,∠B = 40⁰,Find∠C.

Solution:

Given Data:

∠B = 55⁰,∠B = 40⁰,then∠C = ?

We know that

InaΔABC sum of all angles of a triangle is 180⁰

i.e., ∠A+∠B+∠C = 180⁰

⇒ 55⁰+40⁰+∠C = 180⁰

⇒ 95⁰+∠C = 180⁰

⇒ ∠C = 180⁰−95⁰

⇒ ∠C = 85⁰

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 180⁰

x+2x+3x =  180⁰

⇒ 6x =  180⁰

⇒ x =

⇒ x = 30⁰

Since x = 30⁰

2x = 2(30)⁰  =  60⁰

3x = 3(30)⁰  =  90⁰

Therefore, angles are 30⁰, 60⁰, 90⁰

Q3) The angles of a triangle are (x−40⁰),(x−20⁰)and(x−10⁰). Find the value of x.

Solution:

Given that,

The angles of a triangle are

(x−40⁰),(x−20⁰)and(x−10⁰)

We know that,

Sum of all angles of triangle is 180⁰

∴(x−40⁰)+(x−20⁰)+(x−10⁰) = 180⁰

2x+x−70⁰ = 180⁰

x = 180⁰+70⁰

5x = 2(250)⁰

x =

∴x = 100⁰

Q4) The angles of a triangle are arranged in ascending order of magnitude. If the difference between two consecutive angles is 10⁰, find the three angles.

Solution:

Given that,

The difference between two consecutive angles is 10⁰

Let x, x+10⁰, x+20⁰ be the consecutive angles that differ by 10⁰

We know that,

Sum of all angles in a triangle is 180⁰

x+x+10⁰+x+20⁰  =  180⁰

3x+30⁰  =  180⁰

⇒ 3x =  180⁰– 30⁰

⇒ 3x =  150⁰

⇒ x =  50⁰

Therefore, the required angles are

x =  50⁰

x+10⁰  =  50⁰ + 10⁰  =  60⁰

x+20⁰  =  50⁰ + 20⁰  =  70⁰

As the difference between two consecutive angles is 10⁰, the three angles are 50⁰,60⁰,70⁰.

Q5) Two angles of a triangle are equal and the third angle is greater than each of those angles by 30⁰. 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+30⁰ be the angles of a triangle

We know that,

Sum of all angles in a triangle is 180⁰

x + x + x + 30⁰  =  180⁰

3x + 30⁰  =  180⁰

3x =  180⁰−30⁰

3x =  150⁰

x =  50⁰

Therefore, the three angles are 50⁰,50⁰,80⁰.

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 180⁰

⇒ ∠A+∠B+∠C = 180⁰

⇒ ∠B+∠B = 180⁰[∠ B = ∠ A+∠C]

⇒ 2∠B = 180⁰

⇒ ∠B =

⇒ ∠B = 90⁰

Therefore, ABC is a right angled triangle.

Q7) ABC is a triangle in which ∠A = 72⁰, the internal bisectors of angles B and C meet in O. Find the magnitude of ∠BOC.

Solution:

Given,

ABC is a triangle where ∠A = 72⁰ and the internal bisector of angles B and C meeting O.

In ΔABC,

∠A+∠B+∠C = 180⁰

⇒ 720+∠B+∠C = 180⁰

⇒ ∠B+∠C = 180⁰−72⁰

Dividing both sides by ‘2’

⇒ ∠OBC+∠OCB = 54⁰

Now, InΔBOC⇒ ∠OBC+∠OCB+∠BOC = 180⁰

⇒ 54⁰+∠BOC = 180⁰

⇒ ∠BOC = 180⁰−54⁰ = 126⁰

∴∠BOC = 126⁰

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 180⁰

i.e., ∠X+∠Y+∠Z = 180⁰

Dividing both sides by ‘2’

⇒ ∠X+∠Y+∠Z = 180⁰

⇒ ∠X+∠OYZ+∠OYZ = 90⁰              [∵ OY, OZ,∠ Y and ∠ Z]

⇒ ∠OYZ+∠OZY = 90⁰−∠X

Now in ΔYOZ

∴∠YOZ+∠OYZ+∠OZY = 180⁰

⇒ ∠YOZ+90⁰−∠X = 180⁰

⇒ ∠YOZ = 90⁰−∠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 135⁰, prove that the triangle is a right angle.

Solution:

Given the bisectors of the base angles of a triangle enclose an angle of 135⁰

i.e., ∠BOC = 135⁰

But, We know that

∠BOC = 90⁰+∠A

⇒ 135⁰ = 90⁰+∠A

⇒ ∠A = 135⁰−90⁰

⇒ ∠A = 45⁰(2)

⇒ ∠A = 90⁰

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 = 120⁰. Show that ∠A = ∠B = ∠C = 60⁰.

Solution:

Given,

In ΔABC,

∠ABC = ∠ACB

Dividing both sides by ‘2’

∠ABC = ∠ACB

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

Now,

∠BOC = 90⁰+∠A

⇒ 120⁰−90⁰ = ∠A

⇒ 30⁰∗(2) = ∠A

⇒ ∠A = 60⁰

Now in ΔABC

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

⇒ 60⁰+2∠ABC = 180⁰                        [∴∠ABC = ∠ACB]

⇒ 2∠ABC = 180⁰−60⁰

⇒ ∠ABC =  = 60⁰

⇒ ∠ABC = ∠ACB

∴∠ACB = 60⁰

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

∴∠X+∠Y+∠Z

⇒ ∠X+∠X<∠X+∠Y+∠Z

⇒ 2∠X<180⁰                           [Sumofalltheanglesofatriangle]

⇒ ∠X<90⁰

Similarly ∠Y<90⁰and∠Z<90⁰

Hence, the triangles are acute angled.