Triangle And Its Angles

Given, ∠A = 55^o

∠B = 40^o and ∠C =?

We know that, In sum of all angles of triangle is 180^o

∠A + ∠B + ∠C = 180^o

55^o + 40^o + ∠C = 180^o

95^o + ∠C = 180^o

∠C = 85^o

Given that the angles of the triangle are in ratio 1 : 2 : 3

Let, the angles be a, 2a, 3a

Therefore, we know that

Sum of all angles if triangle is 180^o

a + 2a + 3a = 180^o

6a = 180^o

a =

a = 30^o

Since, a = 30^o

2a = 2 (30^o) = 60^o

3a = 3 (30^o) = 90^o

Therefore, angles are a = 30^o, 2a = 60^o and 3a = 90^o

Hence, angles are 30^o, 60^o and 90^o.

Given that,

The angles of the triangle are (x – 40^o), (x – 20^o) and ( – 10^o)

We know that,

Sum of all angles of triangle is 180^o.

Therefore,

x – 40^o + x – 20^o + – 10^o = 180^o

2x + – 70^o = 180^o

= 250^o

5x = 250^o * 2

5x = 500^o

x = 100^o

Therefore, x = 100^o

Given that,

The difference between two consecutive angles is 10^o.

Let, x, x + 10 and x + 20 be the consecutive angles differ by 10^o.

We know that,

x + x + 10 + x + 20 = 180^o

3x + 30^o = 180^o

3x = 180^o – 30^o

3x = 150^o

x = 50^o

Therefore, the required angles are:

x = 50^o

x + 10 = 50^o + 10^o

= 60^o

x + 20 = 50^o + 20^o

= 70^o

The difference between two consecutive angles is 10^o then three angles are 50^o, 60^o and 70^o.

Given that,

Two angles are equal and third angle is greater than each of those angles by 30^o.

Let, x, x, x + 30^o be the angles of the triangle.

We know that,

Sum of all angles of triangle is 180^o

x + x + x + 30^o = 180^o

3x + 30^o = 180^o

3x = 180^o – 30^o

3x = 150^o

x = 50^o

Therefore,

The angles are:

x = 50^o

x = 50^o

x + 30^o = 50^o + 30^o

= 80^o

Therefore, the required angles are 50^o, 50^o, 80^o.

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

i.e. ∠B = ∠A + ∠C

Now, in

Sum of all angles of triangle is 180^o

∠A + ∠B + ∠C = 180^o

∠B + ∠B = 180^o [Therefore, ∠A + ∠C = ∠B]

2∠B = 180^o

∠B = 90^o

Therefore, ABC is right angled triangle.

Given,

ABC is a triangle

∠A = 72^o and internal bisectors of B and C meet O.

In

∠A + ∠B + ∠C = 180^o

72^o + ∠B + ∠C = 180^o

∠B + ∠C = 180^o – 72^o

∠B + ∠C = 108^o

Divide both sides by 2, we get

+ =

+ = 54^o

∠OBC + ∠OCB = 54^o (i)

Now, in

∠OBC + ∠OCB + ∠BOC = 180^o

54^o + ∠BOC = 180^o [Using (i)]

∠BOC = 180^o – 54^o

= 126^o

In sum of all angles of a triangle is 180^o

∠A + ∠B + ∠C = 180^o

Divide both sides by 2, we get

∠A + ∠B + ∠C = 180^o

∠A + ∠OBC + ∠OCB = 90^o [Therefore, OB, OC bisects ∠B and ∠C]

∠OBC + ∠OCB = 90^o - ∠A

Now, in

∠BOC + ∠OBC + ∠OCB = 180^o

∠BOC + 90^o - ∠A = 180^o

∠BOC = 90^o - ∠A

Hence, bisector open base angle cannot enclose right angle.

Given bisector og the base angles of a triangle enclose an angle of 135^o

i.e. ∠BOC = 135^o

But,

135^o = 90^o + ∠A

∠A = 135^o – 90^o

∠A = 45^o (2)

= 90^o

Therefore, is right angled triangle right angled at A.

Given,

∠ ABC = ∠ ACB

Divide both sides by 2, we get

∠ABC = ∠ACB

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

Now,

∠BOC = 90^o + ∠A

120^o – 90^o = ∠A

30^o * 2 = ∠A

∠A = 60^o

Now in

∠A + ∠ABC + ∠ACB = 180^o [Sum of all angles of a triangle]

60^o + 2∠ABC = 180^o [Therefore, ∠ABC = ∠ACB]

2∠ABC = 180^o – 60^o

2∠ABC = 120^o

∠ABC = 60^o

Therefore, ∠ABC = ∠ACB = 60^o

Hence, proved

(i) No, two right angles would up to 180^o so the third angle becomes zero. This is not possible. Therefore, the triangle cannot have two right angles.

(ii) No, a triangle can’t have two obtuse angles as obtuse angle means more than 90^o. So, the sum of the two sides exceeds more than 180^o which is not possible. As the sum of all three angles of a triangle is 180^o.

(iii) Yes, a triangle can have two acute angle as acute angle means less than 90^o.

(iv) No, having angles more than 60^o make that sum more than 180^o which is not possible as the sum of all angles of a triangle is 180^o.

(v) No, having all angles less than 60^o will make that sum less than 180^o which is not possible as the sum of all angles of a triangle is 180^o.

(vi) Yes, a triangle can have three angles equal to 60^o as in this case the sum of all three is equal to 180^o which is possible. This type of triangle is known as equilateral triangle.

Given,

Each angle of a triangle is less than the sum of the other two.

Therefore,

∠A + ∠B + ∠C

∠A + ∠A < ∠A + ∠B + ∠C

2∠A < 180^o [Sum of all angles of a triangle]

∠A = 90^o

Similarly,

∠B < 90^o and ∠C < 90^o

Hence, the triangle is acute angled.

Let, ABC be a triangle and base BC produced to both sides. Exterior angles are ∠ABD and ∠ACE.

∠ABD = 104^o

∠ACE = 136^o

∠ABD + ∠ABC = 180^o (Linear pair)

104^o + ∠ABC = 180^o

∠ABC = 180^o – 104^o

= 76^o

∠ACE + ∠ACB = 180^o

136^o + ∠ACB = 180^o

∠ACB = 180^o – 136^o

= 44^o

In

∠A + ∠ABC + ∠ACB = 180^o

∠A + 76^o + 44^o = 180^o

∠A + 120^o = 180^o

∠A = 180^o – 120^o

= 60^o

Thus, angles of triangle are 60^o, 76^o and 44^o.

Given that ABC is a triangle.

BP and CP are internal bisector of ∠B and ∠C respectively

BQ and CQ are external bisector of ∠B and ∠C respectively.

External ∠B = 180^o - ∠B

External ∠C = 180^o - ∠C

In

∠BPC + ∠B + ∠C = 180^o

∠BPC = 180^o - (∠B + ∠C) (i)

In

∠BQC + (180^o - ∠B) + (180^o - ∠C) = 180^o

∠BQC + 180^o - (∠B + ∠C) = 180^o

∠BPC + ∠BQC = 180^o [From (i)]

Hence, proved

Given,

∠ACD = 105^o

∠EAF = 45^o

∠EAF = ∠BAC (Vertically opposite angle)

∠BAC = 45^o

∠ACD + ∠ACB = 180^o (Linear pair)

105^o + ∠ACB = 180^o

∠ACB = 180^o – 105^o

= 75^o

In

∠BAC + ∠ABC + ∠ACB = 180^o

45^o + ∠ABC + 75^o = 180^o

∠ABC = 180^o – 120^o

= 60^o

Thus, all three angles of a triangle are 45^o, 60^o and 75^o.

(i) ∠DAC + ∠BAC = 180^o (Linear pair)

120^o + ∠BAC = 180^o

∠BAC = 180^o – 120^o

= 60^o

And,

∠ACD + ∠ACB = 180^o

112^o + ∠ACB = 180^o

∠ACB = 68^o

In ABC,

∠BAC + ∠ACB + ∠ABC = 180^o

60^o + 68^o + x = 180^o

128^o + x = 180^o

x = 180^o – 128^o

= 52^o

(ii) ∠ABE + ∠ABC = 180^o (Linear pair)

120^o + ∠ABC = 180^o

∠ABC = 60^o

∠ACD + ∠ACB = 180^o (Linear pair)

110^o + ∠ACB = 180^o

∠ACB = 70^o

In

∠A + ∠ACB + ∠ABC = 180^o

x + 70^o + 60^o = 180^o

x + 130^o = 180^o

x = 50^o

(iii) AB ‖ CD and AD cuts them so,

∠BAE = ∠EDC (Alternate angles)

∠EDC = 52^o

In

∠EDC + ∠ECD + ∠CEO = 180^o

52^o + 40^o + x = 180^o

92^o + x = 180^o

x = 180^o – 92^o

= 88^o

(iv) Join AC

In

∠A + ∠B + ∠C = 180^o

(35^o + ∠1) + 45^o + (50^o + ∠2) = 180^o

130^o + ∠1 + ∠2 = 180^o

∠1 + ∠2 = 50^o

In

∠1 + ∠2 + ∠D = 180^o

50^o + x = 180^o

x = 180^o – 50^o

= 130^o

Given,

AB divides ∠DAC in the ratio 1: 3

∠DAB: ∠BAC = 1: 3

∠DAC + ∠EAC = 180^o

∠DAC + 108^o = 180^o

∠DAC = 180^o – 108^o

= 72^o

∠DAB = * 72^o = 18^o

∠BAC = * 72^o = 54^o

In

∠DAB + ∠ADB + ∠ABD = 180^o

18^o + 18^o + ∠ABD = 180^o

36^o + ∠ABD = 180^o

∠ABD = 180^o – 36^o

= 144^o

∠ABD + ∠ABC = 180^o (Linear pair)

144^o + ∠ABC = 180^o

∠ABC = 180^o – 144^o

= 36^o

In

∠BAC + ∠ABC + ∠ACB = 180^o

54^o + 36^o + x = 180^o

90^o + x = 180^o

x = 180^o – 90^o

= 90^o

Thus, x = 90^o

Exterior ∠B = (180^o - ∠B)

Exterior ∠C = (180^o - ∠C)

In

∠A + ∠B + ∠C = 180^o

(∠A + ∠B + ∠C) = 180^o

(∠B + ∠C) = 180^o - ∠A (i)

In

∠D + ∠DBC + ∠DCB = 180^o

∠D + {180^o - (180^o - ∠B) - ∠B} + {180^o - (180^o - ∠C) - ∠C} = 180^o

∠D + 360^o – 90^o – 90^o – (∠B + ∠C) = 180^o

∠D + 180^o – 90^o - ∠A = 180^o

∠D = ∠A

Hence, proved

Given,

AC is perpendicular to CE

∠A: ∠B: ∠C = 3: 2: 1

Let,

∠A = 3k

∠B = 2k

∠C = k

∠A + ∠B + ∠C = 180^o

3k + 2k + k = 180^o

6k = 180^o

k = 30^o

Therefore,

∠A = 3k = 90^o

∠B = 2k = 60^o

∠C = k = 30^o

Now,

∠C + ∠ACE + ∠ECD = 180^o (Linear pair)

30^o + 90⁰ + ∠ECD = 180^o

∠ECD = 180^o – 120^o

= 60^o

Given,

AM perpendicular to BC

AN is bisector of ∠A

Therefore, ∠NAC = ∠NAB

In

∠A + ∠B + ∠C = 180^o

∠A + 65^o + 33^o = 180^o

∠A = 180^o – 98^o

= 82^o

∠NAC = ∠NAB = 41^o (Therefore, AN is bisector of ∠A)

In

∠AMB + ∠MAB + ∠ABM = 180^o

90^o + ∠MAB + 65^o = 180^o

∠MAB + 155^o = 180⁰

∠MAB = 25^o

Therefore,

∠MAB + ∠MAN = ∠BAN

25^o + ∠MAN = 41^o

∠MAN = 41^o – 25^o

= 16^o

Given,

AB bisects ∠A (∠DAB = ∠DAC)

∠C > ∠B

In ,

∠ADB + ∠DAB + ∠B = 180^o (i)

In ,

∠ADC + ∠DAC + ∠C = 180^o (ii)

From (i) and (ii), we get

∠ADB + ∠DAB + ∠B = ∠ADC + ∠DAC + ∠C

∠ADB > ∠ADC (Therefore, ∠C > ∠B)

Hence, proved

Given,

BD perpendicular to AC

And,

CE perpendicular to AB

In

∠E + ∠B + ∠ECB = 180^o

90^o + ∠B + ∠ECB = 180^o

∠B + ∠ECB = 90^o

∠B = 90^o - ∠ECB .....(i)

In

∠D + ∠C + ∠DBC = 180^o

90^o + ∠C + ∠DBC = 180^o

∠C + ∠DBC = 90^o

∠C = 90^o - ∠DBC .....(ii)

Adding (i) and (ii), we get

∠B + ∠C = 180^o (∠ECB + ∠DBC)

∠180^o - ∠A = 180^o (∠ECB + ∠DBC)

∠A = ∠ECB + ∠DBC

∠A = ∠OCB + ∠OBC


(Therefore, ∠ECB = ∠OCB and ∠DCB = ∠OCB) .... (iii)

In

∠BOC + (∠OBC + ∠OCB) = 180^o

∠BOC + ∠A = 180^o [From (iii)]

∠BOC = 180^o - ∠A

Hence, proved

Given,

AE bisects ∠CAD

∠B = ∠C

In

∠CAD = ∠B + ∠C

∠CAD = ∠C + ∠C

∠CAD = 2∠C

∠1 + ∠2 = 2∠C (Therefore, ∠CAD = ∠1 + ∠2)

∠2 + ∠2 = 2∠C (Therefore, AE bisects ∠CAD)

2∠2 = 2∠C

∠2 = ∠C (Alternate angles)

Therefore, AE ‖ BC

Hence, proved

Since,

AB ‖ DE

∠ABC = ∠CDE (Alternate angles)

∠ABC = 40^o

In

∠A + ∠B + ∠ACB = 180^o

30^o + 40^o + ∠ACB = 180^o

∠ACB = 180^o – 70^o

= 110^o (i)

Now,

∠ACD + ∠ACB = 180^o (Linear pair)

∠ACD + 110^o = 180^o [From (i)]

∠ACD = 180^o – 110^o

= 70^o

Hence, ∠ACD = 70^o.

(i) True

(ii) False

(iii) False

(iv) False

(v) True

(vi) False

(vii) True

(viii) True

(ix) False

(x) True

(xi)True

(i) 180^o

(ii) Interior

(iii) Greater

(iv) One

(v) One

Let,

A, B and C be the angles of

A = B = C (Given)

We know that,

∠A + ∠B + ∠C = 180^o

∠A + ∠A + ∠A = 180^o

3∠A = 180^o

∠A = 60^o

Therefore,

∠A = ∠B = ∠C = 60^o

Thus, each angle is equal to 60^o.

A plane figure with three straight sides and three angles.

Given that the triangle is acute.

So, ∠1, ∠2 and ∠3 be the angles of the triangle.

∠1 = 90^o (Given)

∠2 = ∠3

We know that,

∠1 + ∠2 + ∠3 = 180^o

90^o + ∠2 + ∠2 = 180^o

2∠2 = 180^o – 90^o

∠2 = 45^o

Therefore, ∠2 = ∠3 = 45^o

Thus, each acute angle is equal to 45^o.

A triangle where one of the internal angles is obtuse (greater than 90 degrees) is called an obtuse triangle. The sum of angles of obtuse triangle is also 180°.

In BOC,

∠BOC + ∠OCB + ∠OBC = 180°

∠BOC + 1/2 × (80) + 1/2 × (40) = 180°

∠BOC = 180° -70^o

∠BOC = 110°

Let, ∠1 and ∠2 be two opposite interior angles and ∠3 be exterior angle.

According to question,

∠1 + ∠2 = ∠3

∠1 + ∠1 = 100^o

2∠1 = 100^o

∠1 = 50^o

Therefore, ∠1 = ∠2 = 50^o

Thus, of these angles is equal to 50^o.

A right triangle

By exterior angle theorem:

∠ACD = ∠A + ∠B

120^o = ∠A + ∠A

120^o =

240^o = 3∠A

∠A = 80^o

Given,

In ABC,

B - A = C - B

B + B = A + C

2B = A + C (i)

Now,

A + B + C = 180°

B = 180 - (A + C) (ii)

Using (i) in (ii), we get

B = 180 - 2B

3B = 180°

B = 60°

In

∠BOC + ∠OBC + ∠OCB = 180^o

120^o + ∠B + ∠C = 180^o

(∠B + ∠C) = 60^o

∠B + ∠C = 120^o (i)

In

∠A + ∠B + ∠C = 180^o

∠A + 120^o = 180^o [From (i)]

∠A = 60^o

AX bisects ∠DAC

∠CAD = 2 * ∠DAC

∠CAD = 2 * 70^o

= 140^o

By exterior angle theorem,

∠CAD = ∠B + ∠C

140^o = ∠C + ∠C (Therefore, ∠B = ∠C)

140^o = 2∠C

∠C = 70^o

Therefore, ∠C = ∠ACB = 70^o

If a side of a triangle is produced, the exterior angle so formed is equal to the sum of the two interior opposite angles.

We know that,

In a triangle an exterior angle is equal to sum of two interior opposite angle.

Let, the required interior opposite angle be x.

x + 55^o = 95^o

x = 95^o – 55^o

= 40^o

Thus, other interior angle is 40^o.

Let, ABC be a triangle and AB, BC and AC produced to D, E and F respectively.

∠A + ∠B + ∠C = 180^o (i)

∠CBD = ∠C + ∠A (Exterior angle theorem) (ii)

∠ACE = ∠A + ∠B (Exterior angle theorem) (iii)

∠BAF = ∠B + ∠C (Exterior angle theorem) (iv)

Adding (ii), (iii) and (iv) we get

∠CBD + ∠ACE + ∠BAF = 2∠A + 2∠B + 2∠C

∠CBD + ∠ACE + ∠BAF = 2 (∠A + ∠B + ∠C)

∠CBD + ∠ACE + ∠BAF = 2 * 180^o

∠CBD + ∠ACE + ∠BAF = 360^o

Thus, sum of all three exterior angles is 360^o.

Given that,

BC produced on both sides

We know that,

∠A + ∠ABC + ∠ACB = 180^o (i)

∠ABD = ∠A + ∠ACB (Exterior angle theorem) (ii)

∠ACE = ∠A + ∠ABC (Exterior angle theorem) (iii)

Adding (ii) and (iii), we get

∠ABD + ∠ACE = ∠A + (∠A + ∠ACB + ∠ACB)

∠ABD + ∠ACE = ∠A + 180^o

(∠ABD + ∠ACE) - ∠A = 180^o

Thus, between the sum of the exterior angles so formed and ∠A is 180^o.

Given,

AD perpendicular to BC

∠A = 100^o

In ,

∠ADB + ∠B + ∠DAC = 180^o

90^o + ∠B + ∠A = 180^o

∠B + * 100^o = 180^o – 90^o

∠B + 50^o = 90^o

∠B = 40^o

Given,

AB = AC and,

BD = BC

∠2 = ∠3 (Since, AB = AC)

∠4 = ∠5 (Since, BD = BC)

= (i)

In

∠2 = ∠4 + ∠5

∠2 = 2∠4 (Since, ∠4 = ∠5)

∠3 = 2∠4 (Since, ∠3 = ∠2)

=

=

Thus, ∠ACD: ∠ADC = 3: 1

Let,

∠1, ∠2 and ∠3 be the angles of a triangle.

∠1 + ∠2 = ∠3 (Given) (i)

We know that,

∠1 + ∠2 + ∠3 = 180^o

∠3 + ∠3 = 180^o [From (i)]

2∠3 = 180^o

∠3 = 90^o

Thus, third angle is 90^o.

Let ∠1, ∠2 and ∠3 be the angles of the triangle and ∠4 be its exterior angle.

∠4 = 108⁰ (Given)

∠1: ∠2 = 4: 5 (Given)

Let, ∠1 = 4k

∠2 = 5k

Now,

∠1 + ∠2 = 108^o (Exterior angle theorem)

4k + 5k = 108^o

9k = 108^o

k = 12^o

Thus,

∠1 = 4 * 12 = 48^o

∠2 = 5 * 12 = 60^o

We know that,

∠1 + ∠2 + ∠3 = 180^o

48^o + 60^o + ∠3 = 180^o

108^o + ∠3 = 180^o

∠3 = 180^o – 108^o

= 72^o

Thus, angles of triangle are 48^o, 60^o, 72^o.

Given,

AB ‖ CD and,

EF ‖ BC

∠BAC = 65^o and ∠DHF = 35^o

∠BAC = ∠ACD (Alternate angles)

∠ACD = 65^o

∠DHF = ∠GHC (Vertically opposite angles)

∠GHC = 35^o

In

∠GCH + ∠GHC + ∠HGC = 180^o

65^o + 35^o + ∠HGC = 180^o

∠HGC = 80^o

∠AGH + ∠HGC = 180^o (Linear pair)

∠AGH + 80^o = 180^o

∠AGH = 100^o

In

∠A + ∠B + ∠C = 180^o

60^o + ∠B + ∠C = 180^o

∠B + ∠C = 120^o

∠B + ∠C = 60^o (i)

∠BOC + ∠OBC + ∠OCB = 180^o

∠BOC + ∠B + ∠C = 180^o

∠BOC + (∠B + ∠C) = 180^o

∠BOC + 60^o = 180^o [From (i)]

∠BOC = 120^o

Given,

AB || DE and,

BD || FG

∠FGH + ∠FGE = 180^o (Linear pair)

125^o + y = 180^o

y = 55^o

∠ABC = ∠BDE (Alternate angles)

∠BDF = ∠EFG = 55^o (Alternate angles)

∠EFG + ∠FEG = 125^o (By exterior angle theorem)

55^o + ∠FEG = 125^o

∠FEG = x = 70^o

Thus, x = 70^o and y = 55^o.

Let ABC is an acute angled triangle.

∠B = 90^o

We know that,

∠A + ∠B + ∠C = 180^o

∠A + 90^o + ∠C = 180^o

∠A + ∠C = 90^o (i)

In

∠AOC + ∠ACD + ∠CAD = 180^o

∠AOC + ∠C + ∠A = 180^o

∠AOC + (∠A + ∠C) = 180^o

∠AOC + * 90^o = 180^o [From (i)]

∠AOC + 45^o = 180^o

∠AOC = 180^o – 45^o

= 135^o

Thus, the angle at O between two bisectors is equal to 135^o.

AC ‖ BD

∠CAD = 45^o

∠CDB = 55^o

∠2 = ∠CAD (Alternate angle)

∠2 = 45^o

In

∠BOD + ∠2 + ∠CDB = 180^o

∠BOD + 45^o + 55^o = 180^o

∠BOD + 100^o = 180^o

∠BOD = 180^o – 100^o

= 80^o

By exterior angle theorem,

∠ACD = ∠A + ∠B

∠ACD = 68^o + ∠B

∠ACD = 34^o + ∠B

34^o = ∠ACD - ∠EBC (i)

Now,

In

∠ECD = ∠EBC + ∠E

∠E = ∠ECD - ∠EBC

∠E = ∠ACD - ∠EBC (ii)

From (i) and (ii), we get

∠E = 34^o

Let,

∠1 = 2k, ∠2 = k and ∠3 = 3k

∠1 + ∠2 + ∠3 = 180°

6k = 180

k = 30^o

Therefore, minimum angle be ∠2 = k = 30^o.

∠OBC = 180^o - ∠B - (180^o - ∠B)

∠OBC = 90^o - ∠B

And,

∠OCB = 180^o - ∠C - (180^o - ∠C)

∠OCB = 90^o - ∠C

In

∠BOC + ∠OCB + ∠OBC = 180^o

∠BOC + 90^o - ∠C + 90^o - ∠B = 180^o

∠BOC = (∠B + ∠C)

∠BOC = (180^o - ∠A) [From ]

∠BOC = 90^o - ∠A

∠BOC = 90^o -

In Δ ABC

∠A + ∠B + ∠C = 180^o

50^o + ∠B + ∠C = 180^o

∠B + ∠C = 180^o – 50^o

∠B + ∠C = 10^o (i)

In

∠E + ∠BCE + ∠EBC = 180^o

∠E + 180^o – ( ∠ACD) + ∠B = 180^o (ii)

By exterior angle theorem,

∠ACD = 50^o + ∠B

Putting value of ∠ACD in (ii), we get

∠E + 180^o - (50^o + ∠B) + ∠B = 180^o

∠E – 25^o - ∠B + ∠B = 0

∠E – 25^o = 0

∠E = 25^o

Given,

∠ABC = 30^o

∠ACD = 115^o

By exterior angle theorem,

∠ACD = ∠A + ∠B

115^o = ∠A + 30^o

∠A = 85^o

∠ACD + ∠ACL = 180^o (Linear pair)

∠ACL = 65^o

In

∠ALC + ∠LAC + ∠ACL = 180^o

∠ALC + ∠A + 65^o = 180^o

∠ALC = 72.5^o

Given,

EC ‖ AB

∠ECD = 70^o

∠BDO = 20^o

Since,

EC ‖ AB

And, OC cuts them so

∠ECD = ∠1 (Alternate angle)

∠1 = 70^o

∠1 + ∠3 = 180^o (Linear pair)

∠3 = 110^o

In

∠BOD + ∠OBD + ∠BDO = 180^o

∠3 + ∠ODB + 20^o = 180^o

∠ODB = 50^o

By exterior angle theorem,

In

∠OCA + ∠AOC = x

x = 80^o + 40^o

= 120^o

∠AOC = ∠DOB (Vertically opposite angle)

∠DOB = 40^o

By exterior angle theorem,

In

y = ∠BOD + ∠ODB

= 40^o + 70^o

= 110^o

Now, x + y = 230^o

Let,

∠1, ∠2 and ∠3 be the angles of the triangle which are in the ratio 3: 4: 5 respectively.

∠1 = 3k

∠2 = 4k

∠3 = 5k

We know that,

∠1 + ∠2 + ∠3 = 180^o

3k + 4k + 5k = 180^o

k = 15^o

So,

∠1 = 3 * 15^o = 45^o

∠2 = 4 * 15^o = 60^o

∠3 = 5 * 15^o = 75^o

Thus, smallest angle is 45^o.

Given,

AB is perpendicular to BC so ∠B = 90^o

∠CED = 32^o (Vertically opposite angles)

In

∠BDE + ∠BED + ∠DBE = 180^o

x + 14^o + 32^o + x + 90^o = 180^o

2x = 44^o

x = 22^o

In given that,

x = ∠A + ∠B (Exterior angles)

z = ∠A (Vertically opposite angles)

y = ∠A + ∠C (Exterior angles)

We know that,

∠A + ∠B + ∠C= 180^o

z + x - ∠A + y - ∠A = 180^o

-z = 180^o – x – y

z = x + y – 180^o

Since,

l₁‖ l₂

And,

AB cuts them so,

∠DBA = ∠BAE = 78^o

∠BAC + 35^o = 78^o

∠BAC = 43^o

In

∠BAC + ∠ABC + ∠ACB = 180^o

43^o + x + 90^o = 180^o

x = 47^o

In

x + 2x + ∠ACB = 180^o

∠ACB = 180^o – 3x (i)

In

y + 180^o – 3y + ∠ECD = 180^o

y + 180^o – 3y + 180^o - ∠ACB = 180^o

y = x

Since,

l₁‖ l₂

And PQ cuts them

∠DPQ + ∠PQE = 180^o (Consecutive interior angles)

a + a + b + b = 180^o

2 (a + b) = 180^o

a + b = 90^o (i)

In

∠PAQ + a + b = 180^o

∠PAQ = 90^o

∠PAQ + x + x = 180^o (Linear pair)

90^o + 2x = 180^o

x = 45^o

In

∠ABC + ∠ACB + ∠CAB = 180^o

5y + 3y + x = 180^o

8y + x = 180^o (i)

∠ABC + ∠CBD = 180^o (Linear pair)

5y + 7y = 180^o

y = 15^o

Putting values of y in (i), we get

8 * 15 + x^o = 180^o

x = 60^o

In

∠ROT + ∠RTO + ∠TRO = 180^o

140^o+ b + a = 180^o

a + b = 40^o (i)

In

∠RST + ∠SRT + ∠STR = 180^o

x + a + a + b + b = 180^o

x + 2 (a + b) = 180^o

x + 80^o = 180^o

x = 100^o

In

∠A + ∠ABD + ∠BDA = 180^o

∠ABD = 100^o

In

∠EBC + ∠ECB + ∠CEB = 180^o

-100^o + 40^o + ∠CEB = 0⁰

∠CEB = 60^o

∠CEB + ∠CED = 180^o (Linear pair)

60^o + x = 180^o

x = 120^o

Given,

BP CQ

And,

AC ‖ BC

∠A = ∠ABC (Since, AC = BC)

In

∠A + ∠B + ∠C = 180^o

∠A + ∠A + ∠C = 180^o

2∠A + ∠C = 180^o (i)

∠ACB + ∠ACQ + ∠QCD = 180^o (Linear pair)

∠ACB + x = 110^o (ii)

∠PBC + ∠BCQ = 180^o (Co. interior angle)

20^o + ∠A + ∠ACB + x = 180^o

∠A = 50^o (iii)

Using (iii) in (i), we get

2 * 50^o + ∠ACB = 180^o

∠ACB = 80^o

Using value of ∠ACB in (ii)l we get

80^o + x = 110^o

x = 30^o

Given,

AB ‖ CD

And, EF cuts them

So, 30^o + 65^o + ∠PQR = 180^o

95^o + ∠PQR = 180^o

∠PQR = 85^o

∠APQ + ∠PQC = 180^o (Co. interior angle)

25^o + y + 85^o + 30^o = 180^o

y = 40^o

In

∠PQR + ∠PRQ + ∠QPR = 180^o

85^o + x + y = 180^o

x = 55^o

Thus, x = 55^o and y = 40^o

Given,

∠ABD = 94^o and

∠ACE = 126^o

∠ABD + ∠ABC = 180^o (Linear pair)

∠ABC = 86^o (i)

∠ACE + ∠ACB = 180^o (Linear pair)

∠ACB = 54^o (ii)

In

∠ABC + ∠ACB + ∠BAC = 180^o

86^o + 54^o + ∠BAC = 180^o

∠BAC = 40^o