Angles, Lines And Triangles

(i) Angle – A shape formed by two lines or rays diverging from a common vertex.

Types of angle: (a) Acute angle (less than 90°)

(b) Right angle (exactly 90°)

(c) Obtuse angle (between 90° and 180°)

(d) Straight angle (exactly 180°)

(e) Reflex angle (between 180° and 360°)

(f) Full angle (exactly 360°)

(ii) Interior of an angle – The area between the rays that make up an angle and extending away from the vertex to infinity.

The interior angles of a triangle always add up to 180°.

(iii) Obtuse angle – It is an angle that measures between 90 to 180 degrees.

(iv) Reflex angle – It is an angle that measures between 180 to 360 degrees.

(v) Complementary angles – Two angles are called complementary angles if the sum of two angles is 90°.

(vi) Supplementary angles – Angles are said to be supplementary if the sum of two angles is 180°.

65°11’25’

+ = 36°27’46’’ + 28°43’39’’

= 64°70’85’’

∵ 60’ = 1° ⇒ 70’ = 1°10’

60’’ = 1’ ⇒ 85’’ = 1’ 25’’

∵ + = 65°11’25’’

11°31’30’’

36° - 24°28’30’’ = 35°59’60’’ - 24°28’30’’

= 11°31’30’’

(i) 32°

Complement of angle = 90° – θ

Complement of 58° = 90° - 58°

= 32°

(ii) 74°

Complement of angle = 90° – θ

Complement of 58° = 90° - 16°

= 74°

(iii) 30°

Right angle = 90°

of a right angle = × 90°

= 60°

Complement of 60° = 90° - 60°

= 30°

(iv) 43°30’

Complement of angle = 90° – θ

Complement of 46°30’ = 90° - 46°30’

= 89°60’ - 46°30’

(v) 37°16’40’’

Complement of angle = 90° – θ

Complement of 52°43’20’’ = 90° - 52°43’20’’

= 89°59’60’’ - 52°43’20’’

= 37°16’40’’

(vi) 21°24’15’’

Complement of angle = 90° – θ

Complement of 68°35’45’’ = 90° - 68°35’45’’

= 89°59’60’’ - 68°35’45’’

= 68°35’45’’

(i) 117°

Supplement of angle = 180° – θ

Supplement of 58° = 180° - 63°

= 117°

(ii) 42°

Supplement of angle = 180° – θ

Supplement of 58° = 180° - 138°

= 42°

(iii) 126°

Right angle = 90°

of a right angle = × 90°

= 54°

Supplement of 54° = 180° - 54°

= 126°

(iv) 104°24’

Supplement of angle = 180° – θ

Supplement of 75°36’ = 180° - 75°36’

= 179°60’ - 75°36’

= 104°24’

(v) 55°39’20’’

Supplement of angle = 180° – θ

Supplement of 124°20’40’ = 180° - 124°20’40’’

= 179°59’60” - 124°20’40’’

= 55°39’20’’

(vi) 71°11’28’’

Supplement of angle = 180° – θ

Supplement of 108°48’32’’ = 180° - 108°48’32’’

= 179°59’60” - 108°48’32’’

= 71°11’28’’

(i) 45°

Let, measure of an angle = X

Complement of X = 90° – X

Hence,

⇒ X = 90° – X

⇒ 2X = 90°

⇒ X = 45°

Therefore measure of an angle = 45°

(ii) 90°

Let, measure of an angle = X

Supplement of X = 180° – X

Hence,

⇒ X = 180° – X

⇒ 2X = 180°

⇒ X = 90°

Therefore measure of an angle = 90°

63°

Let, measure of an angle = X

Complement of X = 90° – X

According to question,

⇒ X = (90° – X) + 36°

⇒ X + X = 90° + 36°

⇒ 2X = 126°

⇒ X = 63°

Therefore measure of an angle = 63°

(77.5)°

Let, measure of an angle = X

Supplement of X = 180° – X

According to question,

⇒ X = (180° – X) - 25°

⇒ X + X = 180° - 25°

⇒ 2X = 155°

⇒ X = (77.5)°

Therefore measure of an angle = (77.5)°

72°

Let the angle = X

Complement of X = 90° – X

According to question,

⇒ X = 4(90° – X)

⇒ X = 360° - 4X

⇒ X + 4X = 360°

⇒ 5X = 360°

⇒ X = 72°

Therefore angle = 72°

150°

Let the angle = X

Supplement of X = 180° – X

According to question,

⇒ X = 5(180° – X)

⇒ X = 900° - 4X

⇒ X + 5X = 900°

⇒ 6X = 900°

⇒ X = 150°

Therefore angle = 150°

60°

Let the angle = X

Complement of X = 90° – X

Supplement of X = 180° – X

According to question,

⇒ 180° – X = 4(90° – X)

⇒ 180° – X = 360° – 4X

⇒ – X + 4X = 360° – 180°

⇒ 3X = 180°

⇒ X = 60°

Therefore angle = 60°

180°

Let the angle = X

Complement of X = 90° – X

Supplement of X = 180° – X

According to question,

⇒ 90° – X = 4(180° – X)

⇒ 180° – X = 720° – 4X

⇒ – X + 4X = 720° – 180°

⇒ 3X = 540°

⇒ X = 180°

Therefore angle = 180°

108°, 72°

Let angle = X

Supplementary of X = 180° – X

According to question,

X : 180° – X = 3 : 2

⇒ X / (180° – X) = 3 / 2

⇒ 2X = 3(180° – X)

⇒ 2X = 540° – 3X

⇒ 2X + 3X = 540°

⇒ 5X = 540°

⇒ X = 108°

Therefore angle = 108°

And its supplement = 180° – 108° = 72°

40°, 50°

Let angle = X

Complementary of X = 90° – X

According to question,

X : 90° – X = 4 : 5

⇒ X / (90° – X) = 4 / 5

⇒ 5X = 4(90° – X)

⇒ 5X = 360° – 4X

⇒ 5X + 4X = 360°

⇒ 9X = 360°

⇒ X = 40°

Therefore angle = 40°

And its supplement = 90° – 40° = 50°

25°

Let the measure of an angle = X

Complement of X = 90° – X

Supplement of X = 180° – X

According to question,

⇒ 7(90° – X) = 3(180° – X) - 10°

⇒ 630° – 7X = 540° – 3X - 10°

⇒ – 7X + 3X = 540° – 10° – 630°

⇒ - 4X = 100°

⇒ X = 25°

Therefore measure of an angle = 25°

118°

AOB is a straight line

Therefore, ∠AOB = 180°

⇒ ∠AOC + ∠BOC = 180°

⇒ 62° + x = 180°

⇒ x = 180° – 62°

= 118°

X=27.5, =77.5°=47.5°

AOB is a straight line

Therefore, + ∠COD + = 180°

⇒ (3x - 5)° + 55° + (x + 20)° = 180°

⇒ 3x - 5° + 55° + x + 20° = 180°

⇒ 4x = 180° - 70°

⇒ 4x = 110°

⇒ x = 27.5°

= (3x - 5)°

= 3×27.5 – 5 = 77.5°

= (x + 20)°

= 27.5 + 20 = 47.5°

X=32, =103°, ∠COD =45°=32°

AOB is a straight line

Therefore, + ∠COD + = 180°

⇒ (3x + 7)° + (2x - 19)° + x° = 180°

⇒ 3x + 7° + 2x - 19° + x° = 180°

⇒ 6x = 180° + 12°

⇒ 6x = 192°

⇒ x = 32°

= (3x + 7)°

= 3×32° + 7 = 103°

∠COD = (2x - 19)°

= 2×32° – 19 = 45°

= x

= 32°

X=60, Y=48, Z=72

AOB is a straight line

Therefore, ∠XOP + ∠POQ + ∠YOQ = 180°

Given, x: y: z =5: 4: 6

Let ∠XOP = x° = 5a, ∠POQ = y° = 4a, ∠YOQ = z° = 6a

⇒ 5a + 4a + 6a = 180°

⇒ 15a = 180°

⇒ a = 12°

Therefore,

x = 5a = 5×12° = 60°

y = 4a = 4×12° = 48°

z = 6a = 6×12° = 72°

X=28°

AOB is a straight line

Therefore, ∠AOB = 180°

⇒ (3x +20)° + (4x -36)° = 180°

⇒ 3x + 20° + 4x - 36° = 180°

⇒ 7x - 16° = 180°

⇒ 7x = 196°

⇒ x = 28°

=130°,=50°, =130°

Given AB and CD intersect a O

Therefore, = _____________________ (i)

And ∠BOC = ∠AOD _____________________ (ii)

=50°

Therefore, =50° from equation (i)

AOB is a straight line,

⇒ + = 180°

⇒ 50° + = 180°

⇒ = 180° - 50°

⇒ = 130°

∠AOD = ∠BOC = 130° from equation (ii)

X=4, Y=4, Z=50, t=90

Given, coplanar lines AB, CD and EF intersect at a point O.

Therefore, ∠AOF = ∠BOE ________________________ (i)

∠BOD = ∠AOC ________________________ (ii)

∠DOF = ∠COE ________________________ (iii)

x = y from equation (i)

t = 90 from equation (ii)

z = 50 from equation (iii)

∠AOF + ∠DOF + ∠BOD = 180° (from AOB straight line)

⇒ x + 50° + 90° = 180°

⇒ x = 180° - 140°

⇒ x = 40°

x = y = 40° from equation (i)

)

∠AOD + ∠DOF + ∠BOF + ∠BOC + ∠COE + ∠AOE = 360°

⇒ 2x + 5x + 3x + 2x + 5x + 3x = 360°

⇒ 20x = 360°

⇒ x = 18°

∠AOD = 2x = 2× 18° = 36°

∠COE = 3x = 3× 18° = 54°

∠AOE = 4x = 4× 18° = 72°

100°, 80°

Explanation:

EOF is a straight line and its adjacent angles are ∠EOB and ∠FOB.

Let ∠EOB = 5a, and ∠FOB = 4a

∠EOB + ∠FOB = 180° (EOF is a straight line)

⇒5a + 4a = 180°

⇒9a = 180°

⇒ a = 20°

Therefore, ∠EOB = 5a

= 5 × 20° = 100°

And ∠FOB = 4a

= 4 × 20° = 80°

Proof

Given lines AB and CD intersect each other at point O and ∠AOC = 90°

∠AOC = ∠BOD (Opposite angles)

Therefore, ∠BOD = 90°

⇒ ∠BOD + ∠AOC = 180°

⇒ ∠BOC + 90° = 180°

⇒ ∠BOC = 90°

Now, ∠AOD = ∠BOC (Opposite angles)

Therefore,

∠AOD = 90°

Proved each of the remaining angles measures 90°.

=140°,=40°, = 140°, ∠BOD = 40°

Given lines AB and Cd intersect at a point O and + =280°

= (Opposite angle)

⇒ + = 280°

⇒ + = 280°

⇒ 2 = 280°

⇒ = 140°

= = 140°

Now,

+ = 180° (Because AOB is a straight line)

⇒ + 140° = 180°

⇒ = 40°

= ∠BOD = 40°

Proof

Given OC is the bisector of

Therefore, ∠AOC = ∠COB ______________________ (i)

DOC is a straight line,

+= 180° _______________ (ii)

Similarly, += 180° ________________ (iii)

From equations (i) and (ii)

⇒ += +

⇒+ = + (from equation (i))

⇒ = Proved

34°

Angle of incidence =angle of reflection.

Therefore, ∠PQA = ∠BQR _____________________ (i)

⇒ ∠BQR + ∠PQR + ∠PQA = 180°[Because AQB is a straight line]

⇒ ∠BQR + 112° + ∠PQA = 180°

⇒ ∠BQR + ∠PQA = 180° - 112°

⇒ ∠PQA + ∠PQA = 68° [from equation (i)]

⇒ 2 ∠PQA = 68°

⇒ ∠PQA = 34°

Given, lines AB and CD intersect each other at point O.

OE is the bisector of ∠ BOD.

TO prove: OF bisects ∠AOC.

Proof:

AB and CD intersect each other at point O.

Therefore, ∠ AOC = ∠ BOD

∠ 1 = ∠ 2 [OE is the bisector of ∠BOD] _________________ (i)

∠ 1 = ∠ 3 and ∠ 2 = ∠ 4 [Opposite angles] ____________ (ii)

From equations (i) and (ii)

∠ 3 = ∠ 4

Hence, OF is the bisector of ∠AOC.

Given, ∠AOC and ∠BOC are supplementary angles

OE is the bisector of ∠BOC and

OD is the bisector of ∠AOC

Therefore, ∠1 = ∠2 and ∠3 = ∠ 4 ________________ (i)

∠BOC + ∠AOC = 180° [Because AOB is a straight line]

⇒ ∠1 + ∠2 + ∠3 + ∠4 = 180°

⇒ ∠1 + ∠1 + ∠3 + ∠3 = 180° [From equation (i)]

⇒ 2 ∠1 +2 ∠3 = 180°

⇒ 2(∠1 + ∠3) = 180°

⇒ ∠1 + ∠3 = 90°

Hence, ∠EOD = 90° proved.

=110°^,=70°,=110°, =70°,=110°,=70°, =110°

Given AB ||CD are cut by a transversal t at E and F respectively.

And ∠1 = 70°

∠1 = ∠3 = 70° [Opposite angles]

∠5 = ∠1 = 70° [Corresponding angles]

∠3 = ∠7 = 70° [Corresponding angles]

∠1 + ∠2 = 180° [Because AB is a straight line]

⇒ 70° + ∠2 = 180°

⇒ ∠2 = 110°

∠4 = ∠2 = 110° [Opposite angles]

∠6 = ∠2 = 110° [Corresponding angles]

∠8 = ∠4 = 110° [Corresponding angles]

=100°, =80°, =100°=80°, =100°, =80°, =100°, =80°

Given AB ||CD are cut by a transversal t at E and F respectively.

And = 5:4

Let ∠1 = 5a and ∠2 = 4a

∠1 + ∠2 = 180° [Because AB is a straight line]

⇒ 5a + 4a = 180°

⇒ 9a = 180°

⇒ a = 20°

Therefore, ∠1 = 5a

∠1 = 5 × 20° = 100°

∠2 = 4a

∠2 = 4 × 20° = 80°

∠3 = ∠1 = 100° [Opposite angles]

∠4 = ∠2 = 80° [Opposite angles]

∠5 = ∠1 = 100° [Crossponding angles]

∠6 = ∠4 = 80° [Crossponding angles]

∠7 = ∠5 = 100° [Opposite angles]

∠8 = ∠6 = 80° [Opposite angles]

Given AB||DC and AD||BC

Therefore, +=180° _____________ (i)

+=180° _____________ (ii)

From equations (i) and (ii)

+ = +

= Proved.

(i) x = 100

Given AB||CD, ∠ABE = 35° and ∠EDC = 65°

Draw a line PEQ||AB or CD

∠1 = ∠ABE = 35°[AB||PQ and alternate angle] _______________ (i)

∠2 = ∠EDC = 65°[CD||PQ and alternate angle] _______________ (ii)

From equations (i) and (ii)

∠1 + ∠2 = 100°

⇒ x = 100°

(ii) x=280

Given AB||CD, ∠ABE = 35° and ∠EDC = 65°

Draw a line POQ||AB or CD

∠1 = ∠ABO = 55°[AB||PQ and alternate angle] _______________ (i)

∠2 = ∠CDO = 25°[CD||PQ and alternate angle] _______________ (ii)

From equations (i) and (ii)

∠1 + ∠2 = 80°

Now,

∠BOD + ∠DOB = 360°

⇒ 80° + x° = 360°

⇒ x = 280°

(iii) x=120

Given AB||CD, ∠BAE = 116° and ∠DCE = 124°

Draw a line EF||AB or CD

∠BAE + ∠PAE = 180° [Because PAB is a straight line]

⇒ 116° + ∠3 = 180°

⇒ ∠3 = 180° - 116°

⇒ ∠3 = 64°

Therefore,

∠1 = ∠3 = 64° [Alternate angles] ____________________ (i)

Similarly, ∠4 = 180° - 124°

∠4 = 56°

Therefore,

∠2 = ∠4 = 56° [Alternate angles] ____________________ (ii)

From equations (i) and (ii)

⇒ ∠1 + ∠2 = 64° + 56°

⇒ x = 120°

X=20

Given AB||CD||EF, ∠ABC = 70° and ∠CEF = 130°

AB||CD

Therefore,

∠ABC = ∠BCD = 70° [Alternate angles] ________________ (i)

EF||CD

Therefore,

∠DCE + ∠CEF = 180°

⇒ ∠DCE + 130° = 180°

⇒ ∠DCE = 50°

Now,

∠BCE + ∠DCE = ∠BCD

⇒ x + 50° = 70°

⇒ x = 20°

X=110

Given AB||CD, ∠DCE = 130° and ∠AEC = 20°

Draw a line EF||AB||CD

CD||EF

Therefore, ∠DCE + ∠CEF = 180°

⇒ 130° + ∠1 = 180°

⇒ ∠1 = 180° - 130°

⇒ ∠1 = 50°

AB||EF

Therefore, ∠BAE + ∠AEF = 180°

⇒ x + ∠1 + 20° = 180°

⇒ x + 50° + 20° = 180°

⇒ x = 180° - 70°

⇒ x = 110°

Draw a line EF||AB||CD.

+=180° [Because AB||EF and AE is the transversal] __________________ (i)

+=180° [Because DC||EF and CE is the transversal] __________________ (ii)

From equations (i) and (ii)

⇒ +=+

⇒ -=-

⇒ - = Proved.

X=105

Given AB||CD and BC||ED.

AB||CD

Therefore, ∠BCF = ∠EDC = 75° [Crossponding angles]

∠ABC + ∠BCF = 180° [Because AB||DCF]

⇒ x + 75° = 180°

⇒ x = 105°

Given AB||CD, ∠AEF = P°, ∠EFG = q°, ∠FGD = r°

Draw a line FH||AB||CD

∠HFG = ∠FGD = r° [Because HF||CD and alternate angles] ___________ (i)

∠EFH = ∠EFG - ∠HFG

⇒ ∠EFH = q – r ______________________ (i)

∠AEF + ∠EFH = 180° [Because AB||HF]

⇒ ∠AEF + ∠EFH = 180°

⇒ p + (q – r) = 180°

⇒ p + q – r = 180°Proved.

x=70, y=50

Given AB||PQ

∠GEF + 20° + 75° = 180°[Because EF is a straight line]

⇒ ∠GEF = 180° - 95°

⇒ ∠GEF = 85°________________ (i)

In triangle EFG,

⇒ X + 25° + 85° = 180° [∠GEF = 85°]

⇒ X = 60°

Now,

⇒ ∠BEF + ∠EFQ = 180°[Interior angles on same side of transversal]

⇒ (20° + 85°) + (25° + Y) = 180°

⇒ Y = 180° - 130°

⇒ Y = 50°

Given AB||CD

Therefore, ∠BAC + ∠ACD = 180°

⇒ 75° + ∠ACD = 180°

⇒ ∠ACD = 105°_________________ (i)

∠ACD = ∠ECF = 105°[Opposite angles]

In triangle CEF,

⇒ ∠CEF + ∠EFC + ∠FCE = 180°

⇒ x + 30° + 105° = 180°

⇒ x + 135° = 180°

⇒ x = 45°

x=20

Given AB||CD

Therefore,

= [Crossponding angles]

= 95° ___________________ (i)

In CD straight line,

⇒ +=180°

⇒ 115° + = 180°

⇒ = 65°

In triangle GHQ,

⇒ ∠QGH + ∠GHQ + ∠GQH = 180°

⇒ 95° + 65° + x = 180°

⇒ x = 20°

Z=75, x=35, y=70

Given AB||CD

Therefore,

X = 35°[Alternate angles]

In triangle AOB,

⇒ x + 75° + y = 180°

⇒ 35° + 75° + y = 180°

⇒ y = 70°

⇒ ∠COD = y = 70°[Opposite angles]

In triangle COD,

⇒ z + 35° + ∠COD = 180°

⇒ z + 35° + 70° = 180°

⇒ z = 75°

x=105, y=75, z=50

Given AB||CD

Therefore,

⇒ ∠AEF = ∠EFG = 75°[Alternate angles]

⇒ y = 75°

For CD straight line,

⇒ x + y = 180°

⇒ x + 75° = 180°

⇒ x = 105°

Again,

⇒ ∠EGF + 125° = 180°

⇒ ∠EGF = 155°

In triangle EFG,

⇒ y + z + ∠EGF = 180°

⇒ 75°+ z + 155°= 180°

⇒ z + 130°= 180°

⇒ z = 50°

X=60, y=60, z=70, t=70

Given AB||CD and EF||GH

x = 60° [Opposite angles]

y = x = 60°[Alternate angles]

∠PQS = ∠APR = 110°[Crossponding angles]

∠PQS = ∠PQR + y = 110°_____________ (i)

For AB straight line,

⇒ y + z + ∠PQR = 180°

⇒ z + 110° = 180°[From equation (i)]

⇒ z = 70°

AB||CD

Therefore,

t = z = 70°[Because alternate angles]

(i) x=30

Given l||m

Therefore,

3x – 20° = 2x + 10° [Crossponding angles]

⇒ 3x – 2x = 10° + 20°

⇒ x = 30°

(ii) x=25

Given l||m

Therefore,

(3x + 5)° + 4x° = 180°

⇒ 7x + 5° = 180°

⇒ 7x = 175°

⇒ x = 25°

AB⊥PQ,

Therefore, ∠ABD = 90° _____________ (i)

CD⊥PQ,

Therefore, ∠CDQ = 90° _____________ (ii)

From equations (i0 and (ii)

∠ABD = ∠CDQ = 90°

Hence, AB||CD because Cross ponding angles are equal.

=56°

In ΔABC,

∠A + ∠B + ∠C = 180° [Sum of angles]

⇒ ∠A + 76° + 48° = 180°

⇒ ∠A + 124° = 180°

⇒ ∠A = 56°

40°, 60°, 80°

Let the angles of triangle are 2a, 3a and 4a.

Therefore,

2a + 3a + 4a = 180°[Sum of angles]

⇒ 9a = 180°

⇒ a = 20°

Angles of triangle are,

2a = 2 × 20° = 40°

3a = 3 × 20° = 60°

4a = 4 × 20° = 80°

=80°, =60°, =40°

Let 3= 4=6 = a

Therefore,

∠A = a/3, ∠B = a/4, ∠C = a/6 ____________________ (i)

∠A + ∠B + ∠C = 180° [Sum of angles]

⇒ a/3 + a/4 + a/6 = 180°

⇒ 9a/12 = 180°

⇒ a = 240°

⇒ ∠A = a/3 = 240° /3 = 80°

⇒ ∠B = a/4 = 240° /4 = 60°

⇒ ∠C = a/6 = 240° /6 = 40°

=50°, =58°^,=72°

Given,

+= 108° ___________________ (i)

+ = 130° ___________________ (ii)

We know that sum of angles of triangle = 180°

∠A + ∠B + ∠C = 180° [Sum of angles]

∠A + 130°= 180° [From equation (ii)]

⇒ ∠A = 50°

Value of ∠A = 50° put in equation (i),

+ = 108°

⇒ 50° + = 108°

⇒ = 58°

Value of ∠B = 58° put in equation (ii),

+ = 130°

⇒ 58° + = 130°

⇒ = 72°

=67°, =41°, =89°

Given,

+= 125° ___________________ (i)

+ = 113° ___________________ (ii)

We know that sum of angles of triangle = 180°

∠A + ∠B + ∠C = 180° [Sum of angles]

∠A + 113°= 180° [From equation (ii)]

⇒ ∠A = 67°

Value of ∠A = 50° put in equation (i),

+ = 125°

⇒ 67° + = 108°

⇒ = 41°

Value of ∠B = 41° put in equation (ii),

+ = 130°

⇒ 41° + = 130°

⇒ = 89°

=95°,=53°,=32°

Given,

-= 42° ___________________ (i)

-=21° ___________________ (ii)

= 42° + [From equation (i)] __________________ (iii)

= - 21° [From equation (ii)] __________________ (iv)

We know that sum of angles of triangle = 180°

∠P + ∠Q + ∠R = 180° [Sum of angles]

⇒ 42° + ∠Q + ∠Q + ∠Q - 21° = 180° [From equation (iii) and (iv)]

⇒ 3 ∠Q + 21°= 180°

⇒ 3 ∠Q = 159°

⇒ ∠Q = 53°

Value of ∠Q = 53° put in equation (iii),

= 42° +

⇒ = 42° + 53°

⇒ = 95°

Value of ∠Q = 53° put in equation (iv),

= - 21°

⇒ = 53° - 21°

⇒ = 32°

70°^, 46°, 64°

Let ∠P, ∠Q and ∠R are three angles of triangle PQR.

Now,

∠P + ∠Q = 116° ___________________ (i)

∠P - ∠Q = 24° ___________________ (i)

Adding equation (i) and (ii),

2 ∠P = 140°

⇒ ∠P = 70° ______________________ (iii)

Subtracting equation (i) and (ii),

2 ∠Q = 92°

⇒ ∠Q = 46° _______________________ (iv)

We know that sum of angles of triangle = 180°

∠P + ∠Q + ∠R = 180° [Sum of angles]

⇒ 70° + 46° + ∠R = 180° [From equation (iii) and (iv)]

⇒ ∠R = 64°

54°, 54°, 72°

Let ∠P, ∠Q and ∠R are three angles of triangle PQR,

And ∠P = ∠Q = a ________________ (i)

Then, ∠R = a + 18°________________ (ii)

We know that sum of angles of triangle = 180°

∠P + ∠Q + ∠R = 180° [Sum of angles]

⇒ a + a + a + 18°= 180° [From equation (i) and (ii)]

⇒ 3a= 162°

⇒ a= 54°

Therefore,

∠P = ∠Q =54° [from equation (i)]

∠R = 54° + 18° [from equation (i)]

= 72°

60°, 90°, 30°

Let ∠P, ∠Q and ∠R are three angles of triangle PQR,

And ∠P is the smallest angle.

Now,

∠Q = 2 ∠P ________________ (i)

∠R = 3 ∠P ________________ (ii)

We know that sum of angles of triangle = 180°

∠P + ∠Q + ∠R = 180° [Sum of angles]

⇒ ∠P + 2 ∠P + 3 ∠P = 180° [From equation (i) and (ii)]

⇒ 6 ∠P = 180°

⇒ ∠P = 30°

Therefore,

⇒ ∠Q = 2 ∠P = 60° [from equation (i)]

⇒ ∠R = 3 ∠P = 90° [from equation (ii)]

53°, 37°, 90°

Let PQR be a right angle triangle.

Right angle at P, then

∠P = 90° and ∠Q = 53° ^{____________________________________} (i)

We know that sum of angles of triangle = 180°

∠P + ∠Q + ∠R = 180° [Sum of angles]

⇒ 90° + 53° + ∠R = 180° [From equation (i)]

⇒ ∠R = 37°

Proof

Let PQR be a right angle triangle,

Now,

∠P = ∠Q + ∠R __________________ (i)

We know that sum of angles of triangle = 180°

∠P + ∠Q + ∠R = 180° [Sum of angles]

⇒ ∠P + ∠P = 180° [From equation (i)]

⇒ 2 ∠P = 180°

⇒ ∠P = 90°

Hence, PQR is a right angle triangle Proved.

proof

We know that the sum of two acute angles of a right triangle is 90°.

Therefore,

+=90°

= 90°-

= 90°-____________________ (i)

+=90°

= 90°-____________________ (ii)

From equation (i) and (ii),

= Proved.

Proof

Let ABC be a triangle,

Now,

< +___________________ (i)

< +___________________ (ii)

< +___________________ (iii)

2 < ++ [From equation (i)]

2 < 180° [Sum of angles of triangle]

< 90° ________________ (a)

Similarly,

< 90°__________________ (b)

< 90°__________________ (c)

From equation (a), (b) and (c), each angle is less than 90°

Therefore triangle is an acute angled Proved.

Proof

Let ABC be a triangle,

Now,

> +___________________ (i)

> +___________________ (ii)

> +___________________ (iii)

2 > ++ [From equation (i)]

2 > 180° [Sum of angles of triangle]

> 90° ________________ (a)

Similarly,

> 90°__________________ (b)

> 90°__________________ (c)

From equation (a), (b) and (c), each angle is less than 90°

Therefore triangle is an acute angled Proved.

∠BAC = 85°, ∠ACB = 52°

Given, =128° and =43°

In triangle ABC,

∠ACB + ∠ACD = 180° [Because BCD is a straight line]

⇒ ∠ACB + 128° = 180°

⇒ ∠ACB = 52°

∠ABC + ∠ACB + ∠BAC = 180° [Sum of angles of triangle ABC]

⇒ 43° + 52° + ∠BAC = 180°

⇒ ∠BAC = 85°

74°, 62°, 44°

Given, =106° and =118°

∠ABD + ∠ABC = 180° [Because DC is a straight line]

⇒ 106° + ∠ABC = 180°

⇒ ∠ABC = 74°_______________ (i)

∠ACB + ∠ACE = 180° [Because BE is a straight line]

⇒ ∠ACB + 118° = 180°

⇒ ∠ACB = 62°_______________ (ii)

Now, triangle ABC

∠ABC + ∠ACB + ∠BAC = 180° [Sum of angles of triangle]

⇒ 74° + 62° + ∠BAC = 180° [From equation (i) and (ii)]

⇒ ∠BAC = 44°

(i) 50°

Given, ∠BAE = 110° and ∠ACD = 120°

∠ACB + ∠ACD = 180° [Because BD is a straight line]

⇒ ∠ACB + 120° = 180°

⇒ ∠ACB = 60°_______________ (i)

In triangle ABC,

∠BAE = ∠ABC + ∠ACB

⇒ 110° = x + 60°

⇒ x = 50°

(ii) 120°

In triangle ABC,

∠A + ∠B + ∠C = 180° [Sum of angles of triangle ABC]

⇒ 30° + 40° + ∠C = 180°

⇒ ∠C = 110°

∠BCA + ∠DCA = 180° [Because BD is a straight line]

⇒ 110° + ∠DCA = 180°

⇒ ∠DCA = 70°_________________ (i)

In triangle ECD,

∠AED = ∠ECD + ∠EDC

⇒ x = 70°+ 50°

⇒ x = 120°

(iii) 55°

Explanation:

∠BAC = ∠EAF = 60°[Opposite angles]

In triangle ABC,

∠ABC + ∠BAC = ∠ACD

⇒ X°+ 60°= 115°

⇒ X°= 55°

(iv) 75°

Given AB||CD

Therefore,

∠BAD = ∠EDC = 60°[Alternate angles]

In triangle CED,

∠C + ∠D + ∠E = 180°[Sum of angles of triangle]

⇒ 45° + 60° + x = 180°[∠EDC = 60°]

⇒ x = 75°

(v) 30°

Explanation:

In triangle ABC,

∠BAC + ∠BCA + ∠ABC = 180°[Sum of angles of triangle]

⇒ 40° + 90° + ∠ABC = 180°

⇒ ∠ABC = 50°________________ (i)

In triangle BDE,

∠BDE + ∠BED + ∠EBD = 180°[Sum of angles of triangle]

⇒ x° + 100° + 50° = 180°[∠EBD = ∠ABC = 50°]

⇒ x° = 30°

(vi) x=30

Explanation:

In triangle ABE,

∠BAE + ∠BEA + ∠ABE = 180°[Sum of angles of triangle]

⇒ 75° + ∠BEA + 65° = 180°

⇒ ∠BEA = 40°

∠BEA = ∠CED = 40°[Opposite angles]

In triangle CDE,

∠CDE + ∠CED + ∠ECD = 180°[Sum of angles of triangle]

⇒ x° + 40° + 110° = 180°

⇒ x° = 30°

x=130

Explanation:

In triangle ACD,

∠3 = ∠1 + ∠C __________________ (i)

In triangle ABD,

∠4 = ∠2 + ∠B __________________ (ii)

Adding equation (i) and (ii),

∠3 + ∠4 = ∠1 + ∠C + ∠2 + ∠B

⇒ ∠BDC = (∠1 + ∠2) + ∠C + ∠B

⇒ x°= 55°+ 30°+ 45°

⇒ x°= 130°

X=90

Explanation:

∠BAC + ∠CAE = 180°[Because BE is a straight line]

⇒ ∠BAC + 108° = 180°

⇒ ∠BAC = 72°

Now,
AD = DB

=

∠BAD = ( �)72°= 18°

∠DAC = ( �)72°= 54°

In triangle ABC,

∠A + ∠B + ∠C = 180°[Sum of angles of triangle]

⇒ 72° + 18° + x = 180°

⇒ x = 90°

Proof

In triangle ABC,

=+ ________________ (i)

=+________________ (ii)

=+________________ (iii)

Adding equation (i), (ii) and (iii),

++= 2(++)

⇒ ++= 2(180°) [Sum of angles of triangle]

⇒ ++= 360°Proved.

Proof

In triangle BDF,

++ = 180°[Sum of angles of triangle] _______________ (i)

In triangle BDF,

++ = 180°[Sum of angles of triangle] _______________ (ii)

From equation (i) and (ii),

(++) + (++) = (180°+180°)

⇒ +++ ++ = 360°Proved.

125°

Given, bisector of and meet at O.

If OB and OC are the bisector of and meet at point O .

Then,

= 90°+

⇒= 90°+ 70°

⇒= 125°

70°

Given, bisector of and meet at O.

If OB and OC are the bisector of and meet at point O .

Then,

= 90°-

⇒= 90°- 40°

⇒= 70°

60°

Given, ::= 3:2:1 and AC CD

Let, ∠A = 3a

∠B = 2a

∠C = a

In triangle ABC,

∠A + ∠B + ∠C = 180°[Sum of angles of triangle]

⇒ 3a + 2a + a = 180°

⇒ 6a = 180°

⇒ a = 30°

Therefore, ∠C = a = 30°

Now,

∠ACB + ∠ACD + ∠ECD = 180°[Sum of angles of triangle]

⇒ 30° + 90° + ∠ECD = 180°

⇒ ∠ECD = 60°

17.5°

Given, AMBC and “AN” is the bisector of.

Therefore,

= (-)

⇒ = (65° -30°)

⇒ = 17.5°

(i) False

Because, sum of angles of triangle equal to 180°.In a triangle maximum one right angle.

(ii) True

Because, obtuse angle measures in 90° to 180° and we know that the sum of angles of triangle is equal to 180°.

(iii) False

Because, in an obtuse triangle is one with one obtuse angle and two acute angles.

(iv) False

If each angles of triangle is less than 180° then sum of angles of triangle are not equal to 180°.

Any triangle,

∠1 + ∠2 + ∠3 = 180°

(v) True

If value of angles of triangle is same then the each value is equal to 60°.

∠1 + ∠2 + ∠3 = 180°

⇒ ∠1 + ∠1 + ∠1 = 180°[∠1 = ∠2 = ∠3]

⇒ 3 ∠1 = 180°

⇒ ∠1 = 60°

(vi) True

We know that sum of angles of triangle is equal to 180°.

Sum of angles,

= 10° + 80° + 100°

= 190°

Therefore, angles measure in (10°, 80°, 100°) cannot be a triangle.

If two angles are complements of each other, then each angle is an acute angle

An angle which measures more than 180^o but less than 360^o, is called a reflex angle.

As we know that sum of two complementary – angles is 90^o.

So, x + y = 90^o

72°40’ + y= 90

y = 90^o – 72°40’

y = 17^o20’

As we know that sum of two supplementary – angles is 180^o.

So, x + y = 180^o

54°30’ + y= 180

y = 180^o – 54°30’

y = 125^o30’

As we know that sum of two complementary – angles is 90^o.

So, x + y = 90^o

According to question y =5x

x + 5x= 90

6x = 90^o

x = 15^o

y = 75^o

As we know that sum of two complementary – angles is 90^o.

So, x + y = 90^o

Let x be the common multiple.

According to question angles would be 2x and 3x.

2x + 3x= 90

5x = 90^o

x = 18^o

2x = 36^o

3x = 54^o

So, larger angle is 54^o

As we know that sum of adjacent angle on a straight line is 180^o.

∠BOD + ∠BOC = 180°

∠BOC = 180° – 63°

∠BOC = 117°

As we know that sum of adjacent angle on a straight line is 180^o.

As we know that sum of adjacent angle on a straight line is 180^o.

According to question,

,

4x + 5x = 180^o

9x =180^o

X =20^o

As we know that sum of adjacent angle on a straight line is 180^o.

According to question,

∠ AOC = (3x + 10)°

∠ BOC = (4x – 26)°

3x + 10 + 4x – 26 = 180^o

7x – 16 =180^o

7x =196^o

X= 28^o

∠ BOC = (4x – 26)°

∠ BOC = 112° – 26°

∠ BOC = 86°

As we know that sum of all angles on a straight line is 180°

∠ AOC + ∠COD + ∠BOD = 180°

As we know that sum of all angles on a straight line is 180^o.

Through a given point, we can draw infinite number of lines.

Let x be the common multiple.

According to question,

y= 5x

As we know that sum of two supplementary – angles is 180^o.

So, x + y = 180^o

x + 5x= 180

6x = 180^o

x = 30^o

Let n be the common multiple

As we know that sum of all angles on a straight line is 180^o.

4n + 5n + 6n =180^o

15n = 180^o

N = 12^o

Y = 5n = 60^o

As we know that sum of all angles on a straight line is 180^o.

According to question,

AC and BD intersect at O.

As we know that sum of all angles on a straight line is 180^o.

Incident ray makes the same angle as reflected ray.

So,

Draw a line EF such that EF || AB and EF || CD crossing point O.

FOC + OCD = 180^o (Sum of consecutive interior angles is 180^o)

FOC = 180 – 136 = 44^o

EF || AB such that AO is traversal.

OAB + FOA = 180^o(Sum of consecutive interior angles is 180^o)

FOA = 180 – 124 = 56^o

AOC = FOC + FOA

= 56 + 44

=100^o

Draw a line EF such that EF || AB and EF || CD crossing point O.

ABO + EOB = 180^o(Sum of consecutive interior angles is 180^o)

EOB = 180 – 35 = 145^o

EF || AB such that AO is traversal.

CDO + EOD = 180^o(Sum of consecutive interior angles is 180^o)

EOD = 180 – 40 = 140^o

BOD = EOB + EOD

= 145 + 140

= 285^o

According to question,

AB || CD

AF || CD (AB is produced to F, CF is traversal)

DCF=BFC=110^o

Now, BFC + BFO = 180^o(Sum of angles of Linear pair is 180^o)

BFO = 180^o – 110^o = 70^o

Now in triangle BOF, we have

ABO = BFO + BOF

130 = 70 + BOF

BOF = 130 – 70 =60^o

So, BOC = 60^o

According to question,

AB || CD

AB || DF (DC is produced to F)

OCD=110^o

FCD = 180 – 110 = 70^o(linear pair)

Now in triangle FOC, we have

FOC + CFO + OCF = 180^o

FOC + 60 + 70 = 180^o

FOC = 180 – 130

=50^o

So, AOC = 50^o

From O, draw E such that OE || CD || AB.

OE || CD and OC is traversal.

So,

DCO + COE = 180 (co –interior angles)

x + COE = 180

COE = (180 – x)

Now, OE || AB and AO is the traversal.

BAO + AOE = 180 (co –interior angles)

BAO + AOC + COE = 180

100 + 30 + (180 – x) = 180

180 – x = 50

X = 180 – 50 = 130^O

AB || CD

BAC = DCF = 80^o

ECF + DCF = 180^o (linear pair of angles)

ECF =100^o

Now in triangle CFE,

ECF + EFC + CEF = 180^o

CEF = 180^o – 100^o – 25^o

=55^o

APQ =PQR =70^o

Now, in triangle PQR, we have

PQR + PRQ + QPQ =180^o

70 + 60 + QPQ =180^o

QPQ =180^o – 130^o

=50^o

AC is produced to meet OB at D.

OEC = 180 –

So,BEC = 180 – (180 – ) =

Now, x = BEC + CBE (Exterior Angle)

= +

=

Let say

A =x/3

B = x/4

C = x/6

A + B + C = 180

x/3 + x/4 + x/6 = 180

(4x + 3x + 2x)/12 = 180

9x/12 = 180

X= 240

A =x/3 = 240/3 = 80

B = x/4 = 240/4 = 60

C = x/6 = 240/6 = 40

So, A:B:C = 4:3:2

A + B + C = 180

C = 180 – 125 = 55^o

A + C =113^o

A =113 – 55 =58^o

A = B + 42

C = B – 21

A + B + C = 180

B + 42 + B + B – 21 =180

3B + 21 = 180

3B = 159

B =

ACD + ACB = 180 (Linear pair of angles)

ACB = 60^o

ABC = 40^o

As we know that

ACB + ACB + BAC = 180^o

BAC = 180 – 60 – 40

=80^o

ABD + ABC = 180 (Linear pair of angles)

ABC = 180^o – 125^o=55^o

ACE + ACB = 180 (Linear pair of angles)

ACB = 180^o – 130^o=50^o

As we know that

ACB + ABC + BAC = 180^o

BAC = 180 – 55 – 50

=75^o

ACB + ABC + BAC =180

ACB = 180 – 50 – 30 = 100^o(Sum of angles of triangle is 180)

ACB + ACD = 180 (linear pair of angles)

ACD = 180 – 100 = 80^o

In triangle ECD,

ECD + CDE + DEC = 180

DEC = 180 – 80 – 40

= 60^o

DEC + AED = 180^o(linear pair of angles)

AED = 180^o – 60^o

= 120^o

In triangle AEF,

BED = EFA + EAF

EFA = 100 – 40 = 60^o

CFD = EFA (vertical opposite angles)

= 60^o

In triangle CFD, we have

CFD + FCD + CDF = 180^o

CDF = 180^o – 90^o – 60^o

= 30^o

So, BDE = 30^o

In ∆ABC,

∠A + ∠B + ∠C=180°

50° + ∠B + ∠C=180°

∠B + ∠C=180°−50°=130°

∠B = 65°

∠C = 65°

Now in ∆OBC,

∠OBC + ∠OCB + ∠BOC=180°

∠BOC = 180° – 65° (∠OBC + ∠OCB = 65 because O is bisector of ∠B and ∠C)

= 115°

AB || CD and BC is traversal.

So, ∠DCB = ∠ABC = 60^o

Now in triangle AEB, we have

∠ABE + ∠BAE + ∠AEB =180^o

∠AEB =180^o – 60^o – 50^o

= 70^o

In triangle AOB,

∠AOB =180^o – 75^o – 55^o

= 50^o

∠AOB = ∠COD = 50^o(Opposite angles)

Now in triangle COD,

∠ODC =180^o – 100^o – 50^o

= 30^o

As per question,

So,

∠A = 90^o

∠B = 60^o

∠C = 30^o

∠ACB + ∠ACD + ∠ECD = 180^o (sum of angles on straight line)

∠ECD = 180^o – 90^o – 30^o

= 60^o

∠BOA = 100^o (Opposite pair of angles)

So,

∠BAO = 180^o – 100^o – 45^o

=35^o

∠BAO = ∠CDO =35^o (Corresponding Angles)

∠BCE = ∠ABC =65^o (Alternate Angles)

∠ABC = ∠ABD + ∠DBC

65^o = ∠ABD + 28^o

∠ABD = 65 – 28

= 37^o

X + 20 = 2x – 30(Corresponding Angles)

2x –x = 30 + 20

X = 50^o

4x + 3x + 5 = 180^o (Interior angles of same side of traversal)

7x + 5 = 180^o

7x = 175

X = 25^o

∠ABC = 180 – 110 = 70⁰ (Linear pair of angles)

∠BAC = 180 – 135 = 45⁰ (Linear pair of angles)

So,

In Triangle ABC, we have

∠ABC + ∠BAC + ∠ACB = 180^o

∠ACB = 180 – 70 – 45 = 65⁰

In triangle BDC,

∠B= 40, ∠D = 90

So, ∠C = 180 –(90 + 40)

= 50°

Now in triangle AEC,

∠C= 50, ∠A = 30

So, ∠E = 180 – (50 + 30)

= 100°

Thus, ∠AEB = 180 – 100 (Sum of linear pair is 180°)

= 80°

Let n be the common multiple.

Y + Z = 180

3n + 7n = 180

N =18

So, y = 3n = 54^o

z = 7n = 126^o

x = z (Pair of alternate angles)

So, x = 126^o

According to question

AB || CD || EF and

So, ∠D = ∠B (Corresponding angles)

According to question CD || EF and BE is the traversal then,

∠D + ∠E = 180 (Interior angle on the same side is supplementary)

So, ∠D = 180 – 55 = 125^o

And ∠B = 125^o

Now, AB || EF and AE is the traversal.

So, ∠BAE + ∠FEA = 180 (Interior angle on the same side of traversal is supplementary)

90 + x + 55 = 180

X + 145 = 180

X= 180 – 145 = 35^o

In triangle ABC,

∠B = 70^o

∠C = 20^o

So, ∠A = 180^o – 70^o – 20^o = 90^o

According to question, AN is bisector of ∠A

So, ∠BAN = 45^o

Now, in triangle BAM,

∠B = 70^o

∠M = 90^o

∠BAM = 180^o – 70^o – 90^o = 20^o

Now, ∠MAN = ∠BAN – ∠BAM

= 45^o – 20^o

= 25^o

Exterior angle formed when the side of a triangle is produced is equal to the sum of the interior opposite angles.

Exterior angle = 110°

One of the interior opposite angles = 45°

Let the other interior opposite angle = x

110° = 45° + x

x = 110° – 45°

x = 65°

Therefore, the other interior opposite angle is 65°.

In Δ ABC,

we have CBF = 1 + 3 ...(i) [exterior angle is equal to the sum of opposite interior angles] Similarly, ACD = 1 + 2 ...(ii)

and BAE = 2 + 3 ...(iii)

On adding Eqs. (i), (ii) and (iii),

we get CBF + ACD + BAE =2[ 1 + 2 + 3] = 2 × 180° = 4 × 90°

[by angle sum property of a triangle is 180°] CBF + ACD + BAE = 4 right angles

Thus, if the sides of a triangle are produced in order, then the sum of exterior angles so formed is equal to four right angles = 360°

Let x be the common multiple.

3x + 5x + 7x = 180

15x = 180

x = 180/15

x = 12
3x = 3 X 12 = 36

5x = 5 X 12 = 60

7x = 7 X 12 = 84

Since, all the angles are less than 90^o. So, it is acute angled triangle.

Let x and y be the bisected angles.

So in the original triangle, sum of angles is

130 + 2x + 2y = 180

2(x + y) = 50

x + y= 25

In the smaller triangle consisting of the original side opposite 130 and the 2 bisectors,

x + y + Base Angle = 180

25 + Base Angle = 180

Base Angle = 155^o

BAC = 35^o (opposite pair of angles)

BCD = 180 – 110 = 70^o (linear pair of angles)

Now, in Triangle ABC we have,

A + B + C = 180^o

35 + B + 70 = 180

B = 180 – 105 = 75^o

x + y + 90 = 180 (sum of angles on a straight line)

x + y = 90 ………………….(i)

3x + 72 = 180 (sum of angles on a straight line)

3x = 108

x = 108/3 =36^O

Putting this value in eq (i), we get

x + y = 90

36 + y = 90

Y = 90 – 36 = 54^O

Sum of triangle is = 180°

And 70 + 60 + 50 = 180°

According to linear pair of angle, sum of angles on straight line is 180

And 90 + 90 = 180^o

No, this is not linked with the given reason.

Because when two lines intersect each other, then vertically opposite angles are always equal.

3 and 5 are pair of consecutive interior angles. It is not necessary to be always equal.

(a) – (r), (b) – (s), (c) – (p), (d) – (q)

(a) – (r)

X + y = 90

X + 2x/3 = 90

5x/3 = 90

X = 270/5

= 54

(b) – (s)

X + y = 180 (according to question x =y)

X + x = 180

2x = 180

X = 90

(c) – (p)

X + y = 90 (according to question x =y)

X + x = 90

2x = 90

X = 45

(d) – (q)

X + y = 180 (linear pair of angles) ……………………….(i)

X – y =60 (according to question) …………………….. (ii)

Adding (i) and (ii) we get,

2x = 240

X = 120

Now putting this in (ii) we get,

Y = 120 – 60 = 60

(a) – (r), (b) – (p), (c) – (s), (d) – (q)

(a) – (r)

2x + 3x = 180 (linear pair of angles)

5x =180

X = 36

2x = 2 X 36 = 72

(b) – (p)

2x – 10 + 3x – 10 = 180 (linear pair of angles)

5x – 20 =180

5x = 200

x = 40

AOD = 3x – 10 (opposite angles are equal)

= 120 – 10

= 110

(c) – (s)

C = 180 – (A + B) (sum of angles triangle is 180)

= 180 – (60 + 65)

= 55

ACD = 180 – 55 (sum of linear pair of angles is 180)

= 180 – 55

= 125

(d) – (q)

B = D) (alternate interior angles)

= 55

ACB = 180 – (55 + 40) (sum of angles of triangle is 180)

= 180 – 95

= 85

Let x be the common multiple.

3x + 2x + 7x = 180

12x = 180

X = 15

3x = 45^o

2x = 30^o

7x = 105^o

A = B + 40

C = B – 10

A + B + C = 180

B + 40 + B + B – 10 = 180

3B + 30 = 180

3B = 180 – 30 = 150

B = 50^O

So, A = B + 40 = 90^O

C = B – 10 = 40^O

B = 180 – 105 (sum of linear pair of angles is 180)

= 75

C = 180 – 110 (sum of linear pair of angles is 180)

= 70

So, A = 180 – (B + C) (sum of angles of triangle is 180)

= 180 – (70 + 75)

= 35^O

In Δ ABC,

Let ∠𝐴 = x, ∠B = y and ∠C = z

∠𝐴 + ∠B + ∠C = 180 (sum of angles of triangle is 180)

x + y + z = 180 ……………i)

According to question,

x = y + z ………….(ii)

Adding eq (i) and (ii), we get

x + x = 180

2x = 180

X = 90

Hence, It is a right angled triangle.

3x – 5 + 2x + 10 = 180 (linear pair of angles)

5x + 5 =180

5x = 175

X = 175/ 5 = 35

40 + 4x + 3x = 180 (sum of angles on a straight line)

7x + 40 =180

7x = 180 – 40

X = 140/ 7 = 20

Let x be the angle then,
complement = 90 – x
supplement = 180 – x

According to question,
180 – x = 6(90 – x)
180 – x = 540 – 6x
180 + 5x = 540
5x = 360
x = 72^O

According to question,

AB || EF

EF || CD (AB is produced to F, CF is traversal)

FEC=130^o

Now, BFC + BFO = 180^o(Sum of angles of Linear pair is 180^o)

BFO = 180^o – 130^o = 50^o

Now in triangle BOF, we have

ABO = BFO + BOF

85 = 50 + BOF

BOF = 85 – 50 =35^o

So, x = ^o

A = D (Pair of alternate angles)

= 30^O

Now, in triangle EDC we have

D = 30^O and C = 50^O

So,

CED = 180 – (C + D)

= 180 – 30 – 50

=100^O

According to question EF || BAD

Producing E to O, we get

EFA + AEO = 180 (Linear pair of angles)

AEO = 180 – 55

= 125

Now, in triangle ABC we get,

A = 125 and C = 25

So, ABC = 180 – (A + C)

= 180 – (125 + 25)

= 180 – 150

= 30^O

In triangle BEC we have,

B = 40^O and E = 90^O

So,C = 180^O – (90 + 40)

=50^O

Therefore,ACB = 50^O

Now intriangle ADC we have,

A = 30^O and C = 50^O

So,D = 180^O – (30 + 50)

=100^O

Therefore,

ADB + ADC = 180 (sum of angles on straight line)

ADB + 100 = 180

ADB = 180 – 100

= 80^O

EGB = QHP (Alternate Exterior Angles) = 35^O

QPH = 90^O

So, in triangle QHP we have,

QPH + QHP + PQH = 180^O

90^O + 35^O + PQH = 180^O

PQH = 180^O – 90^O – 35^O

= 55^O

GEC = 180 – 130 = 50^O (linear pair of angles)

According to question,

AB || CD and EF is perpendicular to AB.

GEC = EGF (pair of alternate interior angles)

= 50^O

(a) – (q), (b) – (r), (c) – (s), (d) – (p)

(a) – (q)

x + x + 10 = 90

2x + 10 = 90

2x = 80

x = 40

x + 10 = 50^O

(b) – (r)

A + B + C =180

65 + B + B – 25 = 180

2B + 40 = 180

2B = 140

B = 70^O

(d) – (p)

A + B + C + D =360

2x + 3x + 5x + 40 = 360

10x + 40 = 360

10x = 320

X = 32^O

5x = 32 X 5 = 160^O

According to question,

In the given figure CD is a straight line.

As we know, Sum of angle on a straight line is 180^O

S0,

AOD + BOD + BOC =300

AOD + 180 =300

AOD =300 – 180

= 120^O

According to question,

PRD = 120^O

PRD = APR (Pair of alternate interior angles)

So,

APR = 120

APQ + QPR = 120

50 + QPR = 120

QPR = 120 – 50

= 70^O

In triangle ABC we have,

A + B + C = 180

Let B = x and C = y then,

A + 2x + 2y = 180 (BE and CE are the bisector of angles B and C respectively.)

x + y + A = 180

A = 180 – (x + y) ………….(i)

Now, in triangle BEC we have,

B = x/2

C = y + ((180 – y) / 2)

= (180 + y) / 2

B + C + BEC = 180

x/2 + (180 + y) / 2 + BEC = 180

BEC = (180 – x – y) /2 ………..(ii)

From eq (i) and (ii) we get,

BEC = A/2

Here BO, CO are the angle bisectors of ∠DBC &∠ECB intersect each other at O.

∴∠1 = ∠2 and ∠3 = ∠4

Side AB and AC of ΔABC are produced to D and E respectively.

∴ Exterior of ∠DBC = ∠A + ∠C ………… (1)

And Exterior of ∠ECB = ∠A + ∠B ………… (2)

Adding (1) and (2) we get

∠DBC + ∠ECB = 2 ∠A + ∠B + ∠C.

2∠2 + 2∠3 = ∠A + 180°

∠2 + ∠3 = (1 /2)∠A + 90° ………… (3)

But in a ΔBOC = ∠2 + ∠3 + ∠BOC = 180° ………… ( 4)

From eq (3) and (4) we get

(1 /2)∠A + 90° + ∠BOC = 180°

∠BOC = 90° – (1 /2)∠A

Let x be the common multiple.

So, angles will be x, 2x and 3x

X + 2x + 3x = 180

6x = 180

X =30

2x = 2 X 30 = 60

3x = 3 X 30 = 90

So, Angles are 30^O,60^O and 90^O

Let ∠ABD = x and ∠ACB = y

According to question,

∠B = 90^O

In triangle BDC, we have,

∠BDC = 90^O

∠DBC = (90 – x)^O

∠BDC + ∠DBC + ∠DCB = 180^O

90^O + (90 – x)^O + y = 180^O

180^O – x + y = 180^O

x = y

So,

∠ABD = ∠ACB