Lines And Angles

It is given: ∠AOC + ∠BOE = 70^o

And, ∠BOD = 40^o

Now, according to the question,

∠AOC + ∠BOE + ∠COE = 180^o (Sum of linear pair is always 180^o )

⇒ 70^o + ∠COE = 180^o

⇒ ∠COE = 110^o ...........(1)

And,

∠COE + ∠BOD + ∠BOE = 180^o (Sum of linear pair is always 180^o )

putting the value of ∠COE, we get,

110^o + 40^o + ∠BOE = 180^o

150^o + ∠BOE = 180^o

∴ ∠BOE = 30^o

Now, reflex ∠COE = 360^o - ∠COE

⇒ reflex ∠COE = 360^o - 110^o

∴ reflex ∠COE = 250^o
Note:Reflex here means the reflex angles which are greater than 180° but less than 360°.

Given: ∠POY = 90^O

a : b = 2 : 3

Now, according to the question,

∠POY + a + b = 180^o (Angles made on a straight line are supplementary or sum of angles equals to 180°)

90^o + a + b = 180^o

a + b = 90^o .............eq(1)

Let a be 2 x and b be 3 x
So putting this value in eq(1),

2x + 3x = 90^o

5x = 90^o

x = 18^o

Hence,

a = 2 x 18^o = 36^o

And,

b = 3 x 18^o = 54^o

And,

b + c = 180^o (Linear pair)

54^o + c = 180^o

c = 126^o

Given: ∠PQR = ∠PRQ

To prove: ∠PQS = ∠PRT
Proof:

Now,

∠PQR + ∠PQS = 180^o (Angles on a straight line are supplementary)

⇒ ∠PQR = 180^o - ∠PQS ......(i)

Similarly,

∠PRQ + ∠PRT = 180^o (Angles on a straight line are supplementary)

⇒ ∠PRQ = 180^o - ∠PRT .....(ii)

Now, ∠PQR = ∠PRQ (Given)

Therefore, (i) and (ii) will be equal

180^o - ∠PQS = 180^o - ∠PRT

∠PQS = ∠PRT

Hence, Proved.

Given: x + y = w + z

To prove,: AOB is a line or x + y = 180^° (Linear pair)

Now, according to the question,

x + y + w + z = 360^o (Angles at a point)

(x + y) + (w + z) = 360°

(Given, x + y = w + z)

2 (x + y) = 360^o

(x + y) = 180^o

Therefore, x + y makes a linear pair.

And, AOB is a straight line

Hence, Proved.

Given: OR is perpendicular to line PQ

To prove: ∠ROS = (∠QOS - ∠POS)

Now, according to the question,

∠POR = ∠ROQ = 90° ( ∵ OR is perpendicular to line PQ)

∠QOS = ∠ROQ + ∠ROS = 90° + ∠ROS ............eq(i)

∠POS = ∠POR - ∠ROS = 90° - ∠ROS ...............eq(ii)

Subtracting (ii) from (i), we get

∠QOS - ∠POS = 90^o + ∠ROS – (90° - ∠ROS)

∠QOS - ∠POS = 90^o + ∠ROS – 90° + ∠ROS

∠QOS - ∠POS = 2∠ROS

Hence, proved

Given: ∠XYZ = 64^o

YQ bisects ∠ZYP, therefore ∠QYP = ∠QYZ
∠XYZ + ∠ZYP = 180^o (Sum of angles made on a straight line = 180º)

64^o + ∠ZYP = 180^o

∠ZYP = 116^o

And,

∠ZYP = ∠ZYQ + ∠QYP

∠ZYQ = ∠QYP (YQ bisects ∠ZYP)

∠ZYP = 2∠ZYQ

2∠ZYQ = 116^o

∠ZYQ = 58^o = ∠QYP

Now,

∠XYQ = ∠XYZ + ∠ZYQ

∠XYQ = 64^o + 58^o

∠XYQ = 122^o

And,

Reflex of ∠QYP = 180^o + ∠XYQ

∠QYP = 180^o + 122^o

∠QYP = 302^o

From the figure:

x + 50^o = 180^o (Linear pair)

x = 130^o

And,

y = 130^o (Vertically opposite angles)

Now,

x = y = 130^o (Alternate interior angles)

Hence,

AB || CD

It is given in the question that:

AB || CD,
Thus,
∠x + ∠y = 180° ( Interior angles on same side of transversal ) (1)
Also,
AB || CD and CD || EF

Thus, AB || EF
⇒ ∠x = ∠z ( Alternate interior angles) (2)
From (1) and (2) we can say that,
∠z + ∠y = 180° (3) [Angles will be supplementary]
It is given that,
∠y : ∠z = 3 : 7 (4)
Let ∠y = 3 a and ∠z = 7 a
Putting these values in (2)
3 a + 7 a = 180°
10 a = 180°
a = 18°
∠z = 7 a
∠z = 7 x 18°
∠z = 126°
As ∠z = ∠x
∠x = 126°

It is given in the question that:

AB || CD

EF perpendicular CD

∠GED = 126^o

Now, according to the question,

∠FED = 90^o (EF perpendicular CD)

Now,

∠AGE = ∠GED (Since, AB parallel CD and GE is transversal, hence alternate interior angles)

Therefore,

∠AGE = 126^o

And,

∠GEF = ∠GED - ∠FED

∠GEF = 126^o – 90^o

∠GEF = 36^o

Now,

∠FGE + ∠AGE = 180^o (Linear pair)

∠FGE = 180^o – 126^o

∠FGE = 54^o

It is given in the question that:

PQ parallel ST,

∠PQR = 110^o and,

∠RST = 130^o

Construction: Draw a line XY parallel to PQ and ST

∠PQR + ∠QRX = 180^o (Angles on the same side of transversal)

110^o + ∠QRX = 180^o

∠QRX = 70^o

And,

∠RST + ∠SRY = 180^o (Angles on the same side of transversal)

130^o + ∠SRY = 180^o

∠SRY = 50^o

Now,

∠QRX + ∠SRY + ∠QRS = 180^o ( Angle made on a straight line)

70^o + 50^o + ∠QRS = 180^o

∠QRS = 60^o

It is given in the question that:

AB parallel CD,

∠APQ = 50^o and,

∠PRD = 127^o

According to question,

x = 50^o (Alternate interior angle)

∠PRD + ∠PRQ = 180^o (Angles on the straight line are supplementary)

127^o + ∠PRQ = 180^o

∠PRQ = 53^o

Now,

y + 50^o + ∠ PRQ = 180^o

y + 50^o + 53^o = 180^o

y + 103^o = 180^o

y = 77^o

Given: Two mirrors are parallel to each other, PQ||RS
To Prove: AB || CD
Proof:

Take two perpendiculars BE and CF, As the mirrors are parallel to each other their perpendiculars will also be parallel thus BE || CF

According to laws of reflection, we know that:

Angle of incidence = Angle of reflection

∠1 = ∠2 and,

∠3 = ∠4 (i)

And,

∠2 = ∠3 (Alternate interior angles, since BE, is parallel to CF and a trasversal line BC cuts them at B and C respectively) (ii)
We need to prove that ∠ABC = ∠DCB
∠ABC = ∠1 + ∠2 and ∠DCB = ∠3 + ∠4

From (i) and (ii), we get

∠1 + ∠2 = ∠3 + ∠4

⇒ ∠ABC = ∠DCB

⇒ AB parallel CD (Alternate interior angles)
Hence, Proved.

Method 1:

It is given in the question that:

∠SPR = 135^O

And,

∠PQT = 110^o

Now, according to the question,

∠SPR + ∠QPR = 180^O (SQ is a straight line)

135^o + ∠QPR = 180^O

∠QPR = 45^O

And,

∠PQT + ∠PQR = 180^O (TR is a straight line)

110^o + ∠PQR = 180^O

∠PQR = 70^O

Now,

∠PQR + ∠QPR + ∠PRQ = 180^O (Sum of the interior angles of the triangle)

70^o + 45^o + ∠PRQ = 180^O

115^O + ∠PRQ = 180^O

∠PRQ = 65^O

It is given in the question that:

∠SPR = 135^O

And,

∠PQT = 110^o

∠PQT + ∠PQR = 180^O (TR is a straight line)

110^o + ∠PQR = 180^O

∠PQR = 70^O

Given: ∠X = 62^o, ∠ XYZ = 54^o

YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively.
To Find: ∠ OZY and ∠ YOZ.

Now, according to the question,

∠X + ∠XYZ + ∠XZY = 180^o (Sum of the interior angles of a triangle = 180°)

62^o + 54^o + ∠XZY = 180^o

116^o + ∠XZY = 180^o

∠XZY = 64^o

Now,

∠OZY = 1/2 ∠ XZY

∠OZY = 32^o

And,

∠OYZ = 1/2 ∠XYZ

∠OYZ = 27^o

Now,

∠OZY + ∠OYZ + ∠O = 180^o (Sum of the interior angles of the triangle = 180°)

32^o + 27^o + ∠O = 180^o

59^o + ∠O = 180^o

∠O = 121^o

Given: AB parallel DE, ∠BAC = 35^o, ∠CDE = 53^o


To Find: ∠DCE

According to question,

∠BAC = ∠CED (Alternate interior angles)

Therefore,

∠CED = 35^o

Now, In Δ DEC,

∠DCE + ∠CED + ∠CDE = 180^o (Sum of the interior angles of the triangle)

∠DCE + 35^o + 53^o = 180^o

∠DCE + 88^o = 180^o

∠DCE = 92^o

Given:

∠PRT = 40^o

∠RPT = 95^o and,

∠TSQ = 75^o

Now according to the question,

∠PRT + ∠RPT + ∠PTR = 180^o (Sum of interior angles of the triangle)

40^o + 95^o + ∠PTR = 180^o

40^o + 95^o + ∠PTR = 180^o

135^o + ∠PTR = 180^o

∠PTR = 45^o

∠PTR = ∠STQ = 45^o (Vertically opposite angles)

Now,

∠TSQ + ∠PTR + ∠SQT = 180^o (Sum of the interior angles of the triangle)

75^o + 45^o + ∠SQT = 180^o

120^o + ∠SQT = 180^o

∠SQT = 60^o

To Find: Values of x and y

Given: PQ is perpendicular to PS, PQ parallel SR

∠SQR = 28^o

And, ∠QRT = 65^o

Now according to the question,

x + ∠SQR = ∠QRT (Alternate angles are equal as QR is transversal)

x + 28^o = 65^o

x = 37^o



Now, in Δ PQS, Sum of interior angles of a triangle = 180°

∠ PQS + ∠ PSQ + ∠ QPS = 180°

Therefore,

To Prove: ∠QTR = 1/2 ∠QPR

Given: Bisectors of ∠PQR and ∠PRS meet at point T

Proof:

In ΔQTR,

∠TRS = ∠TQR + ∠QTR (Exterior angle of a triangle equals to the sum of the two opposite interior angles)

∠QTR = ∠TRS - ∠TQR -----------(i)

Similarly in ΔQPR,

∠SRP = ∠QPR + ∠PQR

2∠TRS = ∠QPR + 2∠TQR (∵ ∠TRS and ∠TQR are the bisectors of ∠SRP and ∠PQR respectively.)

∠QPR = 2 ∠TRS - 2 ∠TQR

.............(ii)

From (i) and (ii), we get

Hence, proved