In Fig. 6.13, lines AB and CD intersect at O. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.
Answer
Given,
∠AOC + ∠BOE = 70° and ∠BOD = 40°
A/q,
∠AOC + ∠BOE +∠COE = 180° (Forms a straight line)
⇒ 70° +∠COE = 180°
⇒ ∠COE = 110°
also,
∠COE +∠BOD + ∠BOE = 180° (Forms a straight line)
⇒ 110° +40° + ∠BOE = 180°
⇒ 150° + ∠BOE = 180°
⇒ ∠BOE = 30°
Given,
∠POY = 90° and a : b = 2 : 3
A/q,
∠POY + a + b = 180°
⇒ 90° + a + b = 180°
⇒ a + b = 90°
Let a be 2x then will be 3x
2x + 3x = 90°
⇒ 5x = 90°
⇒ x = 18°
∴ a = 2×18° = 36°
and b = 3×18° = 54°
also,
b + c = 180° (Linear Pair)
⇒ 54° + c = 180°
⇒ c = 126°
Given,
∠PQR = ∠PRQ
To prove,
∠PQS = ∠PRT
A/q,
∠PQR +∠PQS = 180° (Linear Pair)
⇒ ∠PQS = 180° - ∠PQR --- (i)
also,
∠PRQ +∠PRT = 180° (Linear Pair)
⇒ ∠PRT = 180° - ∠PRQ
⇒ ∠PRQ = 180° - ∠PQR --- (ii) (∠PQR = ∠PRQ)
From (i) and (ii)
∠PQS = ∠PRT = 180° - ∠PQR
Therefore, ∠PQS = ∠PRT
Given,
x + y = w + z
To Prove,
AOB is a line or x + y = 180° (linear pair.)
A/q,
x + y + w + z = 360° (Angles around a point.)
⇒ (x + y) + (w + z) = 360°
⇒ (x + y) + (x + y) = 360° (Given x + y = w + z)
⇒ 2(x + y) = 360°
⇒ (x + y) = 180°
Hence, x + y makes a linear pair. Therefore, AOB is a staright line.
Given,
OR is perpendicular to line PQ
To prove,
∠ROS = 1/2(∠QOS – ∠POS)
A/q,
∠POR = ∠ROQ = 90° (Perpendicular)
∠QOS = ∠ROQ + ∠ROS = 90° + ∠ROS --- (i)
∠POS = ∠POR - ∠ROS = 90° - ∠ROS --- (ii)
Subtracting (ii) from (i)
∠QOS - ∠POS = 90° + ∠ROS - (90° - ∠ROS)
⇒ ∠QOS - ∠POS = 90° + ∠ROS - 90° + ∠ROS
⇒ ∠QOS - ∠POS = 2∠ROS
⇒ ∠ROS = 1/2(∠QOS – ∠POS)
Hence, Proved.
Given,
∠XYZ = 64°
YQ bisects ∠ZYP
∠XYZ +∠ZYP = 180° (Linear Pair)
⇒ 64° +∠ZYP = 180°
⇒ ∠ZYP = 116°
also, ∠ZYP = ∠ZYQ + ∠QYP
∠ZYQ = ∠QYP (YQ bisects ∠ZYP)
⇒ ∠ZYP = 2∠ZYQ
⇒ 2∠ZYQ = 116°
⇒ ∠ZYQ = 58° = ∠QYP
Now,
∠XYQ = ∠XYZ + ∠ZYQ
⇒ ∠XYQ = 64° + 58°
⇒ ∠XYQ = 122°
also,
reflex ∠QYP = 180° + ∠XYQ
∠QYP = 180° + 122°
⇒ ∠QYP = 302°
x + 50° = 180° (Linear pair)
⇒ x = 130°
also,
y = 130° (Vertically opposite)
Now,
x = y = 130° (Alternate interior angles)
Alternate interior angles are equal.
Therefore, AB || CD.
Given,
AB || CD and CD || EF
y : z = 3 : 7
Now,
x + y = 180° (Angles on the same side of transversal.)
also,
∠O = z (Corresponding angles)
and, y + ∠O = 180° (Linear pair)
⇒ y + z = 180°
A/q,
y = 3w and z = 7w
3w + 7w = 180°
⇒ 10 w = 180°
⇒ w = 18°
∴ y = 3×18° = 54°
and, z = 7×18° = 126°
Now,
x + y = 180°
⇒ x + 54° = 180°
⇒ x = 126°
Given,
AB || CD
EF ⊥ CD
∠GED = 126°
A/q,
∠FED = 90° (EF ⊥ CD)
Now,
∠AGE = ∠GED (Since, AB || CD and GE is transversal. Alternate interior angles.)
∴ ∠AGE = 126°
Also, ∠GEF = ∠GED - ∠FED
⇒ ∠GEF = 126° - 90°
⇒ ∠GEF = 36°
Now,
∠FGE +∠AGE = 180° (Linear pair)
⇒ ∠FGE = 180° - 126°
⇒ ∠FGE = 54°
Given,
PQ || ST, ∠PQR = 110° and ∠RST = 130°
Construction,
A line XY parallel to PQ and ST is drawn.
∠PQR + ∠QRX = 180° (Angles on the same side of transversal.)
⇒ 110° + ∠QRX = 180°
⇒ ∠QRX = 70°
Also,
∠RST + ∠SRY = 180° (Angles on the same side of transversal.)
⇒ 130° + ∠SRY = 180°
⇒ ∠SRY = 50°
Now,
∠QRX +∠SRY + ∠QRS = 180°
⇒ 70° + 50° + ∠QRS = 180°
⇒ ∠QRS = 60°
Given,
AB || CD, ∠APQ = 50° and ∠PRD = 127°
A/q,
x = 50° (Alternate interior angles.)
∠PRD + ∠RPB = 180° (Angles on the same side of transversal.)
⇒ 127° + ∠RPB = 180°
⇒ ∠RPB = 53°
Now,
y + 50° + ∠RPB = 180° (AB is a straight line.)
⇒ y + 50° + 53° = 180°
⇒ y + 103° = 180°
⇒ y = 77°
In Fig. 6.33, PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD. Prove that AB || CD.
ANSWER:
Let us draw BE ⟂ PQ and CF ⟂ RS.
As PQ || RS
So, BE || CF
By laws of reflection we know that,
Angle of incidence = Angle of reflection
Thus, ∠1 = ∠2 and ∠3 = ∠4 --- (i)
also, ∠2 = ∠3 (alternate interior angles because BE || CF and a transversal line BC cuts them at B and C) --- (ii)
From (i) and (ii),
∠1 + ∠2 = ∠3 + ∠4
⇒ ∠ABC = ∠DCB
⇒ AB || CD (alternate interior angles are equal)
1. In Fig. 6.39, sides QP and RQ of ΔPQR are produced to points S and T respectively. If ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ.
Answer
Given,
∠SPR = 135° and ∠PQT = 110°
A/q,
∠SPR +∠QPR = 180° (SQ is a straight line.)
⇒ 135° +∠QPR = 180°
⇒ ∠QPR = 45°
also,
∠PQT +∠PQR = 180° (TR is a straight line.)
⇒ 110° +∠PQR = 180°
⇒ ∠PQR = 70°
Now,
∠PQR +∠QPR + ∠PRQ = 180° (Sum of the interior angles of the triangle.)
⇒ 70° + 45° + ∠PRQ = 180°
⇒ 115° + ∠PRQ = 180°
⇒ ∠PRQ = 65°
In Fig. 6.40, ∠X = 62°, ∠XYZ = 54°. If YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of Δ XYZ, find ∠OZY and ∠YOZ.
Answer
Given,
∠X = 62°, ∠XYZ = 54°
YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively.
A/q,
∠X +∠XYZ + ∠XZY = 180° (Sum of the interior angles of the triangle.)
⇒ 62° + 54° + ∠XZY = 180°
⇒ 116° + ∠XZY = 180°
⇒ ∠XZY = 64°
Now,
∠OZY = 1/2∠XZY (ZO is the bisector.)
⇒ ∠OZY = 32°
also,
∠OYZ = 1/2∠XYZ (YO is the bisector.)
⇒ ∠OYZ = 27°
Now,
∠OZY +∠OYZ + ∠O = 180° (Sum of the interior angles of the triangle.)
⇒ 32° + 27° + ∠O = 180°
⇒ 59° + ∠O = 180°
⇒ ∠O = 121°
In Fig. 6.41, if AB || DE, ∠BAC = 35° and ∠CDE = 53°, find ∠DCE.
Answer
Given,
AB || DE, ∠BAC = 35° and ∠ CDE = 53°
A/q,
∠BAC = ∠CED (Alternate interior angles.)
∴ ∠CED = 35°
Now,
∠DCE +∠CED + ∠CDE = 180° (Sum of the interior angles of the triangle.)
⇒ ∠DCE + 35° + 53° = 180°
⇒ ∠DCE + 88° = 180°
⇒ ∠DCE = 92°
In Fig. 6.42, if lines PQ and RS intersect at point T, such that ∠PRT = 40°, ∠RPT = 95° and ∠TSQ = 75°, find ∠SQT.
Answer
Given,
∠PRT = 40°, ∠RPT = 95° and ∠TSQ = 75°
A/q,
∠PRT +∠RPT + ∠PTR = 180° (Sum of the interior angles of the triangle.)
⇒ 40° + 95° + ∠PTR = 180°
⇒ 40° + 95° + ∠PTR = 180°
⇒ 135° + ∠PTR = 180°
⇒ ∠PTR = 45°
∠PTR = ∠STQ = 45° (Vertically opposite angles.)
Now,
∠TSQ +∠PTR + ∠SQT = 180° (Sum of the interior angles of the triangle.)
⇒ 75° + 45° + ∠SQT = 180°
⇒ 120° + ∠SQT = 180°
⇒ ∠SQT = 60°
In Fig. 6.43, if PQ ⊥ PS, PQ || SR, ∠SQR = 28° and ∠QRT = 65°, then find the values of x and y.
Answer
Given,
PQ ⊥ PS, PQ || SR, ∠SQR = 28° and ∠QRT = 65°
A/q,
x +∠SQR = ∠QRT (Alternate angles as QR is transveersal.)
⇒ x + 28° = 65°
⇒ x = 37°
also,
∠QSR = x
⇒ ∠QSR = 37°
also,
∠QRS +∠QRT = 180° (Linea pair)
⇒ ∠QRS + 65° = 180°
⇒ ∠QRS = 115°
Now,
∠P + ∠Q+ ∠R +∠S = 360° (Sum of the angles in a quadrilateral.)
⇒ 90° + 65° + 115° + ∠S = 360°
⇒ 270° + y + ∠QSR = 360°
⇒ 270° + y + 37° = 360°
⇒ 307° + y = 360°
⇒ y = 53°
In Fig. 6.44, the side QR of ΔPQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at point T, then prove that ∠QTR = 1/2∠QPR.
AnswerGiven,
Bisectors of ∠PQR and ∠PRS meet at point T.
To prove,
∠QTR = 1/2∠QPR.
Proof,
∠TRS = ∠TQR +∠QTR (Exterior angle of a triangle equals to the sum of the two interior angles.)
⇒ ∠QTR = ∠TRS - ∠TQR --- (i)
also,
∠SRP = ∠QPR + ∠PQR
⇒ 2∠TRS = ∠QPR + 2∠TQR
⇒ ∠QPR = 2∠TRS - 2∠TQR
⇒ 1/2∠QPR = ∠TRS - ∠TQR --- (ii)
Equating (i) and (ii)
∠QTR - ∠TQR = 1/2∠QPR
Hence proved.
