PSEB Solutions for Class 9 Maths Chapter 6 Lines and Angles
PSEB 9th Class Maths Solutions Chapter 6 Lines and Angles
Ex 6.1
Question 1.
In the given figure, lines AB and CD intersect at O. If ∠ AOC + ∠ BOE = 70° and ∠ BOD = 40°, find ∠ BOE and reflex ∠ COE.
Answer:
Lines AB and CD intersect at O.
∴ ∠ AOC = ∠ BOD (Vertically opposite angles)
Now, ∠ BOD = 40° (Given)
∴ ∠ AOC = 40°
Moreover, ∠ AOC + ∠ BOE = 70° (Given)
∴ 40° + ∠ BOE = 70°
∴ ∠BOE = 70° – 40°
∴ ∠ BOE = 30°
∠ BOD and ∠ BOE are adjacent angles with common arm ray OB.
∴ ∠ DOE = ∠ BOD + ∠ BOE = 40° + 30° = 70°
Reflex ∠ COE = ∠ COD + ∠ DOE
= 180° + 70° (∠ COD is a straight angle as ray OA stands on line CD.)
= 250°
Thus, ∠ BOE = 30° and reflex ∠ COE = 250°.
Question 2.
In the given figure, lines XY and MN intersect at O. If ∠ POY = 90° and a : b = 2 : 3, find c.
Answer:
Ray OP stands on line XY.
Hence, ∠XOP and ∠POY from a linear pair of angles.
∴ ∠ XOP + ∠ POY = 180°
∴ ∠ XOP + 90° = 180°
∴ ∠ XOP = 90°
∠XOM and ∠MOP are adjacent angles.
∴ ∠ XOM + ∠ MOP = ∠XOP
∴ b + a = 90° …………. (i)
Now, a : b = 2 : 3
If a = 2x, then b = 3x.
∴ 3x + 2x = 90° [From (1)]
∴ 5x = 90°
∴ x = 18°
Then, ∠ XOM = b = 3x = 3 × 18° = 54°
and ∠ MOP = a = 2x = 2 × 18° = 36°
Now, ∠ MOY = ∠ MOP + ∠ POY (Adjacent angles)
∴ ∠ MOY = 36° + 90° = 126°
Lines XY and MN intersect at O.
∴ ∠ XON and ∠ MOY are vertically opposite angles.
∴ ∠ XON = ∠ MOY
∴ c = 126°
Question 3.
In the given figure, ∠ PQR = ∠ PRQ, then prove that ∠ PQS = ∠ PRT.
Answer:
Ray QP stands on line ST.
∴ ∠ PQR and ∠ PQS form a linear pair of angles.
∴ ∠ PQR + ∠ PQS = 180°
Ray RP stands on line ST.
∴ ∠ PRQ and ∠ PRT form a linear pair of angles.
∴ ∠ PRQ + ∠ PRT = 180°
∴ ∠ PQR + ∠ PRT = 180° (Given : ∠ PQR = ∠ PRQ)
Then, ∠ PQR + ∠ PQS = ∠ PQR + ∠ PRT = 180°
∴ ∠ PQS = ∠ PRT
Question 4.
In the given figure, if x + y = w + z, then prove that AOB is a line.
Answer:
We know that sum of all the angles round any given point is 360°.
∴ x + y + z + w = 360°
∴ x + y + x + y = 360° (Given : x + y = w + z)
∴2x + 2y = 360°
∴ 2 (x + y) = 360°
∴ x + y = 180°
∴ ∠ COB + ∠ COA = 180°
But, ∠ COB and ∠ COA are adjacent angles and their sum is 180°.
∴ ∠ COB and ∠ COA are angles of a linear pair.
Hence, AOB is a line.
Question 5.
In the given figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that.
Answer:
∠ ROS = 1/2 (∠ QOS – ∠ POS).
Ray OR is perpendicular to line PQ.
∴ ∠ QOR = ∠ POR = 90°
Now, ∠ QOR and ∠ ROS are adjacent angles with common arm ray OR.
∴ ∠ QOS = ∠ QOR + ∠ ROS
∴ ∠ QOS = 90° + ∠ ROS ……………….. (1)
Similarly, ∠ POS and ∠ ROS are adjacent angles with common arm ray OS.
∴ ∠ POR = ∠ POS + ∠ ROS
∴ 90° = ∠ POS + ∠ ROS
∴ ∠ POS = 90° – ∠ ROS ………………… (2)
Subtracting (2) from (1), we get
∠ QOS – ∠ POS = (90° + ∠ ROS) – (90° – ∠ ROS)
∴ ∠ QOS – ∠ POS = 90° + ∠ ROS – 90° + ∠ ROS
∴ ∠ QOS – ∠ POS = 2∠ ROS
∴ ∠ ROS = 1/2 (∠ QOS – ∠ POS)
Question 6.
It is given that ∠ XYZ = 64° and XY is produced to point P. Draw a figure from the given information. If ray YQ bisects ∠ ZYP, find ∠ XYQ and reflex ∠ QYP.
Answer:
∠ XYZ + ∠ PYZ = 180° [Angles of a linear pair]
∴ 64° + ∠ PYZ= 180° [Given ∠ XYZ = 64°]
∴ ∠ PYZ = 180°-64°
∴ ∠ PYZ = 116°
Ray YQ bisects ∠ PYZ.
∴ ∠ PYQ = ∠ QYZ = 1/2 ∠ PYZ = 1/2 × 116° = 58°
∴ ∠ XYQ = ∠ XYZ + ∠ QYZ [Adjacent angles]
∴ ∠ XYQ = 64° + 58°
∴ ∠ XYQ = 122°
XY is produced to P.
∴ ∠ XYP is a straight angle.
∴ ∠ XYP = 180°
Reflex ∠ QYP = ∠ XYQ + ∠ XYP
= 122° + 180°
= 302°
Ex 6.2
Question 1.
In the given figure, find the values of x and y and then show that AB || C
Answer:
Ray QA stands on line PS.
∴ ∠ PQA and ∠ AQR form a linear pair.
∴ ∠ PQA + ∠ AQR = 180° [Linear pair axiom]
∴ 50° + x = 180°
∴ x = 130° ……………. (1)
Lines PS and CD intersect at R.
∴ ∠ CRS and ∠ QRD are vertically opposite angles.
∴ ∠ QRD = ∠ CRS
∴ y = 130° ……………. (2)
From (1) and (2),
x = y.
But, these angles are alternate interior angles formed by transversal PS for lines AB and CD and they are equal.
Hence, AB || CD.
Question 2.
In the given figure, if AB || CD, CD || EF and y : z = 3 : 7, find x.
Answer:
AB || CD and CD || EF.
∴ AB || EF (Lines parallel to the same line)
∴ x = z (Alternate interior angles)
Now, AB || CD.
∴ x + y = 180° (Interior angles on the same side of transversal)
Now, x = z and x + y = 180°
∴ z + y = 180°
Moreover, y : z = 3 : 7
Sum of the ratios 3 + 7 = 10
Then, z = 7/10 × 180° = 126°
Now, x = z = 126°
Question 3.
In the given figure, if AB || CD, EF ⊥ CD and ∠ GED = 126°, find ∠ AGE, ∠ GEF and ∠ FGE.
Answer:
Here, AB || CD and GE is transversal for them.
∴ ∠ AGE and ∠ GED are equal alternate interior angles.
∴ ∠ GED = ∠ AGE
∴ ∠ AGE = 126° (Given : ∠ GED= 126°)
EF ⊥ CD
∴ ∠ FED = 90°
∠ GEF + ∠ FED = ∠ GED (Adjacent angles)
∴ ∠ GEF + 90° = 126°
∴ ∠ GEF = 126° – 90°
∴ ∠ GEF = 36°
Ray GE stands on line AF.
∴ ∠ AGE + ∠ FGE = 180° (Angles of linear pair)
∴ 126° + ∠ FGE = 180°
∴ ∠ FGE =180°- 126°
∴ ∠ FGE = 54°
Question 4.
In the given figure, if PQ || ST, ∠ PQR = 110° and ∠ RST = 130°, find ∠ QRS.
[Hint: Draw a line parallel to ST through point R.]
Answer:
Draw line RU parallel to line ST.
PQ || ST and ST || RU
∴ PQ || RU
ST || RU and SR is a transversal for them.
∴ ∠ TSR + ∠ SRU = 180° (Interior angles on the same . side of the transversal)
∴ 130° + ∠ SRU = 180°
∴ ∠ SRU = 50° ………………. (1)
PQ || RU and QR is a transversal for them.
∴ ∠ PQR = ∠ QRU (Alternate interior angles)
∴ 110° = ∠ QRU
∴ ∠ QRU =110° ……………… (2)
Now, ∠ QRS and ∠ SRU are adjacent angles.
∴ ∠ QRS + ∠ SRU = ∠ QRU
∴ ∠ QRS + 50° = 110° [by (1) and (2)]
∴ ∠ QRS = 60°
Question 5.
In the given figure, if AB || CD, ∠ APQ = 50° and ∠ P RD = 127°, find x and y.
Answer:
Here, AB || CD and PQ is transversal for them.
∴ ∠ APQ = ∠ PQR (Alternate interior angles)
∴ 50° = x (Given: ∠ APQ = 50°)
∴ x = 50°
Again, AB || CD and PR is transversal for them.
∴ ∠ APR = ∠ PRD (Alternate interior angles) )
∴ ∠ APR = 127° (Given: ∠ PRD = 127°)
∴ ∠ APQ + ∠ QPR = 127°
(∵ ∠ APQ and ∠ QPR are adjacent angles and their non-common arms form ∠ APR)
∴ 50° + y = 127°
∴ y = 127°- 50°
∴ y = 77°
Question 6.
In the given figure, 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:
Construction: Draw ray BE perpendicular to line PQ and ray CF perpendicular to line RS.
We know that for plane mirrors, angle of incidence = angle of reflection.
Hence, at point B,
∠ ABE = ∠ EBC …………. (1)
and ’ at point C,
∠ BCF = ∠ FCD ………….. (2)
Now, PQ || RS, BE ⊥ PQ and CF ⊥ RS.
Hence, BE || CE
∴ ∠ EBC = ∠ BCF (Alternate interior angles) …………… (3)
From (1), (2) and (3), we get
∠ ABE = ∠ EBC = ∠ BCF = ∠ FCD
∴ ∠ ABE + ∠ EBC = ∠ BCF + ∠ FCD
∴ ∠ ABC = ∠ BCD [Adjacent angles: ∠ ABE and ∠ EBC, ∠ BCF and ∠ FCD]
But ∠ ABC and ∠ BCD are alternate interior angles formed by transversal BC for lines AB and CD and they are equal.
∴ AB || CD
Ex 6.3
Question 1.
In the given figure, sides QP and RQ of ∆ PQR are produced to points S and T respectively. If ∠ SPR =135° and ∠ PQT = 110°, find ∠ PRQ.
Answer:
Here, ∠ SPR and ∠ PQT are exterior angles.
Then, by theorem 6.8,
∠ SPR = ∠ PQR + ∠ PRQ and
∠ PQT = ∠ QPR + ∠ PRQ
∴ ∠ PQR + ∠ PRQ = 135° and
∠ QPR + ∠ PRQ = 110°
Adding these two equations,
∠ PQR + ∠ PRQ + ∠ QPR + ∠ PRQ = 135° + 110°
∴ 180° + ∠ PRQ = 245° (Theorem 6.7)
∴ ∠ PRQ = 245° – 180°
∴ ∠ PRQ = 65°
Question 2.
In the given figure, ∠ X = 62°, ∠ XYZ = 54°. If YO and ZO are the bisectors of ∠ XYZ and ∠ XZY respectively of ∆ XYZ, find ∠ OZY and ∠ YOZ.
Answer:
In ∆ XYZ,
∠ X + ∠ XYZ + ∠ XZY = 180° (Theorem 6.7)
∴ 62° + 54° + ∠ XZY = 180°
∴ ∠ XZY = 180° – 62° – 54°
∴ ∠ XZY = 64°
YO and ZO are bisectors of ∠ XYZ and ∠ XZY respectively.
∴ ∠ OYZ = 1/2 ∠ XYZ = 1/2 × 54° = 27° and
∠ OZY = 1/2 ∠ XZY = 1/2 × 64° = 32°.
Now, in ∆ OYZ,
∠ OYZ + ∠ OZY + ∠ YOZ = 180° (Theorem 6.7)
∴ 27° + 32° + ∠ YOZ = 180°
∴ ∠ YOZ = 180° – 27° – 32°
∴ ∠ YOZ = 121°
Question 3.
In the given figure, if AB || DE, ∠ BAC = 35° and ∠ CDE = 53°, find ∠ DCE.
Answer:
AB || DE and AE is transversal for them.
∴ ∠ AED = ∠ BAE (Alternate interior angles)
∴ ∠ CED = ∠ BAC (Point C lies on line AE)
∴ ∠ CED = 35° (Given : ∠ BAC = 35°)
In ∆ CDE, by theorem 6.8
∠ CDE + ∠ CED + ∠ DCE = 180°
∴ 53° + 35° + ∠ DCE = 180°
∴ ∠ DCE = 180° – 53° – 35°
∴ ∠ DCE = 92°
Question 4.
In the given figure, if lines PQ and RS intersect at point T, such that ∠ PRT = 40°, ∠ RPT = 95° and ∠ TSQ = 75°, find ∠ SQT.
Answer:
In ∆ PRT,
∠ RPT + ∠ PRT + ∠ PTR = 180° (Theorem 6.7)
∴ 95° + 40° + ∠ PTR = 180°
∴ 135° + ∠ PTR = 180°
∴ ∠ PTR = 180°- 135°
∴ ∠ PTR = 45°
Lines PQ and RS intersect at point T.
∴ ∠ STQ = ∠ PTR (Vertically opposite angles)
∴ ∠ STQ = 45°
In ∆ STQ,
∠ TSQ + ∠ STQ + ∠ SQT = 180° (Theorem 6.7)
∴ 75° + 45° + ∠ SQT = 180°
∴ 120° + ∠ SQT = 180°
∴ ∠ SQT = 60°
Question 5.
In the given figure, if PQ ⊥ PS, PQ || SR, ∠ SQR = 28° and ∠ QRT = 65°, then find the values of x and y.
Answer:
PQ || SR and QR is transversal for them.
∴ ∠ PQR = ∠ QRT (Alternate interior angles)
∴ ∠ PQR = 65° (Given : ∠ QRT = 65°)
∴ ∠ PQS + ∠ SQR = 65° (Adjacent angles)
∴ x + 28° = 65° (Given : ∠ SQR = 28°)
∴ x = 65° – 28°
∴ x = 37°
PQ ⊥ PS
∴ ∠ SPQ = 90°
In ∆ PSQ,
∠ SPQ + ∠ PQS + ∠ PSQ = 180° (Theorem 6.7)
∴ 90° + 37° + y = 180°
∴ 127° + y = 180°
∴ y = 180°- 127°
∴ y = 53°
Question 6.
In the given figure, the side QR of ∆ PQR is produced to a point S. If the bisectors of ∠ PQR and ∠ PRS meet at point T, then prove that ∠ QTR = 1/2 ∠ QPR.
MCQ
Multiple Choice Questions and Answer
Answer each question by selecting the proper alternative from those given below each question to make the statement true:
Question 1.
The measure of the complementary angle of an angle with measure 40° is ………………….. .
A. 40°
B. 20°
C. 140°
D. 50°
Answer:
D. 50°
Question 2.
The measure of the supplementary angle of an angle with measure 70° is ………………… .
A. 20°
B. 35°
C. 70°
D. 110°
Answer:
D. 110°
Question 3.
∠ ABC and ∠ ABD form a linear pair. If ∠ ABC = 30°, then ∠ ABD = ………………. .
A. 30°
B. 60°
C. 150°
D. 15°
Answer:
C. 150°
Question 4.
∠P and ∠Q are supplementary angles such that ∠P = 2x – 5 and ∠Q = 3x + 10. Then ∠Q = …………….. .
A. 35°
B. 65°
C. 105°
D. 115°
Answer:
D. 115°
Question 5.
The measure of an angle is four times the measure of its complementary angle. Then, the measure of that angle is ………………… .
A. 18°
B. 72°
C. 40°
D. 10°
Answer:
B. 72°
Question 6.
The measures of two supplementary angles differ by 20°. Then, the measure of the acute angle among them is …………….. .
A. 50°
B. 80°
C. 100°
D. 20°
Answer:
B. 80°
Question 7.
The measure of an angle is twice the measure of its supplementary angle. Then, the measure of that angle is ………………. .
A. 60°
B. 120°
C. 50°
D. 100°
Answer:
B. 120°
Question 8.
∠ ACD is an exterior angle of ∆ ABC. If ∠ ACD = 110° and ∠ A = 60°, then ∠ B = ………………….. .
A. 50°
B. 60°
C. 70°
D. 55°
Answer:
A. 50°
Question 9.
In ∆ ABC, ∠ A = 70° and ∠ B = 60°. Then, the measure of an exterior angle of ∆ ABC can be ……………….. .
A. 50°
B. 110°
C. 100°
D. 70°
Answer:
B. 110°
Question 10.
In ∆ ABC, ∠ B = 55° and ∠ C = 65°. Then the measure of an exterior angle of ∆ ABC cannot be ………………. .
A. 125°
B. 120°
C. 115°
D. 110°
Answer:
D. 110°