Important Questions & Solutions For Class 9 Chapter 6 (Lines and Angles)
Q.1: In the figure, lines AB and CD intersect at O. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.
Solution:
From the given figure, we can see;
∠AOC + ∠BOE +∠COE and ∠COE +∠BOD + ∠BOE form a straight line.
So, ∠AOC + ∠BOE +∠COE = ∠COE +∠BOD + ∠BOE = 180°
Now, by substituting the values of ∠AOC + ∠BOE = 70° and ∠BOD = 40° we get
∠COE = 110° and
∠BOE = 30°
Q.2: In the Figure, lines XY and MN intersect at O. If ∠POY = 90° and a : b = 2 : 3, find c.
Solution:
As we know, the sum of linear pair are always equal to 180°
So,
∠POY + a + b = 180°
substituting the value of ∠POY = 90° (as given in the question) we get,
a + b = 90°
Now, it is given that a : b = 2 : 3 so,
Let a be 2x and b be 3x
∴ 2x + 3x = 90°
Solving this we get
5x = 90°
So, x = 18°
∴ a = 2 × 18° = 36°
Similarly b can be calculated and the value will be
b = 3 × 18° = 54°
From the diagram, b + c also forms a straight angle so,
b + c = 180°
=> c + 54° = 180°
∴ c = 126°
Q.3: In the Figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that ∠ROS = 1/2(∠QOS – ∠POS).
Solution:
In the question, it is given that (OR ⊥ PQ) and ∠POQ = 180°
So, ∠POS + ∠ROS + ∠ROQ = 180°
Now, ∠POS + ∠ROS = 180° – 90° (Since ∠POR = ∠ROQ = 90°)
∴ ∠POS + ∠ROS = 90°
Now, ∠QOS = ∠ROQ + ∠ROS
It is given that ∠ROQ = 90°,
∴ ∠QOS = 90° + ∠ROS
Or, ∠QOS – ∠ROS = 90°
As ∠POS + ∠ROS = 90° and ∠QOS – ∠ROS = 90°, we get
∠POS + ∠ROS = ∠QOS – ∠ROS
=>2 ∠ROS + ∠POS = ∠QOS
Or, ∠ROS = ½ (∠QOS – ∠POS) (Hence proved).
Q.4: 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.
Solution:
Here, XP is a straight line
So, ∠XYZ +∠ZYP = 180°
substituting the value of ∠XYZ = 64° we get,
64° +∠ZYP = 180°
∴ ∠ZYP = 116°
From the diagram, we also know that ∠ZYP = ∠ZYQ + ∠QYP
Now, as YQ bisects ∠ZYP,
∠ZYQ = ∠QYP
Or, ∠ZYP = 2∠ZYQ
∴ ∠ZYQ = ∠QYP = 58°
Again, ∠XYQ = ∠XYZ + ∠ZYQ
By substituting the value of ∠XYZ = 64° and ∠ZYQ = 58° we get.
∠XYQ = 64° + 58°
Or, ∠XYQ = 122°
Now, reflex ∠QYP = 180° + ∠XYQ
We computed that the value of ∠XYQ = 122°. So,
∠QYP = 180° + 122°
∴ ∠QYP = 302°
Q.5: In the Figure, if AB || CD, EF ⊥ CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE.
Solution:
Since AB || CD GE is a transversal.
It is given that ∠GED = 126°
So, ∠GED = ∠AGE = 126° (As they are alternate interior angles)
Also,
∠GED = ∠GEF + ∠FED
As
EF ⊥ CD, ∠FED = 90°
∴ ∠GED = ∠GEF + 90°
Or, ∠GEF = 126 – 90° = 36°
Again, ∠FGE + ∠GED = 180° (Transversal)
substituting the value of ∠GED = 126° we get,
∠FGE = 54°
So,
∠AGE = 126°
∠GEF = 36° and
∠FGE = 54°
Q.6: In the Figure, if PQ || ST, ∠PQR = 110° and ∠RST = 130°, find ∠QRS.
[Hint : Draw a line parallel to ST through point R.]
Solution: First, construct a line XY parallel to PQ.
As we know, the angles on the same side of transversal is equal to 180°.
So, ∠PQR + ∠QRX = 180°
Or,∠QRX = 180° – 110°
∴ ∠QRX = 70°
Similarly,
∠RST + ∠SRY = 180°
Or, ∠SRY = 180° – 130°
∴ ∠SRY = 50°
Now, for the linear pairs on the line XY-
∠QRX + ∠QRS + ∠SRY = 180°
substituting their respective values we get,
∠QRS = 180° – 70° – 50°
Or, ∠QRS = 60°
Q.7: 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.
Solution:
First, draw two lines BE and CF such that BE ⟂ PQ and CF ⟂ RS.
Now, since PQ || RS,
So, BE || CF
As we know,,
Angle of incidence = Angle of reflection (By the law of reflection)
So,
∠1 = ∠2 and
∠3 = ∠4
We also know that alternate interior angles are equal. Here, BE ⊥ CF and the transversal line BC cuts them at B and C
So, ∠2 = ∠3 (As they are alternate interior angles)
Now, ∠1 + ∠2 = ∠3 + ∠4
Or, ∠ABC = ∠DCB
So, AB ∥ CD (alternate interior angles are equal)
Q.8: 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.
Solution:
As we know, the sum of the interior angles of the triangle.
So, ∠X +∠XYZ + ∠XZY = 180°
substituting the values as given in the question we get,
62° + 54° + ∠XZY = 180°
Or, ∠XZY = 64°
Now, As we know, ZO is the bisector so,
∠OZY = ½ ∠XZY
∴ ∠OZY = 32°
Similarly, YO is a bisector and so,
∠OYZ = ½ ∠XYZ
Or, ∠OYZ = 27° (As ∠XYZ = 54°)
Now, as the sum of the interior angles of the triangle,
∠OZY +∠OYZ + ∠O = 180°
substituting their respective values we get,
∠O = 180° – 32° – 27°
Or, ∠O = 121°
Q.9: In the figure, if AB || CD || EF, PQ || RS, ∠RQD = 25° and ∠CQP = 60°, then find ∠QRS?
Solution:
According to the given figure, we have
AB || CD || EF
PQ || RS
∠RQD = 25°
∠CQP = 60°
PQ || RS.
As we know,
If a transversal intersects two parallel lines, then each pair of alternate exterior angles is equal.
Now, since, PQ || RS
⇒ ∠PQC = ∠BRS
We have ∠PQC = 60°
⇒ ∠BRS = 60° … eq.(i)
We also know that,
If a transversal intersects two parallel lines, then each pair of alternate interior angles is equal.
Now again, since, AB || CD
⇒ ∠DQR = ∠QRA
We have ∠DQR = 25°
⇒ ∠QRA = 25° … eq.(ii)
Using linear pair axiom,
We get,
∠ARS + ∠BRS = 180°
⇒ ∠ARS = 180° – ∠BRS
⇒ ∠ARS = 180° – 60° (From (i), ∠BRS = 60°)
⇒ ∠ARS = 120° … eq.(iii)
Now, ∠QRS = ∠QRA + ∠ARS
From equations (ii) and (iii), we have,
∠QRA = 25° and ∠ARS = 120°
Hence, the above equation can be written as:
∠QRS = 25° + 120°
⇒ ∠QRS = 145°
