CBSE - Class 9 - Maths - Lines and Angles (Ex. 6.1)

NCERT Excercise 6.1

Q1: In Fig. 6.13, lines AB and CD intersect at O. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.

Answer:  AB is a straight line and rays OC and OE meets at point O on AB. 

    ∠AOC + ∠COE + ∠BOE =  180°

⇒  70° +  ∠COE = 180°                   (∵ ∠AOC + ∠BOE = 70°, Given)

⇒  ∠COE = 180°  - 70° = 110°

reflex  ∠COE = 360° - 110° =  250°

∵ CD is a straight line and rays OE and OB stand on it.

⇒  ∠COE + ∠BOE + ∠BOD =  180° 

⇒  ∠BOE =   180°  - (∠COE + ∠BOD) =  180° - (110° + 40°) 

⇒  ∠BOE = 180° - 150° = 30°

Q2: In Fig. 6.14, lines XY and MN intersect at O. If ∠POY = 90° and a : b = 2 : 3, find c.

Answer:  Let ∠POM = a = 2x   and ∠XOM = b = 3x  such that a:b = 2:3.

∵ XY is a straight line and rays OM and OP stand on it.

⇒ ∠XOM + ∠POM + ∠POY = 180°

⇒  b + a + 90° = 180°

⇒ 3x + 2x =  180° - 90°

⇒ 5x = 90° 

⇒ x = 90° / 5 = 18°

∴ a = 2 × 18° = 36°  and b =  3 ×  18° = 54°

Since MN is a straight line and XY intersects it.  ∠XOM and ∠XON forms a linear pair.

⇒  ∠XOM +  ∠XON = 180°

⇒  b + c = 180° 

⇒ c = 180° - b = 180° - 54° = 126°

Q3: In Fig. 6.15, ∠ PQR = ∠ PRQ, then prove that ∠ PQS = ∠ PRT. 

Answer:  Line ST is a straight line and PQ meets it at point Q. ∴ ∠PQS and ∠PQR form a linear pair.

⇒ ∠PQS + ∠PQR = 180°

⇒ ∠PQR =  180° -  ∠PQS                                                  ...(I)

Similarly, line PR meets  line ST at R. ∴ ∠PRQ and ∠PRT form a linear pair.

⇒ ∠PRQ + ∠PRT = 180°

⇒ ∠PRQ =  180° -  ∠PRT                                                  ...(II)

Since  ∠PQR = ∠PRQ                                      (Given)

From equations I and II

180° -  ∠PQS = 180° -  ∠PRT

⇒ ∠PQS = ∠PRT 

Q4: In Fig. 6.16, if x + y = w + z, then prove that AOB is a line.

Answer: At point O, angles x, y, z and w makes a complete angle i.e. 360°

⇒  x + y + z + w = 360°

Since x + y = z + w

⇒ x + y + x + y = 360°

⇒ 2x + 2y = 360°

⇒ x + y = 180°

Since x and y form a linear pair, thus AOB is a straight line.

Q5: 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 ∠ROS = ½ (∠QOS – ∠POS).

Answer: Since OR ⊥ PQ, 

∴ ∠POR = 90°, ∠QOR = 90° and ∠POS and ∠SOR are supplementary. 

⇒ ∠POS + ∠ROS = 90°

⇒  ∠ROS = 90° - ∠POS                                    ... (I)

Since ∠QOR = 90°

Adding ∠ROS to both sides,

⇒  ∠QOR + ∠ROS =  90° +  ∠ROS

⇒  ∠QOS  =  90° +  ∠ROS 

⇒  ∠ROS  =  ∠QOS - 90°                                 ... (II)

Adding equation I and II,

 ∠ROS +  ∠ROS = 90° - ∠POS  + ∠QOS - 90°

⇒ 2 × ∠ROS = ∠QOS - ∠POS

⇒ ∠ROS = ½ (∠QOS – ∠POS)

Q6: 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: YQ bisects ∠PYZ

⇒ ∠PYQ = ∠QYZ.

PX is a straight line. Ray QY and ZY stand on it.

⇒ ∠PYQ + ∠QYZ + ∠ZYX = 180°

⇒ 2 ∠PYQ + 64° = 180°

⇒ 2 ∠PYQ =  180° - 64° = 116°

⇒ ∠PYQ =  116° / 2 = 58°

∴ ∠PYQ = ∠QYZ = 58°

∠XYQ = ∠QYZ + ∠ZXY =  58° + 64° = 122°

Reflex ∠QYP = 360° -  ∠QYP = 360° - 58° = 302°