EXERCISE 6.1
Introduction- Lines & Angles, Class 9th Maths
Lines & Angles Theorem 6.1 Page No 94, Class 9th Maths
Q 1, Ex 6.1, Page No 96, Lines And Angles, Mathematics Class 9th
Page No 96:
Question 1:
In the given figure, lines AB and CD intersect at O. If $\angle \mathrm{AOC}+\angle \mathrm{BOE}=70^{\circ}$ and $\angle \mathrm{BOD}=40^{\circ}$ , find ∠BOE and reflex ∠COE.
Answer:
AB is a straight line, rays OC and OE stand on it.
∴$\angle \mathrm{AOC}+\angle \mathrm{COE}+\angle \mathrm{BOE}=180^{\circ}$
⇒$(\angle \mathrm{AOC}+\angle \mathrm{BOE})+\angle \mathrm{COE}=180^{\circ}$
⇒$70^{\circ}+\angle \mathrm{COE}=180^{\circ}$
⇒$\angle C O E=180^{\circ}-70^{\circ}=110^{\circ}$
Reflex $\angle \mathrm{COE}=360^{\circ}-110^{\circ}=250^{\circ}$
CD is a straight line, rays OE and OB stand on it.
$\therefore \angle \mathrm{COE}+\angle \mathrm{BOE}+\angle \mathrm{BOD}=180^{\circ}$
$\Rightarrow 110^{\circ}+\angle \mathrm{BOE}+40^{\circ}=180^{\circ}$
$\Rightarrow \angle \mathrm{BOE}=180^{\circ}-150^{\circ}=30^{\circ}$
Q 2, Ex 6.1, Page No 97, Lines And Angles, CBSE Class 9th Maths
Page No 97:
Question 2:
In the given figure, lines XY and MN intersect at O. If ∠POY =
and a:b = 2 : 3, find c.
Answer:
Let the common ratio between a and b be x.∴ a = 2x, and b = 3x
XY is a straight line, rays OM and OP stand on it.
∴ ∠XOM + ∠MOP + ∠POY = 180º
b + a + ∠POY = 180º
3x + 2x + 90º = 180º
5x = 90º
x = 18º
a = 2x = 2 × 18 = 36º
b = 3x= 3 ×18 = 54º
MN is a straight line. Ray OX stands on it.
∴ b + c = 180º (Linear Pair)
54º + c = 180º
c = 180º − 54º = 126º
∴ c = 126º
Q 3, Ex 6.1, Page No 97, Lines And Angles, CBSE Class 9th Maths
Question 3:
In the given figure, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.
Answer:
In the given figure, ST is a straight line and ray QP stands on it.∴ ∠PQS + ∠PQR = 180º (Linear Pair)
∠PQR = 180º − ∠PQS (1)
∠PRT + ∠PRQ = 180º (Linear Pair)
∠PRQ = 180º − ∠PRT (2)
It is given that ∠PQR = ∠PRQ.
Equating equations (1) and (2), we obtain
180º − ∠PQS = 180
− ∠PRT∠PQS = ∠PRT
Q 4, Ex 6.1, Page No 97, Lines And Angles, Mathematics Class 9th
x + y + z + w = 360º (Complete angle)
It is given that,
x + y = z + w
∴ x + y + x + y = 360º
2(x + y) = 360º
x + y = 180º
Since x and y form a linear pair, AOB is a line.
Q 5, Ex 6.1, Page No 97, Lines And Angles, Class 9th Maths
$\angle \mathrm{ROS}=\frac{1}{2}(\angle \mathrm{QOS}-\angle \mathrm{POS})$
∴ ∠POR = 90º
⇒ ∠POS + ∠SOR = 90º
∠ROS = 90º − ∠POS … (1)
∠QOR = 90º (As OR ⊥ PQ)
∠QOS − ∠ROS = 90º
∠ROS = ∠QOS − 90º … (2)
On adding equations (1) and (2), we obtain
2 ∠ROS = ∠QOS − ∠POS
∠ROS = $\frac{1}{2}$(∠QOS − ∠POS)
It is given that line YQ bisects ∠PYZ.
Hence, ∠QYP = ∠ZYQ
It can be observed that PX is a line. Rays YQ and YZ stand on it.
∴ ∠XYZ + ∠ZYQ + ∠QYP = 180º
⇒ 64º + 2∠QYP = 180º
⇒ 2∠QYP = 180º − 64º = 116º
⇒ ∠QYP = 58º
Also, ∠ZYQ = ∠QYP = 58º
Reflex ∠QYP = 360º − 58º = 302º
∠XYQ = ∠XYZ + ∠ZYQ
= 64º + 58º = 122º
Question 4
In the given figure, if x+y=w+z , then prove that AOB is a line.
Answer:
It can be observed that,x + y + z + w = 360º (Complete angle)
It is given that,
x + y = z + w
∴ x + y + x + y = 360º
2(x + y) = 360º
x + y = 180º
Since x and y form a linear pair, AOB is a line.
Q 5, Ex 6.1, Page No 97, Lines And Angles, Class 9th Maths
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$\angle \mathrm{ROS}=\frac{1}{2}(\angle \mathrm{QOS}-\angle \mathrm{POS})$
Answer:
It is given that OR ⊥ PQ∴ ∠POR = 90º
⇒ ∠POS + ∠SOR = 90º
∠ROS = 90º − ∠POS … (1)
∠QOR = 90º (As OR ⊥ PQ)
∠QOS − ∠ROS = 90º
∠ROS = ∠QOS − 90º … (2)
On adding equations (1) and (2), we obtain
2 ∠ROS = ∠QOS − ∠POS
∠ROS = $\frac{1}{2}$(∠QOS − ∠POS)
Q 6, Ex 6.1, Page No 97, Lines And Angles, CBSE Class 9th
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:
Hence, ∠QYP = ∠ZYQ
It can be observed that PX is a line. Rays YQ and YZ stand on it.
∴ ∠XYZ + ∠ZYQ + ∠QYP = 180º
⇒ 64º + 2∠QYP = 180º
⇒ 2∠QYP = 180º − 64º = 116º
⇒ ∠QYP = 58º
Also, ∠ZYQ = ∠QYP = 58º
Reflex ∠QYP = 360º − 58º = 302º
∠XYQ = ∠XYZ + ∠ZYQ
= 64º + 58º = 122º
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