1. In Fig. 6.13, lines AB and CD intersect at O. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.
solution :
∠AOC + ∠COE + ∠BOE = 180°[ Linear pair of angels ]
But,
∠AOC + ∠BOE = 70° [ Given ]
∠BOD = 40° [ Given ]
Therefore,
∠AOC + ∠BOE + ∠COE = 180°
⇒ 70° + ∠COE = 180°[ ∴ ∠AOC + ∠BOC = 70° (Given) ]
⇒ ∠COE = 180° − 70°
⇒ ∠COE = 110°
Now,
∠AOC = ∠BOC [ Vertically opposite angels ]
⇒ ∠AOC = 40°[ ∴ ∠BOC = 40° (Given) ]
Similarly,
∠AOC + ∠BOE + ∠COE = 180°
⇒ 40° + ∠BOE + 110° = 180°
⇒ 150° + ∠BOE = 180°
⇒ ∠BOE = 180° − 150°
⇒ ∠BOE = 30°
So,
∠BOE = 30° and ∠COE = 110°
2. In Fig. 6.14, lines XY and MN intersect at O. If ∠POY = 90° and a : b = 2 : 3, find c.
solution :
b + a + ∠POY = 180°[ Linear pair of angels ]
But,
a : b = 2 : 3[ Given ]
∠POY = 90° [ Given ]
Therefore,
b + a + ∠POY = 180°
⇒ b + a + 90° = 180°
⇒ b + a = 180° − 90°
⇒ b + a = 90°
Then,
a : b = 2 : 3[ Given ]
` \a = \frac{2} {5} × 90° \ `
⇒ ` \a = 2 × 18° \ `
⇒ ` \a = 36° \ `
Similarly,
a : b = 2 : 3[ Given ]
` \b = \frac{3} {5} × 90° \ `
⇒ ` \b = 3 × 18° \ `
⇒ ` \b = 54° \ `
Now,
∠MOY = ` \c \ `[ Vertically opposite angels ]
Therefore,
∠MOY = a + ∠POY
⇒ ∠MOY = 36° + 90°
⇒ ∠MOY = 126°
∴ c = 126°
So,
c = 126°
3. In Fig. 6.15, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.
solution :
∠PQR = ∠PRQ [ Given ]
∠PQR + ∠PQS = 180°--------(i)[ Linear pair of angles ]
Similarly,
∠PRQ + ∠PRT = 180°--------(ii)[ Linear pair of angles ]
From (i) and (ii),
We have,
∠PQR + ∠PQS = ∠PRQ + ∠PRT
But,
∠PQR + ∠PRQ [ Given ]
∴ ∠PQS = ∠PRT
4. In Fig. 6.16, if ` \x+y = w+z\ `, then prove that AOB is a line.
solution :
We know,
The angles around a point are 360°.
` \x+y+z+w=360°\ `
But,
` \x+y = w+z \ `[ Given ]
So,
` \(x+y)+(x+y)=360°\ `
⇒ ` \2(x+y)=360°\ `
⇒ ` \(x+y)=\frac{360°} {2}\ `
⇒ ` \(x+y)=180°\ `
⇒ ` \x+y=180°\ `
Now,
I say AOB is a line.
5. In Fig. 6.17, POQ is a line. Ray OR is
perpendicular to line PQ. OS is another ray lying between rays OP and
OR. Prove that
` \∠ROS=\frac{1} {2}(∠QOS-∠POS)\
`.
solution :
∠POS + ∠ROS + ∠ROQ = 180°[ Linear pairs of angels ]
But,
∠ROQ = 90°[ Given ]
Then,
∠POS + ∠ROS + ∠ROQ = 180°[ Linear pairs of angels ]
∠POS + ∠ROS + 90° = 180°
⇒ ∠POS + ∠ROS = 180° − 90°
⇒ ∠POS + ∠ROS = 90°--------(i)
Now,
∠QOS = ∠ROQ + ∠ROS
⇒ ∠QOS = 90° + ∠ROS
⇒ ∠QOS − ∠ROS = 90°--------(ii)
From (i) and (ii),
We have,
∠POS + ∠ROS = ∠QOS − ∠ROS
⇒ ∠POS + ∠ROS + ∠ROS = ∠QOS
⇒ ∠POS + 2 ∠ROS = ∠QOS
⇒ 2 ∠ROS = ∠QOS − ∠POS
⇒` \ ∠ROS = \frac {∠QOS − ∠POS} {2} \ `
So,
` \∠ROS = \frac{1} {2} (∠QOS − ∠POS)\ `
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.
solution :
∠XYZ + ∠ZYQ + ∠QYP = 180°[ Linear pairs of angels ]
⇒ 64° + ∠ZYQ + ∠QYP = 180°[ ∴ ∠XYZ = 64° (Given) ]
⇒ ∠ZYQ + ∠QYP = 180° − 64°
⇒ ∠ZYQ + ∠QYP = 116°
⇒ ∠ZYQ + ∠ZYQ = 116°[ YQ bisects ∠ZYP So, ∠QYP = ∠ZYQ ]
⇒ 2 ∠ZYQ = 116°
⇒ ∠ZYQ = ` \frac{116°} {2} \ `
⇒ ∠ZYQ = 58°
So,
∴ ∠QYP = 58°
∴ Reflex ∠QYP = 360° − 58° = 302°
Now,
∠XYQ = ∠XYZ + ∠ZYQ
⇒ ∠XYQ = 64° + ∠QYP[ ∠XYZ = 64° (Given) ∠QYP = ∠ZYQ ]
⇒ ∠XYQ = 64° + 58° = 122° [ ∠QYP = 58° ]
Thus,
∠XYQ = 122° and reflex ∠QYP = 302°
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