Question

7 standard exercise 6.1

Answer 1

1-It is given: ∠AOC + ∠BOE = 70o
And, ∠BOD = 40o 
Now, according to the question,
∠AOC + ∠BOE + ∠COE = 180o (Sum of linear pair is always 180o )
⇒ 70o + ∠COE = 180o
⇒ ∠COE = 110o     ...........(1)
And, 
∠COE + ∠BOD + ∠BOE = 180o  (Sum of linear pair is always 180o )
putting the value of ∠COE, we get, 
110o + 40o + ∠BOE = 180o 
150o + ∠BOE = 180o 
∴ ∠BOE = 30o 
Now, reflex ∠COE = 360o - ∠COE
⇒ reflex ∠COE = 360o - 110o
∴ reflex ∠COE = 250o


2-It is given in the question that:
∠POY = 90O
And,
a : b = 2 : 3
Now, according to the question,
∠POY + a + b = 180o
90o + a + b = 180o
a + b = 90o
Let a be 2x and b as 3x
2x + 3x = 90o
5x = 90o
x = 18o
Hence,
a = 2 * 18o = 36o
And,
b = 3 * 18o = 54o
And,
b + c = 180o (Linear pair)
54o + c = 180o
c = 126o



3-∠PQR + ∠PQS = 180o (Linear angles pair)
⇒ ∠PQS = 180o - ∠PQR ...(i)
Similarly,
∠PRQ + ∠PRT = 180o (Linear angles pair)
⇒ ∠PRT = 180o - ∠PRQ 
Now, ∠PQR = ∠PRQ (Given)
∠PRQ = 180o - ∠PQR ..,(ii)
Using (i) and (ii), we get
∠PQS = ∠PRT = 180o - ∠PQR
Therefore,
∠PQS = ∠PRT
Hence, proved



4-It is given in the question that:
x + y = w + z

To prove,

AOB is a line or x + y = 180° (Linear pair)

Now, according to the question,

x + y + w + z = 360o  (Angles at a point)

(x + y) + (w + z) = 360°   

(x + y) + (x + y) = 360o (Given, x + y = w + z)

2 (x + y) = 360o

(x + y) = 180o

Therefore, x + y makes a linear pair.

And, AOB is a straight line



5-It is given in the question that:
OR is perpendicular to line PQ

To prove,

∠ROS =  (∠QOS - ∠POS)

Now, according to the question,

∠POR = ∠ROQ = 90°   ( ∵ OR is perpendicular to line PQ)

∠QOS = ∠ROQ + ∠ROS = 90° + ∠ROS (i)

∠POS = ∠POR - ∠ROS = 90° - ∠ROS (ii)

Subtracting (ii) from (i), we get

∠QOS - ∠POS = 90o + ∠ROS – (90° - ∠ROS)

∠QOS - ∠POS = 90o + ∠ROS – 90° + ∠ROS

∠QOS - ∠POS = 2∠ROS



6-It is given in the question that:
∠XYZ = 64o
YQ bisects ∠ZYP
∠XYZ + ∠ZYP = 180o (Linear pair)
64o + ∠ZYP = 180o
∠ZYP = 116o
And,
∠ZYP = ∠ZYQ + ∠QYP
∠ZYQ = ∠QYP (YQ bisects ∠ZYP)
∠ZYP = 2∠ZYQ
2∠ZYQ = 116o
∠ZYQ = 58o = ∠QYP
Now,
∠XYQ = ∠XYZ + ∠ZYQ
∠XYQ = 64o + 58o
∠XYQ = 122o
And,
Reflex of ∠QYP = 180o + ∠XYQ
∠QYP = 180o + 122o
∠QYP = 302o