1. In Fig. 6.13, lines AB and CD intersect at O. If ∠AOC +∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.
Ans:
From the diagram, we have
(∠AOC +∠BOE +∠COE) and (∠COE +∠BOD +∠BOE) forms a straight line.
So, ∠AOC+∠BOE +∠COE = ∠COE +∠BOD+∠BOE = 180°
Now, by putting the values of ∠AOC + ∠BOE = 70° and ∠BOD = 40° we get
∠COE = 110° and ∠BOE = 30°
So, reflex ∠COE = 360o – 110o = 250o
2. In Fig. 6.14, lines XY and MN intersect at O. If ∠POY = 90° and a : b = 2 : 3, find c.
Ans:
We know that the sum of linear pair is always equal to 180°
So,
∠POY +a +b = 180°
Putting 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°
3. In Fig. 6.15, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.
Solution:
Since ST is a straight line, so
∠PQS+∠PQR = 180° (linear pair) and
∠PRT+∠PRQ = 180° (linear pair)
Now, ∠PQS + ∠PQR = ∠PRT+∠PRQ = 180°
Since ∠PQR =∠PRQ (as given in the question)
∠PQS = ∠PRT. (Hence proved).
4. In Fig. 6.16, if x+y = w+z, then prove that AOB is a line.
To prove AOB is a straight line, we will have to prove x+y is a linear pair
i.e. x+y = 180°
We know that the angles around a point are 360°, so
x+y+w+z = 360°
In the question, it is given that,
x+y = w+z
So, (x+y)+(x+y) = 360°
2(x+y) = 360°
∴ (x+y) = 180° (Hence proved).