Chapter 5: Lines and Angles

Solution:-

(c) 80^o

From the figure POQ is a straight line,

Then, 100 + a = 180^o

a = 180^o – 100

a= 80^o

Solution:-

(d) equal

Solution:-

(a) 20^o

Given, a = 40^o

Then, 2a = 2 × 40 = 80^o

From the figure, angles formed on the straight line are equal to 180^o,

Then, 5b + 2a = 180^o

5b + 80^o = 180^o

5b= 180^o – 80^o

5b = 100^o

b = 100/5

b = 20^o

Solution:-

(a) 100^o

Let us assume the angle be P.

Then, its supplement is 180^o – P

As per the condition in the question,

P = 2(180^o – P) – 60^o

P = 360^o – 2P – 60^o

P + 2P = 300^o

3P = 300^o

P = 300/3

P = 100^o

So, its supplement is 180^o – P = 180^o – 100^o = 80^o

Therefore, the greater angle is 100^o.

Solution: –

(c) (108–b)^o

From the question it is given that, ∠1 = (2a + b)^o and ∠6 = (3a – b)^o

Since ∠5 and ∠6 form a linear pair of angles

Then,

∠5 = (180-3a + b)^o … [equation 1]

∠5 = ∠1 = (180-3a + b)^o [Because Corresponding angles] …equation (2)

From equation (2) we get,

2a + b = 180-3a + b

5a = 180

a = 36⁰

Since ∠1 and ∠2 forms a linear pair so

∠2 = 180⁰– 2a-b

Substituting the value of a

∠2 = 180⁰– 72⁰ – b

∠2 = 108⁰– b

Solution: –

(b) 108^o

From the figure, PQ||RS.

From the question it is given that, a: b = 3: 2

So, let us assume a = 3m and b = 2m

We know that, sum of angles on the straight line is equal to 180^o

Then, ∠a + ∠b = 180^o

3m + 2m = 180^o

5m = 180^o

m = 180^o/5

m = 36^o

So, a = 3m = 3 × 36^o = 108^o

b = 2m = 2 × 36^o = 72

Therefore, ∠a = ∠f = 108^o [because corresponding angles]

Solution:-

(d) ∠4 = ∠8

Because, ∠4 ≠ ∠8

Solution:-

(d) ∠2 + ∠3 = 180^o

We know that, interior opposite angles are equal

∠2 = ∠3

Solution:-

(c) ∠5 = ∠8

From the figure, PQ||RS

∠5 = ∠8 [interior alternate angles are equal]

Solution: –

(b) 135^o

From the figure, PO is a straight line

We know that, sum of angles on the straight is equal to 180^o.

Then,

y + ∠ PQR = 180⁰

y + 130⁰ = 180⁰

y = 50^o

Then,

∠ QOS = ∠ TSO [Co-interior angle]

x = 85⁰

x + y = 135