RS Aggarwal Class 7 Math Thirteenth Chapter Lines and Angles Exercise 13 Solution

RS Aggarwal Class 7 Math Thirteenth Chapter Lines and Angles Exercise 13 Solution

EXERCISE 13

(1) Find the complete of each of the following angles:

(i) 35^o

Ans: let the measure of its complement be x^o. Then,

x + 35 = 90

or, x = (90 – 35)

or, x = 55

(ii) 47^o

Ans: let the measure of its complement be x^o. Then,

x + 47 = 90

or, x = (90 – 47)

or, x = 43

(iii) 60^o

Ans: let the measure of its complement be x^o. Then,

x + 60 = 90

or, x = (90 – 60)

or, x = 30

(iv) 73^o

Ans: let the measure of its complement be x^o. Then,

x + 73 = 90

or, x = (90 – 73)

or, x = 27

(2) Find the supplement of each of the following angles:

(i) 80^o

Ans: Let its supplement be x^o. Then,

x + 80 = 180

or, x = (180 – 80)

or, x = 100

(ii) 54^o

Ans: Let its supplement be x^o. Then,

x + 54 = 180

or, x = (180 – 54)

or, x = 126

(iii) 105^o

Ans: Let its supplement be x^o. Then,

x + 105 = 180

or, x = (180 – 105)

or, x = 75

(iv) 123^o

Ans: Let its supplement be x^o. Then,

x + 123 = 180

or, x = (180 – 123)

or, x = 67

(3) Among two supplementary angles, the measure of the larger angle is 36^o more than the measure of e smaller. Find their measures.

Solution: Let one angle is x and other is (x+36).

∴ x + x + 36 = 180

or, 2x = 180 – 36

or, x =

or, x = 72

Hence, required angles are 72^o and (72 + 36) = 108^o.

(4) Find the angle which is equal to its supplement.

Solution: Let the two angles are x and x.

2x = 180

or, x = 180/2

or, x = 90

(5) Can two angles be supplementary if both of them are:

(i) Acute = No

(ii) Obtuse = No

(iii) Right = Yes

(6) In the given figure, AOB is a straight line and the ray OC stands on it. If ∠AOC = 64^o and ∠BOC = x^o, find the value of x.

Solution: By linear property, we have:

∠AOC + ∠BOC = 180^o

or, 64 + x = 180

or, x = 180 – 64

or , x = 116

Hence, ∠BOC = 116^o.

(7) In the given figure, AOB is a straight line and the ray OC and OD stand on it. If ∠AOC = (2x – 10)^o and ∠BOC = (3x + 20)^o, find the value of x.  Also, find ∠AOC and ∠BOC.

Solution: By linear property, we have,

∠AOC + ∠BOC = 180

or, (2x – 10) + (3x + 20) = 180

or, 2x – 10 + 3x + 20 = 180

or, 5x + 10 = 180

or, x = 170/5

or, x = 34

Therefore, ∠AOC = (2×34 -10) = 68 -10 = 58

∠BOC  = (3×34 + 20 ) = 102 + 20 = 122.

(8) In the given figure, AOB is a straight line and the rays OC and OD stand on it. If ∠AOC = 65^o, ∠BOD = 70^o and ∠COD = x^o, find the value of x.

Solution: By linear property, we have,

∠AOC + ∠BOD + ∠COD = 180

or, 65 + 70 + x = 180

or, x = 180 – 135

or, x = 45

(9) In the given figure, two straight lines AB and CD intersect at a point O. If ∠AOC = 42^o, find the measure of each of the angles:

Solution: Since CD is a line and the ray OA stands on it, we have:

∠AOC + ∠AOD = 180^o

or, 42^o + ∠AOD = 180^o

or, ∠AOD = 180^o – 42^o

or, ∠AOD = 138^o

And ∠BOD = ∠AOC = 42^o

∴ ∠COB = ∠AOD = 138^o

(10) In the given figure, two straight lines PQ and RS intersect at O. If ∠POS = 114^o, find the measure of each of the angles:

Solution: Since PQ is a line and the ray RS stands on it, we have:

∠POS + ∠QOS = 180^o

or, 114^o + ∠QOS = 180^o

or, ∠QOS = 180^o – 114^o

or, ∠QOS = 66^o

And ∠POS = ∠ROQ = 114^o

∴ ∠QOS = ∠ROP = 66^o                                           

(11) In the given figure, rays OA, OB, OC and OD are such that ∠AOB = 56^o, ∠BOC = 100^o, ∠COD = x^o and ∠DOA = 74^o. Find the value of x.

Solution: The sum of all angles around a point is 360^o.

Thus, in the given figure, we have:

∠AOB + ∠BOC + ∠COD + ∠DOA = 360^o

or, 56 + 100 + x + 74 = 360

or, x + 230 = 360

or, x = 360 – 230

or, x = 130