RD Chapter 8- Lines and Angles Ex-8.2 |
RD Chapter 8- Lines and Angles Ex-8.3 |
RD Chapter 8- Lines and Angles Ex-8.4 |
RD Chapter 8- Lines and Angles Ex-VSAQS |

Write thecomplement of each of the following angles:

(i)20^{0}

(ii)35^{0}

(iii)90^{0}

(iv) 77^{0}

(v)30^{0}

**Answer
1** :

(i) Thesum of an angle and its complement = 90^{0}

Therefore, the complement of 20^{0 }= 90^{0} –20^{0} = 70^{0}

(ii) Thesum of an angle and its complement = 90^{0}

Therefore, the complement of 35° = 90° – 35° = 55

(iii) Thesum of an angle and its complement = 90^{0}

Therefore, the complement of 90^{0} = 90^{0} –90^{0} = 0^{0}

(iv) Thesum of an angle and its complement = 90^{0}

Therefore, the complement of 77^{0} = 90° – 77^{0} = 13^{0}

(v) Thesum of an angle and its complement = 90^{0}

Therefore, the complement of 30^{0} = 90^{0} –30^{0} = 60^{0}

Write thesupplement of each of the following angles:

(i) 54^{0}

(ii) 132^{0}

(iii) 138^{0}

**Answer
2** :

(i) Thesum of an angle and its supplement = 180^{0}.

Therefore supplement of angle 54^{0} = 180^{0} –54^{0} = 126^{0}

(ii) Thesum of an angle and its supplement = 180^{0}.

Therefore supplement of angle 132^{0} = 180^{0} –132^{0} = 48^{0}

(iii) Thesum of an angle and its supplement = 180^{0}.

Therefore supplement of angle 138^{0 }= 180^{0 }–138^{0} = 42^{0}

If an angle is 28^{0} lessthan its complement, find its measure?

**Answer
3** :

Let the measure of any angle is ‘ a ‘ degrees

Thus, its complement will be (90 – a)^{ 0}

So, the required angle = Complement of a – 28

a = ( 90 – a ) – 28

2a = 62

a = 31

Hence, the angle measured is 31^{0}.

If an angle is 30°more than one half of its complement, find the measure of the angle?

**Answer
4** :

Let an angle measured by ‘ a ‘ in degrees

Thus, its complement will be (90 – a)^{ 0}

Required Angle = 30^{0} +complement/2

a = 30^{0} + (90 – a )^{ 0} / 2

a + a/2 = 30^{0} + 45^{0}

3a/2 = 75^{0}

a = 50^{0}

Therefore, the measure of required angle is 50^{0}.

Two supplementaryangles are in the ratio 4:5. Find the angles?

**Answer
5** :

Two supplementary angles are in the ratio 4:5.

Let us say, the angles are 4a and 5a (in degrees)

Since angle are supplementary angles;

Which implies, 4a + 5a = 180^{0}

9a = 180^{0}

a = 20^{0}

Therefore, 4a = 4 (20) = 80^{0 }and

5(a) = 5 (20) = 100^{0}

Hence, required angles are 80° and 100^{0}.

Two supplementaryangles differ by 48^{0}. Find the angles?

**Answer
6** :

Given: Two supplementary angles differ by 48^{0}.

Consider a^{0} beone angle then its supplementary angle will be equal to (180 – a)^{ 0}

According to the question;

(180 – a ) – x = 48

(180 – 48 ) = 2a

132 = 2a

132/2 = a

Or a = 66^{0}

Therefore, 180 – a = 114^{0}

Hence, the two angles are 66^{0} and114^{0}.

An angle is equalto 8 times its complement. Determine its measure?

**Answer
7** :

Given: Required angle = 8 times of its complement

Consider a^{0} beone angle then its complementary angle will be equal to (90 – a)^{ 0}

According to the question;

a = 8 times of its complement

a = 8 ( 90 – a )

a = 720 – 8a

a + 8a = 720

9a = 720

a = 80

Therefore, the required angle is 80^{0}.

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