## Chapter 4 – Angles and Pairs of Angles

## Practice Set 15

**1. Observe the figure and complete the table for ∠AWB.**

**Ans:**

**2. Name the pairs of adjacent angles in the figures below.**

**(i)**

**Ans: **

∠ANB and ∠ANC

∠BNA and ∠BNC

∠ANC and ∠BNC

**(ii)**

**Ans:** ∠PQR and ∠PQT

**3. Are the following pairs adjacent angles? If not, state the reason.**

**(i) ∠PMQ and ∠RMQ **

**Ans:** ∠PMQ and ∠RMQ are adjacent angles.

**(ii) ∠RMQ and∠SMR**

**Ans:** ∠RMQ and ∠SMR not adjacent angles since they do not have separate interiors.

**(iii) ∠RMS and ∠RMT **

**Ans:** ∠RMS and ∠RMT not adjacent angles since they do not have separate interiors.

**(iv) ∠SMT and ∠RMS**

**Ans:** ∠SMT and ∠RMS are adjacent angles.

## Practice set 16

**1. The measures of some angles are given below. Write the measures of their complementary angles.**

**(i) 40° **

**Solution:**

Let the measure of the complementary angle be x°.

∴ 40 + x = 90

∴ x = 90 – 40

∴ x = 50

**Ans:** The measure of the complement of an angle of measure 40° is 50°.

**(ii) 63° **

**Solution:**

Let the measure of the complementary angle be x°.

∴ 63 + x = 90

∴ x = 90 – 63

∴ x = 27

**Ans:** The measure of the complement of an angle of measure 63° is 27°.

**(iii) 45° **

**Solution:**

Let the measure of the complementary angle be x°.

∴ 45 + x = 90

∴ x = 90 – 45

∴ x = 45

**Ans:** The measure of the complement of an angle of measure 45° is 45°.

**(iv) 55° **

**Solution:**

Let the measure of the complementary angle be x°.

∴ 55 + x = 90

∴ x = 90 – 55

∴ x = 35

**Ans:** The measure of the complement of an angle of measure 55° is 35°.

**(v) 20° **

**Solution:**

Let the measure of the complementary angle be x°.

∴ 20 + x = 90

∴ x = 90 – 20

∴ x = 70

**Ans:** The measure of the complement of an angle of measure 20° is 70°.

**(vi) 90° **

**Solution:**

Let the measure of the complementary angle be x°.

∴ 90 + x = 90

∴ x = 90 – 90

∴ x = 0

**Ans:** The measure of the complement of an angle of measure 90° is 0°.

**(vii) x°**

**Solution:**

Let the measure of the complementary angle be a°.

∴ x + a = 90

∴ a = (90 – x)⁰

**Ans:** The measure of the complement of an angle of measure x° is (90 – x)°.

**2. (y – 20)° and (y + 30)° are the measures of complementary angles. Find the measure of each angle.**

**Solution:**

(y – 20)° and (y + 30)° are the measures of complementary angles.

∴ (y – 20) + (y + 30) = 90

∴ y – 20 + y + 30 = 90

∴ y + y + 30 – 20 = 90

∴ 2y + 10 = 90

∴ 2y = 90 – 10

∴ 2y = 80

∴ y = \(\large \frac {80}{2}\)

∴ y = 40

∴ (y – 20)° = (40 – 20)° = 20°

and (y + 30)° = (40 + 30)° = 70°

**Ans:** The measure of the two angles is 20° and 70°.

## Practice set 17

**1. Write the measures of the supplements of the angles given below.**

**(i) 15° **

**Solution: **

Let the measure of the supplementary angle be x°.

∴ 15 + x = 180

∴ x = 180 – 15

∴ x = 165

**Ans:** The measure of the supplement of an angle of 15° is 165°.

**(ii) 85° **

**Solution: **

Let the measure of the supplementary angle be x°.

∴ 85 + x = 180

∴ x = 180 – 85

∴ x = 95

**Ans:** The measure of the supplement of an angle of 85° is 95°.

**(iii) 120°**

**Solution: **

Let the measure of the supplementary angle be x°.

∴ 120 + x = 180

∴ x = 180 – 120

∴ x = 60

**Ans:** The measure of the supplement of an angle of 120° is 60°.

**(iv) 37° **

**Solution: **

Let the measure of the supplementary angle be x°.

∴ 37 + x = 180

∴ x = 180 – 37

∴ x = 143

**Ans:** The measure of the supplement of an angle of 37° is 143°.

**(v) 108° **

**Solution: **

Let the measure of the supplementary angle be x°.

∴ 108 + x = 180

∴ x = 180 – 108

∴ x = 72

**Ans:** The measure of the supplement of an angle of 108° is 72°.

**(vi) 0° **

**Solution: **

Let the measure of the supplementary angle be x°.

∴ 0 + x = 180

∴ x = 180

**Ans:** The measure of the supplement of an angle of 0° is 180°.

**(vii) a°**

**Solution: **

Let the measure of the supplementary angle be x°.

∴ a + x = 180

∴ x = (180 – a)⁰

**Ans:** The measure of the supplement of an angle of a° is (180 – a)⁰.

**2. The measures of some angles are given below. Use them to make pairs of complementary and supplementary angles.**

**m∠B = 60° **

**m∠N = 30° **

**m∠Y = 90° **

**m∠J = 150°**

**m∠D = 75° **

**m∠E = 0° **

**m∠F = 15° **

**m∠G = 120°**

**Solution: **

(i) m∠B + m∠N

= 60° + 30°

= 90°

∴ ∠B and ∠N are a pair of complementary angles.

(ii) m∠Y + m∠E

= 90° + 0°

= 90°

∴ ∠Y and ∠E are a pair of complementary angles.

(iii) m∠D + m∠F = 75° + 15°

= 90°

∴ ∠D and ∠F are a pair of complementary angles.

(iv) m∠B + m∠G

= 60° + 120°

= 180°

∴ ∠B and ∠G are a pair of supplementary angles.

(v) m∠N + m∠J

= 30° + 150°

= 180°

∴ ∠N and ∠J are a pair of supplementary angles.

**3. In ∆XYZ, m∠Y = 90°. What kind of a pair do ∠X and ∠Z make?**

**Solution: **

In ∆ XYZ,

m∠X + m∠Y + m∠Z = 180° …*[Sum of the measure of the angles of a triangle is 180°]*

∴ m∠X + 90 + m∠Z = 180

∴ m∠X + m∠Z = 180 – 90

∴ m∠X + m∠Z = 90°

**Ans:** ∠X and ∠Z make a pair of complementary angles.

**4. The difference between the measures of the two angles of a complementary pair is 40°. Find the measures of the two angles.**

**Solution: **

Let the measure of one angle be x°.

∴ Measure of other angle = (90 – x)⁰

According to the given condition,

x – (90 – x) = 40

∴ x – 90 + x = 40

∴ 2x – 90 = 40

∴ 2x = 40 + 90

∴ 2x = 130

∴ x = \(\large \frac {130}{2}\)

∴ x = 65⁰

∴ (90 – x)⁰ = 90 – 65 = 25⁰

**Ans:** The measures of the two angles are 65° and 25°.

**5. □ PTNM is a rectangle. Write the names of the pairs of supplementary angles.**

**Solution: **

Each angle of the rectangle is 90°.

∴ The pairs of supplementary angles are:

(i) ∠P and ∠M

(ii) ∠P and ∠N

(iii) ∠P and ∠T

(iv) ∠M and ∠N

(v) ∠M and ∠T

(vi) ∠N and ∠T

**6. If m∠A = 70°, what is the measure of the supplement of the complement of ∠A?**

**Solution: **

Let the measure of the complement of ∠A be x°

m∠A + x = 90°

∴ 70 + x = 90

∴ x = 90 – 70

∴ x = 20

Let the measure of the supplementary of the complementary angle be y°.

Since, x and y are supplementary angles.

∴ x + y = 180

∴ 20 + y = 180

∴ y = 180 – 20

∴ y = 160⁰

**Ans:** The measure of supplement of the complement of ∠A is 160°.

**7. If ∠A and ∠B are supplementary angles and m∠B = (x + 20)°, then what would be m∠A?**

**Solution: **

∠A and ∠B are supplementary angles.

∴ m∠A + m∠B = 180

∴ m∠A + (x + 20) = 180

∴ m∠A + x = 180 – 20

∴ m∠A + x = 160

∴ m∠A = (160 – x)°

**Ans:** The measure of ∠A is (160 – x)°.

## Practice set 18

**1. Name the pairs of opposite rays in the figure alongside.**

**Ans: **

Ray PN and ray PT

Ray PL and ray PM

**2. Are the ray PM and PT opposite rays? Give reasons for your answer.**

**Ans:** No, Ray PM and PT are not opposite rays. Ray PM and Ray PT do not form a straight line and hence are not opposite rays.

## Practice set 19

#### Draw the pairs of angles as described below. If that is not possible, say why

**(i) Complementary angles that are not adjacent.****Ans:**

**(ii) Angles in a linear pair which are not supplementary. **

**Ans:** Sum of angles in a linear pair is 180° i.e. they are supplementary .

∴ Angles in a linear pair which are not supplementary cannot be drawn.

**(iii) Complementary angles that do not form a linear pair. **

**Ans: **

**(iv) Adjacent angles which are not in a linear pair.**

**Ans:**

**(v) Angles which are neither complementary nor adjacent. ****Ans:**

**(vi) Angles in a linear pair which are complementary.**

**Ans:** Sum of angles in linear pair is 180°, but the sum of complementary angles is 90°.

∴ Angles in a linear pair which are complementary cannot be drawn.

## Practice set 20

**1. Lines AC and BD intersect at point P. m∠APD = 47°. Find the measures of ∠APB, ∠BPC, ∠CPD.**

**Solution: **

∠APD and ∠APB are angles in a linear pair.

∴ m∠APD + m∠APB = 180°

∴ 47 + m∠APB = 180

∴ m∠APB = 180 – 47

∴ m∠APB = 133°

Now,

m∠CPD ≅ m∠APB …*[Vertically opposite angles]*

∵ m∠APB = 133°

∴ m∠CPD = 133⁰

m∠BPC ≅ m∠APD = 47° …*[Vertically opposite angles]*

∵ m∠APD = 47°

∴ m∠BPC = 47⁰

**Ans:** The measures of ∠APB, ∠BPC and ∠CPD are 133°, 47° and 133° respectively.

**2. Lines PQ and RS intersect at point M. m∠PMR = x°. What are the measures of ∠PMS, ∠SMQ and ∠QMR?**

**Solution: **

∠PMR and ∠PMS are angles in a linear pair.

∴ m∠PMR + m∠PMS = 180°

∴ x + m∠PMS = 180

∴ m∠PMS = (180 – x)°

Now,

m∠QMR ≅ m∠PMS …*[Vertically opposite angles]*

∵ m∠PMS = (180 – x)°

∴ m∠QMR = (180 – x)°

m∠SMQ ≅ m∠PMR …*[Vertically opposite angles]*

∵ m∠PMR = x

∴ m∠SMQ = x

**Ans:** The measures of ∠PMS, ∠SMQ and ∠QMR are (180 – x)°, x° and (180 – x)° respectively.

## Practice set 21

**1. ∠ACD is an exterior angle of ∆ABC. The measures of ∠A and ∠B are equal. If m∠ACD = 140°, find the measures of the angles ∠A and ∠B.**

**Solution:**

Let the measure of ∠A be x°.

∴ m∠A = m∠B = x°

∠ACD is the exterior angle of ∆ABC

∴ m∠ACD = m∠A + m∠B

∴ 140 = x + x

∴ 2x = 140

∴ x = \(\large \frac {140}{2}\)

∴ x = 70

∴ m∠A = m∠B = x° = 70⁰

**Ans:** The measures of the angles ∠A and ∠B is 70° each.

**2. Using the measures of the angles given in the figure alongside, find the measures of the remaining three angles.**

**Solution:**

m∠EOD ≅ m∠AOB …*[Vertically opposite angles]*

∵ m∠AOB = 8y

∴ m∠EOD = 8y

m∠AOF ≅ m∠COD …*[Vertically opposite angles]*

∵ m∠COD = 6y

∴ m∠AOF = 6y

m∠BOC ≅ m∠FOE …*[Vertically opposite angles]*

∵ m∠FOE = 6y

∴ m∠BOC = 6y

∠FOL, ∠EOD and ∠COD form a straight angle.

∴ m∠FOE + m∠EOD + m∠COD = 180°

∴ 4y + 8y + 6y = 180

∴ 18y = 180

∴ y = \(\large \frac {180}{18}\)

∴ y = 10

m∠EOD = 8y = 8 × 10 = 80°

m∠AOF = 6y = 6 × 10 = 60°

m∠BOC = 4y = 4 × 10 = 40°

**Ans:** The measures of ∠EOD, ∠AOF and ∠BOC are 80°, 60° and 40° respectively.

**3 . In the isosceles triangle ABC, ∠A and ∠B are equal. ∠ACD is an exterior angle of ∆ABC. The measures of ∠ACB and ∠ACD are (3x – 17)° and (8x + 10)° respectively. Find the measures of ∠ACB and ∠ACD. Also find the measures of ∠A and ∠B.**

**Solution:**

Let the measure of ∠A be y°

∴ m∠A = m∠B = y°

∠ACB and ∠ACD form a pair of linear angles.

∴ m∠ACB + m∠ACD = 180°

∴ (3x – 17) + (8x + 10) = 180

∴ 3x + 8x – 17 + 10 = 180

∴ 11x – 7 = 180

∴ 11x = 180 + 7

∴ 11x = 187

∴ x = \(\large \frac {187}{11}\)

∴ x = 17

m∠ACB = 3x – 17

∴ m∠ACB = (3 × 17) – 17

∴ m∠ACB = 51 – 17

∴ m∠ACB = 34°

m∠ACD = 8x + 10

∴ m∠ACD = 8 × 17 + 10

∴ m∠ACD = 136 + 10

∴ m∠ACD = 146°

Here ∠ACD is the exterior angle of ∆ABC and ∠A and ∠B are its remote interior angles.

∴ m∠ACD = m∠A + m∠B

∴ 146 = y + y

∴ 146 = 2y

∴ 2y = 146

∴ y = \(\large \frac {146}{2}\)

∴ y = 73⁰

**Ans:** The measures of ∠ACB, ∠ACD, ∠A and ∠B are 34°, 146°, 73° and 73° respectively.