Hanna's Teaching Notes on Geometry by Soliman's Mathematics

† Chapter 01, Hanna’s Teaching Notes on Geometry Angles

01 Definition and the method of writing the name of the angle Q•

R When two lines, such as PQ and QR, intersect at point Q then the angle is formed at this point or vertex Q and between the two lines is the angle which is usually named ^ ^ by the point of intersection i.e. angle Q or Q or by three letters PQ R such that the middle letter represents point Q, where the lines PQ and QR intersect.

D # $

" ! • E

If more than two lines intersect at one point then we must name the required angle by three letters or by our own single letter. For example the three lines HD, ED and GD intersect at the point D. ^ ∴ D might be angle α or angle β or angle (α + β) . ^ ∴ α = ED G , ^ β = ED H, ^ Φ = DE G, ^ (α + β) = GD H and ^ θ = DE H

Book 1, Hanna’s Series

† Chapter 01, Hanna’s Teaching Notes on Geometry Angles

02 Types I of Angles Acute Angle • A

^ Angle A or A is an Acute Angle The acute angle is always between 0˚ and 90˚ ^ 0˚ < A < 90˚ Right Angle

•

^ Angle B or B is a Right Angle The right angle is always = 90˚ ^ B = 90˚ Obtuse Angle

•

^ Angle C or C is an Obtuse Angle The obtuse angle is always between 90˚ and 180˚ ^ 90˚ < C < 180˚ Straight Angle

•

D ^ Angle D or D is a Straight Angle The straight angle is always = 180˚ or = 2 right angles ^ D = 180˚

Book 1, Hanna’s Series

† Chapter 01, Hanna’s Teaching Notes on Geometry Angles

03 Types II of Angles Reflex Angles

The above three angles P, Q and R are all Reflex or bent back Angles They are greater than 180˚ but less than 360˚ 180˚ < Reflex Angle < 360˚ Angle of one turn T•

^ Angle T or T is an Angle of one turn The Angle of one turn is always = 360˚ or = 4 right angles ^ T = 360˚

04 Complementary angles 50˚ 40˚

diagram (a)

diagram (b)

y˚

(90!y)˚

diagram (c)

The two angles that add up to 90˚ are called Complementary Angles. For example, in diagram (a) angle 50˚ is the complement of 40˚, because 50˚ + 40˚ = 90˚, in diagram (b) angle α is the complement of angle β, because α + β = 90˚ and in diagram (c) angle y˚ is the complement of angle (90−y) ˚ because y˚ + (90−y) ˚ = 90˚

Book 1, Hanna’s Series

† Chapter 01, Hanna’s Teaching Notes on Geometry Angles

05 Supplementary angles

110˚ 70˚

k˚

(180!k)˚

diagram (i) diagram (ii) diagram (iii) The two angles that add up to 180˚ are called Supplementary Angles. For example in diagram (i) angle 110˚ is the supplement of 70˚, because 110˚ + 70˚ = 180˚, in diagram (ii) angle φ is the supplement of angle θ, because φ + θ = 180˚ and in diagram (iii) angle k˚ is the complement of angle (180−k) ˚ because k˚ + (180−k) ˚ = 180˚

Example 01 Find, for each of the given diagrams, in its simplest form, the relation between α and β. a) !˚ "˚

α˚ + 90˚ + β˚ = 180˚, (straight angle) ∴ α + β = 180 − 90 ⇒ α + β = 90 b)

4"˚

2"˚ !˚ 3!˚

4β˚ + 2β˚ + 3α˚ + α˚ = 360˚, (angles of οne turn) ∴ 6β + 4α = 360, divide by 2 ⇒ 3β + 2α = 180

Book 1, Hanna’s Series

† Chapter 01, Hanna’s Teaching Notes on Geometry Angles

Example 02 Calculate the value of α in each of the given diagrams. a) 6! 48˚

48˚ + 6α + 5α = 180˚, (straight angle) ∴ 11α = 180˚ − 48˚ = 132˚, divide by 11 ⇒ α = 132˚ ÷ 11 = 12˚ b)

30˚

! 2α + α + 30˚ = 360˚, (angles of οne turn) ∴ 3α= 360˚ − 30˚ = 330˚, divide by 3 ⇒ α = 330˚ ÷ 3 = 110˚ c)

4! 3!

3α + 4α + 3α = 90˚, (right angle) ∴ 9α = 90˚, divide by 10 ⇒ α = 90˚ ÷ 10 ⇒ α = 9˚ d) 5! 3! 80˚+2!

5α + 3α + 2α + 80˚ + 90˚ = 360˚, (angles of οne turn) 10α + 170˚ = 360˚, divide by 10 ∴ α + 17˚ = 36˚ ⇒ α = 19˚

Book 1, Hanna’s Series

† Chapter 01, Hanna’s Teaching Notes on Geometry Angles

05 Vertically opposite angles are equal K

d V c •

b L M a = c (vertically opposite angles) and b = d (vertically opposite angles) Proof ^ a + d = 180˚ the straight angle KVL

(

^ and c + d = 180˚ the straight angle MVN

)

(

)

∴ a=c Similarly ^ a + b = 180˚ the straight angle MVN

(

)

^ and a + d = 180˚ the straight angle KVL

(

)

∴ b=d Notice that ^ ^ In the diagram below, KV L and MV N are not straight angles N K M

d V

• a

∴ a ≠ c and b ≠ d

Example 03 Find the values of x, y and z giving the geometrical reasons for all your work. 3x 5z 2y 80˚+ x

3x = 80˚ + x , (vertically opposite angles) , ⇒ 2x = 80˚ ∴ x = 80˚ ÷ 2 = 40˚ 2y +3x = 180˚, (straight angle) , ∴ 2y = 180˚ − 3 x 40˚ = 180˚ − 120 = 60˚ ∴ y = 60˚ ÷ 2 = 30˚ 5z = 2y, (vertically opposite angles) , ⇒ 5z = 2x30˚ = 60˚ ∴ z = 60˚ ÷ 5 = 12˚

Book 1, Hanna’s Series

† Chapter 01, Hanna’s Teaching Notes on Geometry Angles

Example 04 a) b)

Write down the angle 324˚ as a fraction of a full turn angle. Write down the angle 117˚ as a fraction of a right angle.

a) 324 324÷9 36 324˚ = 360 of a full turn angle = 360÷9 = 40 of a full turn angle 9 ∴ 324˚ = 10 of a full turn angle b) 117 117÷9 17 117˚ = 90 of a right angle = 90÷9 = 10 of a right angle 7 ∴ 117˚ = 1 10 of a right angle

Example 05 Calculate in degrees each of the following angles 11 a) 24 of a full turn b) 4.15 of a straight angle, c) the sum of 5y right angles, 3 straight angles and 2y full turns angles. a) 30 11 15 11 11 11  = 2 x 30  = 11 x 15 24 of a full turn = 24 x 360 = 24  x 360 2

11 ∴ 24 of a full turn = 165˚ b) 4.15 of a straight angle = 4.15 x 180˚ = 41.5 x 18˚ = 41.5 x 2 x 9˚ = 83.0 x 9˚ ∴ 4.15 of a straight angle = 747˚ c) 5y right angles = 5y x 90˚ = 450y˚ 3 straight angles = 3 x 180˚ = 540˚ 2y full turns angles = 2y x 360˚ = 720y˚ ∴ Sum of the three angles = (1170y + 540) ˚

Book 1, Hanna’s Series

† Chapter 01, Hanna’s Teaching Notes on Geometry Angles

Example 06 Find the value of a and of b , given that angles a˚ and b˚ are a) complementary and angle a˚ is five times angle b˚ b) supplementary and angle b˚ is four times angle a˚ a) As a and b are complementary ⇒ a˚ + b˚ = 90˚, 90 90÷6 and since a = 5 b ⇒ 5b + b = 90 ⇒ 6b = 90 ⇒ b = 6 = 6÷6 = 15, ∴ a = 5 b = 5 x 15 = 75 ⇒ a = 75 and b = 15 b) As a and b are supplementary ⇒ a˚ + b˚ = 180˚, 180 180÷5 and since b = 4a ⇒ a + 4a = 180 ⇒ 5a = 180 ⇒ a = 5 = 5÷5 = 36, ∴ b = 4a = 4 x 36 = 144 ⇒ a = 36 and b = 144

Example 07

" ! !

For the above diagram, prove that α + β = 90˚ α + α + β + β = 180˚ (straight angle) ∴ 2α + 2β = 180˚, divide each side by 2 ∴ α + β = 90˚

Example 08 Given that “f” represents a full turn angle. Write down in terms of f a) a straight angle b) angle 195˚ 1 2 c) the complementary to 10 f d) the supplementary to 9 f a)

180 1 360 = 2 1 ∴ The straight angle, 180˚ = 2 f

195 195÷5 39 39÷3 360 = 360÷5 = 72 = 72÷3 13 ∴ Angle 195˚ = 24 f

Book 1, Hanna’s Series

† Chapter 01, Hanna’s Teaching Notes on Geometry Angles 90 90÷90 1 360 = 360÷90 = 4 1 Right angle, 90˚ = 4 f c)

1 1 1 5 2 3 ∴ The complementary to 10 f = 4 f − 10 f = 20 f − 20 f = 20 f 180 180÷180 1 360 = 360÷180 = 2 2 1 2 9 4 5 ∴ The supplementary to 9 f = 2 f − 9 f = 18 f − 18 f = 18 f

Example 09 Copy and complete the table below Angle

the complement of the angle = 90˚ − the angle

the supplement of the angle = 180˚ − the angle

90˚ − α

180˚ − α

40˚ + β

50˚ − β

140˚ − β

20˚ − λ

70˚ + λ

160˚ + λ

90˚ − θ

90˚ + θ

Example 10 Given that angles a and b are supplementary and if angle b equals 110˚ to the nearest ten degrees. Find the upper value of angle a. Since angle b equals 110˚ to the nearest ten degrees ∴ 105˚ ≤ b < 115˚, i.e. the lower value of angle b = 105˚ Since angles a and b are supplementary, ⇒ a + b = 180˚, ∴ upper value of angle a = 180˚ − the lower value of angle b = 180˚ − 105˚ = 75˚

Example 11 Given that angles p and q are complementary and if angle q equals 36˚ to the nearest four degrees. Find the upper value of angle p. Since angle q equals 32˚ to the nearest four degrees ∴ 34˚ ≤ q < 38˚, i.e. the lower value of angle q = 34˚ Since angles p and q are complementary ⇒ p + q = 90˚, ∴ The upper value of angle p = 90˚ − the lower value of angle q = 90˚ − 34˚ = 56˚

Book 1, Hanna’s Series

† Chapter 01, Hanna’s Teaching Notes on Geometry Angles

Exercise 9A Find the values of α, β and/or γ giving the geometrical reasons for all your work.

03 !

63˚ !

05 136˚

6! 5!

08 !

4! 2!

3! 75˚

53˚ 4!

63˚ 2! 4!

! 3!

12 !

! 35˚

39˚

2! 6# 3" 60˚+ !

142˚

5# 2! 48˚+ 2"

17 2! 60˚ 2# 3! 45˚+ 3"

75˚+ 2" 3" 7# 2! 41˚

Book 1, Hanna’s Series

3! #

6" 37˚+ 7"

18 5! 42˚ 6" 4# 18˚+ 7!

† Chapter 01, Hanna’s Teaching Notes on Geometry Angles

Write down as a fraction of the straight angle each of the following angles. Simplify your answer as much as possible. a) 135˚ b) 117˚ c) 48˚.

Write down as a fraction of a full turn each of the following angles. Simplify your answer as much as possible. a) 144˚ b) 375˚ c) 315˚.

Write down as a fraction of a right angle each of the following angles. Simplify your answer as much as possible. a) 15˚ b) 36˚ c) 67.5˚

Calculate in degrees each of the following angles 7 11 a) e) 12 of a straight angle 24 of a full turn 29 1 b) f) 1 72 of a full turn 30 of a full turn 5 17 c) 2 9 of a right angle g) 1 18 of a right angle 13 19 d) of a straight angle h) 1 45 36 of a straight angle

Calculate in degrees each of the following angles a) b) c) d)

1.1 of a right angle 0.002 of a full turn 0.55 of a full turn 2.75 of a straight angle

e) f) g) h)

1.25 of a right angle 1.01 of a straight angle 2.05 of a straight angle 1.125 of a full turn

In each of the following, find the sum, in degree, of the given three angles a) 7 right angles, 5 straight angles and 4 full turns angles b) h right angles, 3k straight angles and 2h full turns angles c) 3m right angles, m straight angles and 5 full turns angles d) b right angles, c straight angles and d full turns angles e) (p +2) right angles, (p −1) straight angles and p full turns angles f) (2t +5) right angles, (t −2) straight angles and t full turns angles

Book 1, Hanna’s Series

† Chapter 01, Hanna’s Teaching Notes on Geometry Angles

Given that “r” represents a right angle, write down in terms of r a) angle 30˚ d) a full turn angle r b) angle 45˚ e) the complementary to 6 3r c) a straight angle f) the supplementary to 8

Given that “f” represents a full turn angle, write down in terms of f a) angle 120˚ d) angle 105˚ 2f b) angle 225˚ e) the complementary to 15 5f c) angle 108˚ f) the supplementary to 12

Given that “s” represents a straight angle, write down in terms of s a) angle 9˚ e) angle 315˚ b) angle 75˚ f) angle 198˚ 2 c) angle 72˚ g) the supplementary to 5 s 4s d) angle 135˚ h) the complementary to 9

Given that angles h and k are complementary and if angle k is twice angle h, find h and k.

Given that angles m and n are complementary and if angle m is eight times angle n, find m and n.

Given that angles a and b are supplementary and if angle b is seven times angle a, find a and b.

Given that angles w and y are supplementary and if angle w is four times angle y, find w and y.

Given that angles a and b are supplementary and if angle b equals 80˚ to the nearest ten degrees, find the lower limit value of angle a.

Given that angles x and y are complementary and if angle x equals 23˚ to the nearest degree, find the upper limit value of angle y.

Copy and complete the table below

Book 1, Hanna’s Series

† Chapter 01, Hanna’s Teaching Notes on Geometry Angles Angle

the complement of the angle = 90˚ − the angle

the supplement of the angle = 180˚ − the angle

90˚ − θ

180˚ − θ

30˚ + β

80˚ + λ

70˚ − µ

50˚ + α

90˚ + β

125˚ − φ

α+β

λ−µ

Book 1, Hanna’s Series

† Chapter 01, Hanna’s Teaching Notes on Geometry Angles

Final Answers of Questions of Exercise 9A 01 05 09 13 15 17

α = 27˚ 02 α = 30˚ 03 α = 10˚ 04 α = 44˚ α = 10˚ 06 α = 21˚ 07 α = 307˚ 08 α = 36˚ α = 297˚ 10 α = 35˚ 11 α = 51˚ 12 α = 52˚ α = 60˚, β = 20˚, γ = 10˚ 14 β = 20˚, α = 46˚, γ = 8˚ γ = 11˚, β = 66˚, γ = 38˚ 16 α = 24˚, γ = 36˚, β = 21˚ α = 21˚, γ = 38˚, β = 9˚ 18 β = 12˚, γ = 13˚, α = 19.5˚ 3 19 a) 135˚ = 4 of a straight angle 13 b) 117˚ = 20 of a straight angle 4 c) 48˚ = 15 of a straight angle

20 a)

2 144˚ = 5 of a full turn angle

25 375˚ = 24 of a full turn angle

7 315˚ = 8 of a full turn angle

21 a)

1 15˚ = 6 of a right angle

2 36˚ = 5 of a right angle

3 67.5˚ = 4 of a right angle

22 a) c) e) g)

7 12 of a straight angle = 210˚ 5 2 9 of a right angle = 230˚ 11 24 of a full turn = 165˚ 17 118 of a right angle = 175˚

23 a)

99˚

e) 24 a) c) e)

112.5˚ f) 181.8˚ 2970˚ (450m + 1800) ˚ 630p ˚

0.72˚

b) d) f) h)

29 30 of a full turn = 348˚ 13 45 of a straight angle = 52˚ 1 1 72 of a full turn = 365˚ 19 136 of a straight angle = 275˚

198˚

369˚ b) d) f)

Book 1, Hanna’s Series

495˚

h) 405˚ (810h + 540k ) ˚ (90b + 180c + 360d) ˚ (720t + 90) ˚

† Chapter 01, Hanna’s Teaching Notes on Geometry Angles r angle 30˚ = 3

25 a) c)

r angle 45˚= 2

straight angle = 2 r d) full turn angle = 4 r r 5r angle 6 is the complementary to the angle 6 3r 13r angle 8 is the complementary to the angle 8 f 13f angle 120˚ = 3 b) angle 195˚ = 24 3f 7f angle 108˚ = 10 d) angle 105˚ = 24 2f 7f angle 15 is the complementary to the angle 60 5f f angle 12 is the supplementary to the angle 12 s 5s angle 9˚ = 20 b) angle 75˚ = 12 2s 3s angle 72˚ = 5 d) angle 135˚ = 4 2s 3s angle 72˚ = 5 f) angle 135˚ = 4 2s 3s angle 5 is the supplementary to the angle 5 4s s angle 9 is the complementary to the angle 18

e) f)

26 a) c) e) f)

27 a) c) e) g) h)

28 30 32 33 34

a = 22.5˚ and b = 157.5˚ 29 m = 80˚ and n = 10˚ w = 144˚ and y = 36˚ 31 h = 60˚ and k = 30˚ The minimum value of angle a = 95˚ The maximum value of angle y = 77.5˚ Angle

the complement of the angle = 90˚ − the angle

the supplement of the angle = 180˚ − the angle

90˚ − θ

180˚ − θ

30˚ + β

60˚ − β

150˚ − β

100˚ − λ

λ − 10˚

80˚ + λ

70˚ − µ

20˚ + µ

110˚ + µ

40˚ − α

50˚ + α

140˚ + α

90˚ − β

90˚ + β

φ − 35˚

125˚ − φ

215˚ − φ

α+β

90˚ − α − β

180˚ − α − β

λ−µ

90˚ − λ + µ

180˚ − λ + µ

Book 1, Hanna’s Series

† Chapter 01, Hanna’s Teaching Notes on Geometry Angles

Completed Solutions for Exercise 9A Find the values of α, β and/or γ giving the geometrical reasons for all your work.

63˚ ! α + 63˚ = 90˚, (right angle) ⇒ α = 90˚ − 63˚ = 27˚

α + α + α = 90˚, (right angle) ⇒ 3α = 90˚ ⇒ α = 90˚ ÷ 3 = 30˚

03 3!

4! 2!

2α + 3α + 4α = 90˚, (right angle) ⇒ α = 90˚ ÷ 9 ⇒ α = 10˚

04 136˚

α + 136˚ = 180˚, (straight angle) ⇒ α = 180˚ − 136˚ ⇒ α = 44˚

Book 1, Hanna’s Series

† Chapter 01, Hanna’s Teaching Notes on Geometry Angles

05 7!

6! 5!

7α + 6α + 5α = 180˚, (straight angle) ⇒ 18α = 180˚, divide by 18 ∴ α = 10˚

06 3! 75˚

2α + 3α + 75˚ = 180˚, (straight angle) ⇒ 5α + 75˚ = 180˚, divide by 5 ∴ α + 15˚ = 36˚ ∴ α = 36˚ − 15˚ = 21˚

07 !

53˚

α + 53˚ = 360˚, (angles of οne turn) ∴ α = 360˚ − 53˚ = 307˚

2! ! 3!

4α + 2α + 3α + α = 360˚, (angles of οne turn) ⇒ 10α = 360˚ ∴ α = 360˚ ÷ 10 = 36˚

Book 1, Hanna’s Series

† Chapter 01, Hanna’s Teaching Notes on Geometry Angles

63˚ 2! 4!

5α + 4α + 2α + 63˚ = 360˚, (angles of οne turn) ∴ 11α = 360˚ − 63˚ = 297˚ ⇒ α = 297˚ ÷ 11 ∴ α = 27˚

10 ! 35˚

α = 35˚, (vertically opposite angles) ∴ α = 35˚

11 ! 39˚

α + 39 + 90˚ = 180˚, (straight angle) ⇒ α + 129˚ = 180˚ ∴ α = 51˚

12 ! 142˚ α + 90˚ = 142˚, (vertically opposite angles) ⇒ α = 142˚ − 90˚ ∴ α = 52˚

Book 1, Hanna’s Series

† Chapter 01, Hanna’s Teaching Notes on Geometry Angles

13 2! 6# 3" 60˚+ ! 2α = 60˚ + α, (vertically opposite angles) ∴ α = 60˚ 3β + 2α = 180˚, (straight angle) ⇒ 3β = 180˚ − 2 x 60˚ = 180˚ − 120 = 60˚ ∴ β = 60˚ ÷ 3 = 20˚ 6γ = 3β, (vertically opposite angles) ⇒ 6γ = 60˚ ∴ γ = 60˚ ÷ 6 = 10˚

14 7! 5# 2! 48˚+ 2" 7β + 2β = 180˚, (straight angle) ⇒ 9β = 180˚ ∴ β = 20˚ 48˚ + 2α = 7β, (vertically opposite angles) ∴ 2α = 7 x 20˚ − 48˚ = 140˚ − 48˚ = 92˚ ∴ α = 92˚ ÷ 2 = 46˚ 5γ = 2β, (vertically opposite angles) ⇒ 5γ = 2 x 20˚ = 40˚ ∴ γ = 40˚ ÷ 5 = 8˚

15 3! #

6" 37˚+ 7"

37˚ + 7γ + 6γ = 180˚, (straight angle) ⇒ 13γ = 180˚ − 37˚ = 143˚ ∴ γ = 143˚ ÷ 13 = 11˚ β = 6γ, (vertically opposite angles) ∴ β = 6 x 11˚ = 66˚ 3α + 6γ = 180˚, (straight angle) , divide by 3, ⇒ α + 2γ = 60˚ ∴ α = 60˚ − 2 x 11˚ = 60˚ − 22˚ = 38˚

Book 1, Hanna’s Series

† Chapter 01, Hanna’s Teaching Notes on Geometry Angles

16 2! 60˚ 2# 3! 45˚+ 3"

2α + 60˚ + 3α = 180˚, (straight angle) ⇒ 5α = 180˚ − 60˚ = 120˚ ∴ α = 120˚ ÷ 5 = 24˚ 2γ = 3α, (vertically opposite angles) ⇒ 2γ = 3 x 24˚ = 72˚ ∴ γ = 72˚ ÷ 2 = 36˚ 45˚ + 3β = 2α + 60˚, (vertically opposite angles) , divide by 3 ∴ 15˚ + β = 2 x 8˚ + 20˚ ⇒ β = 16˚ + 20˚ − 15˚ ∴ β = 21˚

17 75˚+ 2" 3" 7# 2! 41˚

75˚ + 2α + 3α = 180˚, (straight angle) ∴ 5α = 180˚ − 75˚ = 105˚ ∴ α = 105˚ ÷ 5 = 21˚ 2γ + 41˚ = 75˚ + 2α, (vertically opposite angles) ∴ 2γ = 75˚ + 42˚ − 41˚ = 75˚ + 1˚ = 76˚ ∴ γ = 76˚ ÷ 2 = 38˚ 7β = 3α, (vertically opposite angles) ⇒ 7β = 3 x 21˚ ∴ β = 3 x 3˚ = 9˚

Book 1, Hanna’s Series

† Chapter 01, Hanna’s Teaching Notes on Geometry Angles

18 5! 42˚ 6" 4# 18˚+ 7!

5β + 42˚ = 18˚ + 7β, (vertically opposite angles) ∴ 42˚ − 18˚ = 7β − 5β ⇒ 2β = 24˚ ⇒ β = 12˚ 5β + 42˚ + 6γ = 180˚, (straight angle) ∴ 6γ = 180˚ − 60˚ − 42˚ = 120˚ − 42˚ = 78˚ ∴ γ = 78˚ ÷ 6 = 13˚ 4α = 6γ, (vertically opposite angles) ⇒ 4α = 6 x 13˚ = 78˚ ∴ α = 78˚ ÷ 4 = 19.5˚

Write down as a fraction of the straight angle each of the following angles. Simplify your answer as much as possible. a) 135˚ b) 117˚ c) 48˚.

a) 135 135÷9 15 15÷5 180 = 180÷9 = 20 = 20÷5 3 ∴ 135˚ = 4 of a straight angle b) 117 117÷9 13 180 = 180÷9 = 20 13 ∴ 117˚ = 20 of a straight angle c) 48 48÷12 4 180 = 180÷12 = 15 4 ∴ 48˚ 15 of a straight angle

Book 1, Hanna’s Series

† Chapter 01, Hanna’s Teaching Notes on Geometry Angles

Write down as a fraction of a full turn each of the following angles. Simplify your answer as much as possible. a) 144˚ b) 375˚ c) 315˚.

a) 144 144÷9 16 16÷8 360 = 360÷9 = 40 = 40÷8 2 ∴ 144˚ = 5 of a full turn angle b) 375 375÷3 125 125÷5 360 = 360÷3 = 120 = 120÷5 25 ∴ 375˚ = 24 of a full turn angle c) 315 315÷15 21 21÷3 360 = 360÷15 = 24 = 24÷3 7 ∴ 315˚ = 8 of a full turn angle

Write down as a fraction of a right angle each of the following angles. Simplify your answer as much as possible. a) 15˚ b) 36˚ c) 67.5˚

a) 15 15÷15 15˚ = 90 of a right angle = 90÷15 1 15˚ = 6 of a right angle b) 36 36÷9 4 4÷2 36˚ = 90 of a right angle = 90÷9 = 10 of a right angle = 10÷2 2 36˚ = 5 of a right angle c) 315

( ) ˚ = ( 2 ) ˚ = 2 x 90 of a right angle (R) 1

67.5˚ = 67 2

135

135÷15 9 3 = 2x90÷15 of (R) = 2x6 of (R) = 4 of a right angle

Book 1, Hanna’s Series

† Chapter 01, Hanna’s Teaching Notes on Geometry Angles

13 45 of a straight angle

19 1 36 of a straight angle

a) 30 7 7 7  = 7 x 30˚ 12 of a straight angle = 12 x 180˚ = 12  x 180˚ 1

7 ∴ 12 of a full turn = 210˚ b) 29 1 30 of a full turn = 1 − 30

(

) of a full turn = 360˚ − 30 x 360˚ = 360˚ − 12˚ 29 ∴ 30 of a full turn = 348˚

c) 5 23 2 9 of a right angle = 9 x 90˚ = 23 x 10˚ 5 ∴ 2 9 of a right angle = 230˚ d) 4 13 13 13  = 13 x 4˚ 45 of a straight angle = 45 x 180˚ = 45  x 180˚ 1

13 ∴ 45 of a straight angle = 52˚ e) 30 11 11 11  = 11 x 15˚ 24 of a full turn = 24 x 360˚ = 24  x 360˚ 2

11 ∴ 24 of a full turn = 165˚ f) 1 1 1 72 of a full turn = 360˚ + 72 x 360˚ = 360˚ + 5˚ 1 ∴ 1 72 of a full turn = 365˚

Book 1, Hanna’s Series

† Chapter 01, Hanna’s Teaching Notes on Geometry Angles g) 17 1 1 18 of a right angle = 2 − 18 of a right angle 1 = 2 x 90˚ − 18 x 90˚ = 180˚ − 5˚

(

)

17 ∴ 1 18 of a right angle = 175˚ h) 5 19 55 55 1 36 of a straight angle = 36 x 180˚ = 36 x 180˚  = 55 x 5˚  1

19 ∴ 1 36 of a straight angle = 275˚

Calculate in degrees each of the following angles a) b) c) d)

1.1 of a right angle 0.002 of a full turn 0.55 of a full turn 2.75 of a straight angle

e) f) g) h)

1.25 of a right angle 1.01 of a straight angle 2.05 of a straight angle 1.125 of a full turn

a) 1.1 of a right angle = 1.1 x 90˚ = 11 x 9˚= 99˚ b) 0.002 of a full turn = 0.002 x 360˚ = 0.02 x 36˚ = 0.72˚ c) 0.55 of a full turn = 0.55 x 360˚ = 5.5 x 36˚ = 5.5 x 2 x 18˚ = 11 x 18˚ = 198˚ e) 1.25 of a right angle = 1.25 x 90˚ = 12.5 x 9˚ = 112.5˚ f) 1.01 of a straight angle = 1.01 x 180˚ = 10.1 x 18˚ = 181.8˚ g) 2.05 of a straight angle = 2.05 x 180˚ = 20.5 x 18˚ = 20.5 x 2 x 9˚ = 41 x 9˚ = 369˚ d) 3 11 2.75 of a straight angle = 2.75 x 180˚ = 2 4 x 180˚ = 4 x 180˚ 45 11 = 4 x 180˚  = 11 x 45˚ = 495˚ 1

Book 1, Hanna’s Series

† Chapter 01, Hanna’s Teaching Notes on Geometry Angles h) 1 9 1.125 of a full turn = 1.125 x 360˚ = 1 8 x 360˚ = 8 x 360˚ 45 9 = 8 x 360˚  = 9 x 45˚ = 405˚ 1

a) 7 right angles = 7 x 90˚ = 630˚ 5 straight angles = 5 x 180˚ = 900˚ 4 full turns angles = 4 x 360˚ = 1440˚ ∴ Sum of the 3 angles = 630 + 900 + 1440 = 2970˚ b) h right angles = h x 90˚ = 90h˚ 3k straight angles = 3k x 180˚ = 540k˚ 2h full turns angles = 2h x 360˚ = 720h˚ ∴ Sum of the 3 angles = (810h + 540k) ˚ c) 3m right angles = 3m x 90˚ = 270m˚ m straight angles = m x 180˚ = 180m˚ 5 full turns angles = 5 x 360˚ = 1800˚ ∴ Sum of the 3 angles = (450m + 1800) ˚ d) b right angles = b x 90˚ = 90b˚ c straight angles = c x 180˚ = 180c˚ d full turns angles = d x 360˚ = 360d˚ ∴ Sum of the 3 angles = (90b + 180c + 360d) ˚

Book 1, Hanna’s Series

† Chapter 01, Hanna’s Teaching Notes on Geometry Angles f) (p + 2) right angles = 90˚ x (p + 2) = (90p + 180) ˚ (p − 1) straight angles = 180˚ x (p − 1) = (180p − 180) ˚ p full turns angles = 360p˚ ∴ Sum of the 3 angles = 630p˚ g) (2t + 5) right angles = 90˚ x (2t + 5) = (180t + 450) ˚ (t − 2) straight angles = 180˚ x (t − 2) = (180t − 360) ˚ t full turns angles = 360t˚ ∴ Sum of the 3 angles = (720t + 90) ˚

Given that “r” represents a right angle, write down in terms of r a) angle 30˚ d) a full turn angle r b) angle 45˚ e) the complementary to 6 3r c) a straight angle f) the supplementary to 8

a) 30 x r r angle 30˚ = 90 = 3 b) 45 x r 45r÷45 r angle 45˚ = 90 = 90÷45 = 2 c) straight angle = 180˚ =

180 x r 90 = 2r

360r full turn angle = 360˚ = 90 = 4r r the right angle, 90˚ = r ⇒ the complementary to 6 r 6r r 5r =r− 6 = 6 − 6 = 6 3r the straight angle = 2r ⇒ the supplementary to 8 3r 16r 3r 13r = 2r − 8 = 8 − 8 = 8

Book 1, Hanna’s Series

† Chapter 01, Hanna’s Teaching Notes on Geometry Angles

Given that “f” represents a full turn angle, write down in terms of f a) angle 120˚ d) angle 105˚ 2f b) angle 225˚ e) the complementary to 15 5f c) angle 108˚ f) the supplementary to 12

a) 120f f angle 120˚ = 360 = 3 b) 225f÷15 15f 5f angle 225˚ = 360÷15 = 24 = 8 c) 108 x f 108f÷36 3f angle 108˚ = 360 = 360÷36 = 10 d)

105 x f 105 x f÷15 7f angle 105˚ = 360 = 360÷15 = 24 f 2f the right angle, 90˚ = 4 ⇒ the complementary to 15 f 2f 15f 8f 7f = 4 − 15 = 60 − 60 = 60 f 5f the straight angle = 2 ⇒ the supplementary to 12 f 5f 6f 5f f = 2 − 12 = 12 − 12 = 12 Given that “s” represents a straight angle, write down in terms of s a) angle 9˚ e) angle 315˚ b) angle 75˚ f) angle 198˚ 2 c) angle 72˚ g) the supplementary to 5 s 4s d) angle 135˚ h) the complementary to 9

a) 9 9÷9 1 angle 9˚ = 180 s = 180÷9 s = 20 s b) 75 75÷15 5 angle 75˚ = 180 s = 180÷15 s = 12 s c) 72 72÷9 8 2 angle 72˚ = 180 s = 180÷9 s = 20 s = 5 s Book 1, Hanna’s Series

† Chapter 01, Hanna’s Teaching Notes on Geometry Angles d) 135 135÷15 9 3 angle 135˚ = 180 s = 180÷15 s = 12 s = 4 s e) 315 315÷15 21 7 angle 315˚ = 180 s = 180÷15 s = 12 s = 4 s f) 198 198÷9 22 11 angle 198˚ = 180 s = 180÷9 s = 20 s = 10 s g)

the straight angle, 180˚ = s

2 5 2 3 ∴ the supplementary to 5 s = 5 s − 5 s = 5 s 1 the right angle, 90˚ = 2 s 4 1 4 9 8 s ∴ the complementary to 9 s = 2 s − 9 s = 18 s − 18 s = 18

28 29

Given that angles h and k are complementary and if angle k is twice angle h, find h and k. Given that angles m and n are complementary and if angle m is eight times angle n, find m and n.

28 Since angles h and k are complementary

k h ∴ h + k = 90˚ and since k = 2h ∴ h + 2h = 90˚ ⇒ 3h = 90˚ ⇒ h = 30˚ and hence k = 2 x 30˚ ∴ h = 60˚ and k = 30˚

29 Since angles m and n are complementary

m n ∴ m + n = 90˚ and since m = 8n ∴ 8n + n = 180˚ ⇒ 9n = 90˚ ⇒ n = 10˚ and hence m = 8 x 10˚ ∴ m = 80˚ and n = 10˚

Book 1, Hanna’s Series

† Chapter 01, Hanna’s Teaching Notes on Geometry Angles

30 31

Given that angles a and b are supplementary and if angle b is seven times angle a, find a and b. Given that angles w and y are supplementary and if angle w is four times angle y, find w and y.

30 Since angles a and b are supplementary

∴ a + b = 180˚, and since b = 7a ∴ a + 7a = 180˚ ⇒ 8a = 180˚ 180˚ 45˚ ∴ a = 8 = 2 = 22.5˚ ⇒ b = 7 x 22.5˚ ∴ a = 22.5˚, b = 157.5˚

31 Since angles w and y are supplementary

∴ w + y = 180˚, and since w = 4 y ∴ 4y + y = 180˚ ⇒ 4y + y = 180˚ ⇒ 5y = 180˚ ∴ y = 36˚ and hence w = 4 y = 4 x 36˚ ∴ w = 144˚, y = 36˚

Given that angles a and b are supplementary and if angle b equals 80˚ to the nearest ten degrees, find the lower limit value of angle a.

Since angle b equals 80˚ to the nearest ten degrees ∴ 75˚ ≤ b < 85˚, i.e. the upper limit value of angle b = 85˚ Since angles a and b are supplementary ⇒ a + b = 180˚, ∴ Lower limit value of angle a = 180˚ − upper limit value of angle b = 180˚ − 85˚ = 95˚

Book 1, Hanna’s Series

† Chapter 01, Hanna’s Teaching Notes on Geometry Angles

Given that angles x and y are complementary and if angle x equals 23˚ to the nearest degree, find the upper limit value of angle y.

Since angle x equals 23˚ to the nearest degree ∴ 22.5˚ ≤ x < 23.5˚, i.e. the lower limit value of angle x = 22.5˚ Since angles x and y are complementary ⇒ x + y = 90˚ ∴ Upper limit value of angle y = 90˚ − the lower limit value of angle x = 90˚ − 22.5˚ = 77.5˚

Copy and complete the table below Angle

the complement of the angle = 90˚ − the angle

the supplement of the angle = 180˚ − the angle

90˚ − θ

180˚ − θ

30˚ + β

60˚ − β

150˚ − β

100˚ − λ

λ − 10˚

80˚ + λ

70˚ − µ

20˚ + µ

110˚ + µ

40˚ − α

50˚ + α

140˚ + α

90˚ − β

90˚ + β

φ − 35˚

125˚ − φ

215˚ − φ

α+β

90˚ − α − β

180˚ − α − β

λ−µ

90˚ − λ + µ

180˚ − λ + µ

Book 1, Hanna’s Series

â&#x20AC; Chapter 01, Hannaâ&#x20AC;&#x2122;s Teaching Notes on Geometry Angles

06 If two parallel lines are cut by a line then corresponding angles are equal Draw two parallel lines with a horizontal line cutting them (transversal) like the diagrams below, a), b) or c). Measure the corresponding angles (a, b) , (c, d) and (e, f) . Write down your observations. a)

a = b, (corresponding angles) b)

c = d, (corresponding angles) c)

e = f, (corresponding angles)

Book 1, Hannaâ&#x20AC;&#x2122;s Series

† Chapter 01, Hanna’s Teaching Notes on Geometry Angles

07 If two parallel lines are cut by a line then alternate angles are equal Draw two parallel lines with a horizontal line cutting them (transversal) like the diagrams below, a), b) or c). Measure the corresponding angles (a, b) , (c, d) and (e, f) . Write down your observations. a)

a b

a = b, (alternate angles) b)

c = d, (alternate angles) c)

f e e = f, (alternate angles)

Example 12 Find the values of the angles α and β giving the geometrical reasons for all your work.

70˚

α = 70˚, (corresponding angles) β = 70˚, (alternate angles) ∴ α = 70˚, β = 70˚

Book 1, Hanna’s Series

† Chapter 01, Hanna’s Teaching Notes on Geometry Angles

08 If two parallel lines are cut by a line then allied angles add up to 180˚ Draw two parallel lines with a horizontal line cutting them (transversal) like the diagrams below, a), b) or c). Measure the corresponding angles (α, β) , (φ, θ) , and (δ, γ) . Write down your observations. a)

α + β = 180˚, (allied angles) b)

φ + θ = 180˚, (allied angles) c)

δ + γ = 180˚, (allied angles)

Example 13 Find the values of the angles α and β giving the geometrical reasons for all your work.

65˚

α + 65˚ = 180˚, (allied angles) ⇒ α = 180˚ − 65˚ = 115˚ β = 65˚, (corresponding angles) ∴ α = 115˚ and β = 65˚

Book 1, Hanna’s Series

† Chapter 01, Hanna’s Teaching Notes on Geometry Angles

Example 14 Find the value of the angle p giving the geometrical reasons for all your work. a)

57˚

9p 2p

28˚ a)

57˚ a b

28˚ a = 57˚, (alternate angles) b = 28˚, (alternate angles) a + b + p = 360˚, (angles of one turn) ∴ 57˚ + 28˚ + p = 360˚ ∴ p = 360˚ − 85˚ ∴ p = 275˚ b)

4p x y

2p 4p + x = 180˚, (allied angles) and 2p + y = 180˚, (allied angles) Add the above two equations ⇒ 4p + x + 2p + y = 180˚ + 180˚ ∴ 6p + x + y = 360˚ Since x + y = 9p ⇒ 6p + 9p = 360˚ ⇒ 15p = 360˚ ∴ p = 360˚ ÷ 15 = 24˚

Book 1, Hanna’s Series

† Chapter 01, Hanna’s Teaching Notes on Geometry Angles

Exercise 9B Find the values of the angles a, b, c, d, e, f and/ or g giving the geometrical reasons for all your work.

a 72˚

b 61˚

64˚

80˚

b 75˚

b a

73˚

70˚

63˚

08 b

65˚

115˚

110˚

a b

55˚ a 120˚

110˚

125˚ a 50˚ b

b 2a a 150˚

Book 1, Hanna’s Series

† Chapter 01, Hanna’s Teaching Notes on Geometry Angles

118˚

130˚ a

99˚ a

120˚

125˚

62˚

125˚ a 50˚

4a a b

3a 69˚

2a 71˚ 3b

33˚

107˚

a 50˚

75˚

80˚

60˚

63˚ a 23˚

Book 1, Hanna’s Series

47˚ a 38˚

† Chapter 01, Hanna’s Teaching Notes on Geometry Angles

26 135˚

140˚ 115˚

65˚

11a

4a 5a

4a a

29 b

5c 2a

c 75˚

80˚

y C

2a 3b

In the given diagram, BA is parallel to CD and AD is parallel to BC. Prove that a = c and b = d C

In the given diagram, AB is parallel to CD . Prove that x + y + z = 360˚

Book 1, Hanna’s Series

† Chapter 01, Hanna’s Teaching Notes on Geometry Angles

01 03 05 07 09 11 13 15 17 19 21 23 25 27 29

Final Answers of Questions of Exercise 9B a = 72˚, b = 61˚ 02 a = 73˚, b = 54˚ a = 80˚, b = 80˚ 04 a = 75˚, b = 105˚ a = 70˚, b = 110˚ 06 a = 63˚, b = 117˚ a = 65˚, b = 115˚ 08 a = 70˚, b = 70˚ a = 65˚, b = 70˚ 10 a = 60˚, b = 125˚ a = 55˚, b = 50˚ 12 a = 30˚, b = 120˚ a = 143˚ 14 a = 110˚ a = 118˚, b = 62˚, c = 118˚ 16 a = 55˚, b = 125˚, c = 55˚ a = 36˚, b = 45˚ 18 a = 30˚, b = 20˚ a = 11˚, b = 51˚ 20 a = 18˚, b = 12˚ a = 55˚, b = 55˚ 22 a = 60˚, b = 40˚ a = 85˚ 24 a = 274˚ a = 60˚ 26 a = 105˚ a = 20˚ 28 a = 24˚ a = 40˚, b = 140˚, c = 8˚ 30 a = 15˚, b = 50˚, c = 30˚

Book 1, Hanna’s Series

† Chapter 01, Hanna’s Teaching Notes on Geometry Angles

Completed Solutions for Exercise 9B Find the values of the angles a, b, c, d, e, f and/ or g giving the geometrical reasons for all your work.

01 a 72˚ b 61˚

a = 72˚, (corresponding angles) & b = 61˚, (corresponding angles)

02 73˚ a b 64˚

a = 73˚, (corresponding angles) & b = 54˚, (corresponding angles)

03 a

80˚

a = 80˚, (alternate angles) & b = 80˚, (corresponding angles)

04 b 75˚

a = 75˚, (corresponding angles) & b + a = 180˚ ⇒ b = 180˚ − 75˚ ∴ a = 75˚ and b = 105˚

Book 1, Hanna’s Series

† Chapter 01, Hanna’s Teaching Notes on Geometry Angles

05 b a

70˚

a = 70˚, (corresponding angles) & b + 70˚ = 180˚, (allied angles) ⇒ b = 180˚ − 70 ∴ a = 70˚ and b = 110˚

06 a

63˚ a = 63˚, (alternate angles) & a + b = 180˚, (allied angles) ⇒ b = 180˚ − 63˚ ∴ a = 63˚ and b = 117˚

65˚

a = 65˚, (corresponding angles) & b + a = 180˚, (allied angles) ∴ b = 180˚ − 65˚ = 115˚ ∴ a = 65˚ and b = 115˚

110˚

a + 110˚ = 180˚, (allied angles) ⇒ a = 180˚ − 110˚ = 70˚ b = a = 70˚, (corresponding angles) ∴ a = 70˚ and b = 70˚

Book 1, Hanna’s Series

† Chapter 01, Hanna’s Teaching Notes on Geometry Angles

09 115˚ a b 110˚ a + 115˚ = 180˚, (allied angles) ⇒ a = 180˚ − 115˚ = 65˚ b + 110˚ = 180˚, (allied angles) ⇒ b = 180˚ − 110˚ ∴ a = 65˚ and b = 70˚

10 b 55˚ a 120˚ a + 120˚ = 180˚, (allied angles) ⇒ a = 180˚ − 120˚ = 60˚ b + 55˚ = 180˚, (allied angles) ⇒ b = 180˚ − 55˚ ∴ a = 60˚ and b = 125˚

11 125˚ a 50˚ b a + 125˚ = 180˚, (allied angles) ⇒ a = 180˚ − 125˚ = 55˚ b + 50˚ = 180˚, (allied angles) ⇒ b = 180˚ − 50˚ ∴ a = 55˚ and b = 130˚

Book 1, Hanna’s Series

† Chapter 01, Hanna’s Teaching Notes on Geometry Angles

12 b 2a a 150˚ a + 150˚ = 180˚, (allied angles) ⇒ a = 180˚ − 150˚ = 30˚ 2a + b = 180˚, (allied angles) ⇒ b = 180˚ − 60˚ ∴ a = 30˚ and b = 120˚

13 118˚ 99˚ h k a h + 118˚ = 180˚, (allied angles) ⇒ h = 180˚ − 118˚ = 62˚ k + h = 99˚, (given) ⇒ k = 99˚ − h = 99˚ − 62˚ = 37˚ a + k = 180˚, (allied angles) ∴ a = 180˚ − k = 180˚ − 37˚ = 143˚

14 130˚ x a y 120˚ x + 130˚ = 180˚, (allied angles) ⇒ x = 180˚ − 130˚ = 50˚ y + 120˚ = 180˚, (allied angles) ⇒ y = 180˚ − 120˚ = 60˚ a = x + y = 50˚ + 60˚ = 110˚

Book 1, Hanna’s Series

† Chapter 01, Hanna’s Teaching Notes on Geometry Angles

15 b

62˚

a + 62˚ = 180˚, (allied angles) ⇒ a = 180˚ − 62˚ = 118˚ b + a = 180˚, (allied angles) ⇒ b = 180˚ − a = 180˚ − 118˚ = 62˚ c + 62˚ = 180˚, (allied angles) ⇒ c = 180˚ − 62˚ ∴ a = 118˚, b = 62˚, c = 118˚ Another method to find b and c b = 62˚, (opposite angles of a parallelogram) Similarly c = a = 118˚, (opposite angles of a parallelogram)

16 a

125˚

a + 125˚ = 180˚, (allied angles) ⇒ a = 180˚ − 125˚ = 55˚ b + a = 180˚, (allied angles) ⇒ b = 180˚ − a = 180˚ − 55˚ = 125˚ c + 125˚ = 180˚, (allied angles) ⇒ c = 180˚ − 125˚ ∴ a = 55˚, b = 125˚, c = 55˚ Another method to find b and c b = 125˚, (opposite angles of a parallelogram) Similarly c = a = 55˚, (opposite angles of a parallelogram)

Book 1, Hanna’s Series

† Chapter 01, Hanna’s Teaching Notes on Geometry Angles

17 4a a b 3b 4a + a = 180˚, (allied angles) ⇒ 5a = 180˚ ∴ a = 180˚ ÷ 5 = 36˚ b + 3b = 180˚, (allied angles) ⇒ 4b = 180˚ ∴ b = 180˚ ÷ 4 = 45˚

18 5a a 2b 7b 5a + a = 180˚, (allied angles) ⇒ 6a = 180˚ ∴ a = 180˚ ÷ 6 = 30˚ 2b + 7b = 180˚, (allied angles) ⇒ 9b = 180˚ ∴ b = 180˚ ÷ 9 = 20˚

19 3a 69˚ 33˚ 2b

3a = 33˚, (alternate angles) ∴ a = 33˚ ÷ 3 = 11˚ 2b = 69˚ + 3a, (corresponding angles) ⇒ 2b = 69˚ + 33˚ = 102˚ ∴ b = 102˚ ÷ 2 = 51˚

Book 1, Hanna’s Series

† Chapter 01, Hanna’s Teaching Notes on Geometry Angles

20 2a 71˚ 3b 107˚

2a + 71˚ = 107˚, (corresponding angles) ⇒ 2a = 107˚ − 71˚ = 36˚ ∴ a = 36˚ ÷ 2 = 18˚ 3b = 2a, (alternate angles) ⇒ 3b = 2 x 18˚ = 36˚ ∴ b = 36˚ ÷ 3 = 12˚

21 x

a 50˚

75˚

a + 50˚ + 75˚ = 180˚, (allied angles) ⇒ a = 180˚ − 125˚ = 55˚ x = a = 55˚, (corresponding angles) b = x = 55˚, (alternate angles) ∴ a = 55˚ and b = 55˚

22 y

60˚

80˚

y = 60˚, (alternate angles) a = y = 60˚, (corresponding angles) a + b + 80˚ = 180˚, (allied angles) ⇒ b = 180˚ − 60˚ − 80˚ ∴ a = 60˚ and b = 40˚

Book 1, Hanna’s Series

† Chapter 01, Hanna’s Teaching Notes on Geometry Angles

23 Method I

47˚ h a k 38˚ h = 47˚, (alternate angles) k = 38˚, (alternate angles) ∴ a = h + k = 47˚ + 38˚ = 85˚ Method II

47˚ x a y 38˚ x + 47˚ = 180˚, (allied angles) ⇒ x = 180˚ − 47˚ = 133˚ y + 38˚ = 180˚, (allied angles) ⇒ y = 180˚ − 38˚ = 142˚ a + x + y = 360˚, (angles of one turn) ⇒ a = 360˚ − x − y ∴ a = 360˚ − 133˚ − 142˚ = 360˚ − 275˚ = 85˚

Book 1, Hanna’s Series

† Chapter 01, Hanna’s Teaching Notes on Geometry Angles

24 Method I

63˚ h a k 23˚ h = 63˚, (alternate angles) k = 23˚, (alternate angles) h + k + a = 360˚, (angles of one turn) ⇒ a = 360˚ − h − k ∴ a = 360˚ − 63˚ − 23˚ = 360˚ − 86˚ ∴ a = 274˚ Method II

63˚ x y

23˚ x + 63˚ = 180˚, (allied angles) ⇒ x = 180˚ − 63˚ = 117˚ y + 23˚ = 180˚, (allied angles) ⇒ y = 180˚ − 23˚ = 157˚ ∴ a = x + y = 117˚ + 157˚ = 274˚

Book 1, Hanna’s Series

† Chapter 01, Hanna’s Teaching Notes on Geometry Angles

135˚ x

115˚ a

x + 135˚ + 115˚ = 360˚, (angles of one turn) ⇒ x = 360˚ − 250˚ = 110˚ ∴ a + x = 180˚, (allied angles) ⇒ a = 180˚ − 110˚ ∴ a = 70˚

140˚ a

65˚

y + 65˚ = 180˚, (allied angles) ⇒ y = 180˚ − 65˚ = 115˚ a + y + 140˚ = 360˚, (angles of one turn) ⇒ a = 360˚ − 140˚ − 115˚ = 360˚ − 255˚ ∴ a = 105˚

Book 1, Hanna’s Series

† Chapter 01, Hanna’s Teaching Notes on Geometry Angles

27 3a 11a

h k 4a h = 3a, (alternate angles) k = 4a, (alternate angles) h + k + 11a = 360˚, (angles of one turn)

360˚ ∴ 3a + 4a + 11a = 360˚ ⇒ 18a = 360˚ ⇒ a = 18 ∴ a = 20˚

28 6a x 4a y 5a 6a + x = 180˚, (allied angles) 5a + y = 180˚, (allied angles) By adding the above two results ∴ x + y + 11a = 360˚, (angles of one turn), where x + y = 4a 360˚ 2 x 360˚ ∴ 4a + 11a = 360˚ ⇒ 15a = 360˚ ⇒ a = 15 = 30 = 2 x 12˚ ∴ a = 24˚

Book 1, Hanna’s Series

† Chapter 01, Hanna’s Teaching Notes on Geometry Angles

29 a b

5c 2a 80˚

2a = 80˚, (alternate angles) ⇒ a = 80˚ ÷ 2 = 40˚ b + a = 180˚, (allied angles) ⇒ b = 180˚ − 40˚ = 140˚ 5c = a = 40˚, (alternate angles) ⇒ c = 40˚ ÷ 5 = 8˚ ∴ a = 40˚, b = 140˚, c = 8˚

30 2a c 75˚

5a 5a = 75˚, (alternate angles) ⇒ a = 75˚ ÷ 5 = 15˚ c = 2a = 2 x 15˚ = 30˚, (alternate angles) 3b + 2a = 180˚, (allied angles) ⇒ 3b = 180˚ − 2a = 180˚ − 30˚ = 150˚ ∴ b = 150˚ ÷ 3 = 50˚ ∴ a = 15˚, b = 50˚, c = 30˚

Book 1, Hanna’s Series

† Chapter 01, Hanna’s Teaching Notes on Geometry Angles

d b

In the given diagram, BA is parallel to CD and AD is parallel to BC. Prove that a = c and b = d C

Since BA is parallel to CD ⇒ a + d = 180˚ (allied angles) Since AD is parallel to BC ⇒ c + d = 180˚ (allied angles) ∴ a=c Since BA is parallel to CD ⇒ b + c = 180˚ (allied angles) Since AD is parallel to BC ⇒ c + d = 180˚ (allied angles) ∴ b=d

32 E

! µ y C

B F

In the given diagram, AB is parallel to CD . Prove that x + y + z = 360˚

Construction draw the line EF parallel to AB and parallel to CD Since AB is parallel to EF ⇒ λ + x = 180˚ (allied angles) Since CD is parallel to BC ⇒ µ + z = 180˚ (allied angles) Add the above two results ∴ λ + x + µ + z = 360˚, where λ + µ = y ∴ x + y + z = 360˚

Book 1, Hanna’s Series