topic: angles 2 - jimmymaths.com · size of angle = 55° [email protected]

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Topic: Angles 2

1) In the diagram, RS // PAQ, CD // ASB, ∠ADC = 62° and ∠QAD = 26°.

Calculate a) x

b) y

2) In the diagram below, ∠AOB = p°.

If ∠BOC is two times ∠AOB, ∠COD is four times ∠AOB and ∠DOA is five times

∠AOB, find the values of all the four angles.

3) Using the figure below, find y in each of the following cases

a) If a = 2y° and c = (y + 30)°

b) If a = (3y + 40)° and c = (y + 60)°

4) In the diagram given,

a) write an equation involving a and d;

b) find the value of ∠HOK;

c) if d = 25°, find a.

C B

D

O A

p°

a c

d

b

O N L

H

M

K

d

d a

a

62°

B

T

R

S

Q D

C

A

P

x°

x°

y°

26°

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5) Find the measure of the complementary angle of each of the following angles:

a) 18°

b) 46°

c) 53°

d) 64°

e) 7°

6) Find the measure of the supplementary angle of each of the following angles:

a) 36°

b) 12°

c) 102°

d) 171°

e) 88°

7) In the figure XYZ is parallel to ST and TYU is a straight line.

Find the value of w.

8) In the figure, AOB, COD, EOF and GFC are straight lines.

Given that AOB // GFC, ∠OCF = 90° and ∠OFC = 51°, calculate

a) ∠BOE,

b) ∠DOE,

c) ∠OFG.

9) Find the value of x, and y in the given figure.

U

Z Y

X

S T

117°

69°

w°

A

G F C

B

E D

O

51°

70°

y°

28° 56°

x°




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10) In the diagram below, BDF is a straight line and BD=DE=EF. ABC is a straight line.

BDE=110°.

Calculate

a) ∠BED

b) ∠DEF

c) ∠BEF

11) In the figure PQ // RS.

Find

a) the value of a,

b) the value of b.

12) In the diagram, PQ is parallel to SR, SP = SR, ∠SPR = 66° and ∠PQS = 22° .

Find the values of x, y and z.

13) Find the value of x in the figure below.

A

B

C

D

F

E

110°

P Q

R S

90 100

20

b a

P Q

R S

60°

z

°

20°

x

°

y

°




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14) Given that BCD is a straight line, BC = AC = CD and ∠CAD = 70°.

Find the values of x, y and z.

15) In the figure, ABK is parallel to CDE,

CA is parallel to DX.

If CAD = 56o, XDE = 62o and

XBK = 42o, calculate

a) ADX,

b) BAD,

c) BXD.

16) In the diagram, ABC is a straight line, AD is parallel to EF,

∠BEG = 104° and ∠BGF = 152°. Find

a) x,

b) y.

C D E

A B K

X

62°

42°

56°

D

A

B

E

G

F

C

x y

152°




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17) In the figure below, OE and OC bisect ∠AOD and ∠DOB respectively, find ∠EOC.

18) The supplement of an angle is 15° less than 4 times its complement.

What is the size of the angle?

19) In the diagram, PQ, RS and TU are parallel. Given that ∠LNM = 115° and LM bisects

∠PMN, find

a) ∠LMN,

b) ∠MLR,

c) ∠LKU.

20) In the diagram, l1 is parallel to l2.

Express x in terms of p, q and r.

O A B

C

D

E

P

R

T K

L N

M

U

S

Q

115°

x

p

q

r l2

l1




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Topic: Angles 2 - Solutions

1) a) ∠BAD = ∠ADC (alternate ∡s)

∠BAQ + 26° = 62°

∠BAQ = 62° – 26°

∠BAQ = 36°

∠RSA = ∠BAQ (alternate ∡s)

∠RSA = 36°

∠BST + ∠TSR + ∠RSA = 180° (Sum of ∡s on a straight line)

x° + x° + 36° = 180°

2x° = 180° – 36°

2x° = 144°

x = 72

b) ∠QAD + ∠DAP = 180° (Supplementary ∡s)

26° + y° = 180°

y° = 180° – 26°

y = 154

2) ∠AOB = p°

∠BOC = 2 × p°

∠BOC = 2p°

∠AOB + ∠BOC = 90° (complementary ∡s)

p° + 2p° = 90°

3p° = 90°

p = 30

∴ ∠AOB = 30°,

∠BOC = 2 × 30°

∠BOC = 60°

∠COD = 4 × 30°

∠COD = 120°

∠DOA = 5 × 30°

∠DOA = 150°




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3) a) a = c (vertically opposite ∡s)

2y° = (y + 30) °

2y° - y° = 30°

y = 30

b) a = c (vertically opposite ∡s)

(3y + 40) ° = (y + 60) °

3y° – y° = 60° – 40°

2y° = 20°

y = 10

4) a) a° + a° + d° + d° = 180° (Sum of ∡s on a straight line)

2a + 2d = 180

a + d = 90

b) ∠HOK = a + d

∠HOK = 90°

c) a + d = 90°

a + 25° = 90°

a = 90° – 25°

a = 65°

5) a) 90° - 18° = 72°

b) 90° - 46° = 44°

c) 90° - 53° = 37°

d) 90° - 64° = 26°

e) 90° - 7° = 83°

6) a) 180° - 36° = 144°

b) 180° - 12° = 168°

c) 180° - 102° = 78°

d) 180° - 171° = 9°

e) 180° - 88° = 92°




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7) ∠XYT + ∠STY = 180° (interior ∡s)

∠XYT + 117° = 180°

∠XYT = 180° - 117°

∠XYT = 63°

∠UYZ = ∠XYT (vertically opposite ∡s)

∠UYZ = 63°

∠UZY + ∠UYZ + ∠YUZ = 180° (Sum of ∡s in a triangle)

w° + 63° + 69° = 180°

w° = 180°- 63°- 69°

w = 48

8) a) ∠FOC + ∠OFC + ∠OCF = 180°

∠FOC + 51° + 90° = 180°

∠FOC = 180° – 90° – 51°

∠FOC = 39°

∠BOC = ∠OCF

∠BOC = 90°

∠BOE + ∠BOC + ∠FOC = 180° (Sum of ∡s on a straight line)

∠BOE + 90° + 39° = 180°

∠BOE = 180° – 39° – 90°

∠BOE = 51°

OR

∠BOE = ∠CFO (corresponding ∡s)

∠BOE = 51°

b) ∠DOE = ∠FOC (vertically opposite ∡s)

∠DOE = 39°

c) ∠OFG = 180° – 51° (supplementary ∡s)

∠OFG = 129°

9) x = 28 (alternate ∡s)

y° + x° = 70° (exterior ∡s)

y° + 28° = 70°

y° = 70° – 28°

y = 42




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10) a) ∠BED = (180° – 110°)÷ 2 (isosceles triangle)

∠BED = 35°

b) ∠EDF + 110° = 180° (supplementary ∡s)

∠EDF = 180° – 110°

∠EDF = 70°

∠DEF = 180° – 70° – 70°

∠DEF = 40°

c) ∠BEF = ∠BED + ∠DEF

∠BEF = 35° + 40°

∠BEF = 75°

11) a) a° + 20° + 100° = 180° (interior ∡s)

a° = 180° – 100° – 20°

a = 60

b) b° + a° + 90° = 180° (Sum of ∡s in a triangle)

b° + 60° + 90° = 180°

b° = 180° – 60° – 90°

b = 30

12) ∠QSR = ∠PQS (vertically opposite ∡s)

x = 20

∠PSR + ∠PRS + ∠SPR = 180° (Sum of ∡s in a triangle)

x° + y° + 60° + 60° = 180°

20° + y° + 60° + 60° = 180°

y° = 180° – 60° – 60° – 20°

y = 40

z° = y° + x° (alternate s)

z° = 40° + 20°

z = 60




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13)

180° – 125° = 55° (interior ∡s)

180° – 34° – 55° = 91° (interior ∡s)

x = 360° – 91° (∡s at a point)

x = 269

14) x = 70° + 70° (exterior s)

x = 140

y = (180° – 140°) ÷ 2 (Sum of s in an isosceles triangle)

y = 20

z = 180° – 20° – 90° (Sum of s in a triangle)

z = 70




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15) a) ∠ADX = ∠CAD (alternate ∡s)

∠ADX = 56°

b) ∠CDA + ∠ADX + ∠XDE = 180° (Sum of ∡s on a straight line)

∠CDA + 56° + 62° = 180°

∠CDA = 180° – 56° – 62°

∠CDA = 62°

c)

∠BXD = 42° + 62°

∠BXD = 104°

16) a) x = 180° – 152° (interior ∡s)

x = 28°

b)

y = 28° + 104°

y = 132°

C D E

A B K

X

62°

42°

56°

D

A

B

E

G

F

C

x y

152°




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17) ∠AOE + ∠EOD + ∠DOC + ∠COB = 180° (Sum of s on a straight line)

If ∠AOE = ∠EOD, ∠DOC = ∠COB,

∠EOD + ∠EOD + ∠DOC + ∠DOC = 180°

2∠EOD + 2∠DOC = 180°

∠EOD + ∠DOC = 90°

∠EOC = ∠EOD + ∠DOC

∠EOC = 90°

18) Let x be the size of the angle

Supplement of angle = 180 – x

Complement of angle = 90 – x

(180 – x) = 4 x (90 – x) – 15

180 – x = 360 – 4x – 15

-x + 4x = 360 – 180 – 15

3x = 165

x = 55

Size of angle = 55°




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19) a) LM bisects PMN -> ∠PML = ∠LMN

∠PMN + 115° = 180° (interior ∡s)

∠PML + ∠LMN + 115° = 180°

2∠LMN = 180° – 115°

2∠LMN = 65°

∠LMN = 32.5°

b) ∠PML = ∠LMN

∠PML = 32.5°

∠PML + ∠MLR = 180° (interior ∡s)

32.5° + ∠MLR = 180°

∠MLR = 180° – 32.5°

∠MLR = 147.5°

c) ∠LKU = ∠PML (alternate ∡s)

∠LKU = 32.5°

20)

(p – x)° + (q – r ) ° = 180° (interior ∡s)

p° – x° + q° – r° = 180°

- x° = 180° – p° – q° + r°

x = p + q – r – 180

x

p

q

r l2

l1

x

r



topic: angles 2 - jimmymaths.com · size of angle = 55° [email protected]

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