1

Additional Mathematics SPM Chapter 10

© Penerbitan Pelangi Sdn. Bhd.

1. (a) Using sine rule,

sin ∠BAC–––––––––

BC =

sin ∠ABC–––––––––AC

sin ∠BAC = sin 50°–––––––10 × 12

∠BAC = 66°49′

(b) ∠ACB = 180° – 50° – 66°49′ = 63°11′

AB–––––––––

sin ∠ACB =

AC–––––––––sin ∠ABC

AB = 10–––––––sin 50° × sin 63°11′

= 11.65 cm

2. (a) ∠PRQ = 180° − 50° − 60° = 70°

(b) PQ–––––––

sin ∠R = PR–––––––

sin ∠Q

PQ–––––––

sin 70° = 14–––––––

sin 50°

PQ = 14––––––– sin 50°

× sin 70°

= 17.17 cm

(c) RQ––––––– sin ∠P

= PR––––––– sin ∠Q

RQ–––––––

sin 60° = 14–––––––

sin 50°

RQ = 14––––––– sin 50°

× sin 60°

= 15.83 cm

3. (a) sin ∠BAC–––––––––

BC =

sin ∠ACB–––––––––AB

sin ∠BAC–––––––––9

= sin 20°––––––– 5

sin ∠BAC = sin 20°––––––– 5

× 9

∠BAC = 38°

(b) ∠ABC = 180° − 20° − 38° = 122°

(c) AC––––––– sin ∠B

= AB–––––––

sin ∠C

AC–––––––

sin 122° = 5–––––––

sin 20°

AC = 5 sin 122°–––––––––sin 20°

= 12.4 cm

4. (a) PQ–––––––

sin ∠R =

QR––––––– sin ∠P

PQ–––––––

sin 50° = 4–––––––

sin 30°

PQ = 4––––––– sin 30°

× sin 50°

= 6.128 cm

(b) ∠PQR = 180° − 30° − 50° = 100°

(c) PR–––––––

sin ∠Q =

RQ––––––– sin ∠P

PR–––––––

sin 100° = 4–––––––

sin 30°

PR = 4 sin 100°––––––––– sin 30°

= 7.878 cm

5. (a)

8 cm10 cm

50°A P

B

C1 C2

In ∆ABC2,

sin ∠AC2B–––––––––

AB =

sin ∠A–––––––BC2

sin ∠AC2B–––––––––

10 =

sin 50°–––––––8

CHAPTER

10 Solution of Triangles

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Additional Mathematics SPM Chapter 10

© Penerbitan Pelangi Sdn. Bhd.

sin ∠AC2B = 10–––8 × sin 50°

∠AC2B = 73°15′

∠AC1B = 180° − 73°15′ = 106°45′

Therefore, the two possible angles of ACB are 73°15′ and 106°45′.

(b) In ∆ABC1, ∠ABC1 = 180° − 50° − 106°45′ = 23°15′

AC1–––––––––

sin 23°15′ = 8–––––––

sin 50°

AC1 = 8––––––– sin 50°

× sin 23°15′

= 4.122 cm

In ∆ABC2, ∠ABC2 = 180° − 50° − 73°15′ = 56°45′

AC2–––––––––

sin ∠ABC2

= BC2–––––––––

sin ∠BAC2

AC2–––––––––

sin 56°45′ = 8–––––––

sin 50°

AC2 = 8––––––– sin 50°

× sin 56°45′

= 8.734 cm

6. (a)

8 cm7 cm

70°A

B

C1 C2

∠BC1C2 = ∠BC2C1 = 70°

∠AC1B = 180° − 70° = 110°

(b) In ∆ABC1,

sin ∠BAC1–––––––––

BC1

= sin ∠BC1A–––––––––

AB

sin ∠BAC1–––––––––

7 = sin 110°–––––––

8

sin ∠BAC1 = 7 × sin 110°––––––––––– 8

∠BAC1 = 55°19′

(c) ∠ABC2 = 180° − ∠BAC1 − ∠BC2A = 180° − 55°19′ − 70° = 54°41′

AC2–––––––––

sin ∠ABC2

= AB––––––––––

sin ∠AC2B

AC2–––––––––

sin 54°41′ =

8––––––– sin 70°

AC2 = 8–––––––

sin 70° × sin 54°41′

= 6.947 cm

7. (a)

13 cm

30°A P

B

C

In ∆ABC, ∠BCA = 90°

sin 30° = BC––– AB

= BC––– 13

BC = 13 × sin 30° = 6.5 cm

(b) 6.5 cm , BC , 13 cm

(c) sin ∠BPA–––––––––13

= sin 30°––––––– 12

sin ∠BPA = 13 × sin 30°––––––––––– 12

∠BPA = 32°48′

∠ABP = 180° − 32°48′ – 30° = 117°12′

8. (a) ∠ABE = 1—2

∠ABC

= 1—2

× 60°

= 30°

In ∆ABD,

sin ∠ADB–––––––––7

= sin 30°––––––– 12

sin ∠ADB = 7–––12

× sin 30°

∠ADB = 16°57′

(b) ∠DEA = 90° In ∆ADE, cos ∠ADE = DE––––

DA

cos 16°57′ = DE–––– 12

DE = 12 × cos 16°57′ = 11.48 cm

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Additional Mathematics SPM Chapter 10

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(c) ∠BAD = 180° − ∠ABE – ∠ADB = 180° − 30° − 16°57′ = 133°3′

9. BC2 = 102 + 122 − 2(10)(12) cos 48° = 83.41 BC = 9.133 cm

10. 102 = 8.22 + 7.52 − 2(8.2)(7.5) cos ∠PQR

cos ∠PQR = 8.22 + 7.52 − 102––––––––––––––

2(8.2)(7.5) ∠PQR = 78°59′

11. cos ∠ABC = 72 + 82 − 142–––––––––––

2(7)(8) = −0.7411 ∠ABC = 137°49′

12. RQ2 = 52 + 72 − 2(5)(7) cos 150° = 134.6 RQ = 11.6 cm

13. (a) ∠BDC = 180° − 40°––––––––––2 = 70°

∠ADB = 180° − 70° = 110°

(b) AB2 = 102 + 82 − 2(10)(8) cos 110° = 218.7 AB = 14.79 cm

14. In ∆ABC, BC2 = AB2 + AC2

= 122 + 52

BC = 13 cm

DC = 13–––2

= 6.5 cm

In ∆ACD, cos ∠ADC = 72 + 6.52 − 52

––––––––––––2(7)(6.5)

∠ADC = 43°17′

15. (a) In ∆ABD, BD2 = 82 + 32 − 2(8)(3) cos 120° = 97 BD = 9.849 cm

(b) In ∆BCD,

sin ∠BDC–––––––––BC

= sin ∠BCD–––––––––BD

sin ∠BDC–––––––––12

= sin 50°––––––– 9.849

sin ∠BDC = sin 50°––––––– 9.849

× 12

∠BDC = 68°58′

16. (a) In ∆ABC, ∠C = 180° − 30° − 80° = 70°

AB–––––––

sin 70° = 10–––––––

sin 80°

AB = 10––––––– sin 80°

× sin 70°

= 9.542 cm

(b) In ∆ABC, BC2 = 9.5422 + 102 − 2(9.542)(10) cos 30° BC = 5.077 cm

(c) CE = 1—2

BC

= 1—2

× 5.077

= 2.539 cm

cos ∠CDE = 52 + 62 − 2.5392––––––––––––––

2(5)(6) ∠CDE = 24°36′

17. Area of ∆ABC = 1—2

× 5 × 8 × sin 30°

= 10 cm2

18. Area of ∆PQR = 1—2

× 12 × 10 × sin 120°

= 51.96 cm2

19. Area of ∆ABC = 20 cm2

1—2

× 8 × 13 × sin ∠BAC = 20

sin ∠BAC = 20 × 2––––––8 × 13

∠BAC = 22°37′

20. Area of ∆ABC = 1—2

× 5 × 8 × sin 25°

Volume of the prism = 1—2

× 5 × 8 × sin 25° × 10

= 84.52 cm3

21. (a) In ∆ABC,

AC–––––––

sin 70° = 10–––––––

sin 50°

AC = sin 70°1 10––––––– sin 50° 2

= 12.27 cm

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Additional Mathematics SPM Chapter 10

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(b) In ∆ACV, tan ∠ACV = 12–––––

12.27 ∠ACV = 44°22′

Therefore, the angle between the line VC and the plane ABC is 44°22′.

22. (a)

6 cm

8 cmA

D

B

In ∆ABD, DB2 = 62 + 82

DB = 10 cm

In ∆VOB,

tan ∠VBO = VO–––– BO

tan 30° = VO–––– 5

BO = 1—2

DB

= 1—2

(10)

= 5 VO = 5 × tan 30° = 2.887 cm

(b)

2.887 cm

4 cmO

V

E

Let E be the midpoint of BC.

tan ∠VEO = 2.887––––––4

∠VEO = 35°49′

Therefore, the angle between the plane VBC and the plane ABCD is 35°49′.

1. (a) In ∆ABC, given the area of ∆ABC = 20 cm2

1—2

× 5 × AC × sin 30° = 20

AC = 20 × 2––––––––––5 × sin 30°

= 16 cm

(b) In ∆ACD,

cos ∠ADC = 62 + 122 − 162––––––––––––

2(6)(12) ∠ADC = 121°51′

(c) In ∆ABC, ∠BCA = 180° − (30° + 130°) = 20°

BC–––––––

sin 30° = 5–––––––

sin 20°

BC = 5––––––– sin 20°

× sin 30°

= 7.31 cm

(d) Area of ∆ACD = 1—2

× 6 × 12 × sin 121°51′

= 30.58 cm2

2. (a) (i) In ∆ABC,

sin ∠BAC––––––––– 7

= sin 30°––––––– 5.2

sin ∠BAC = sin 30°–––––––

5.2 × 7

∠BAC = 42°18′

(ii) In ∆ACD, DC2 = 82 + 9.92 − 2(8)(9.9) cos 50° = 60.19 DC = 7.758 cm

(iii) Area of quadrilateral ABCD = Area of ∆ABC + Area of ∆ACD

= 1—2

× 7 × 9.9 × sin 30° +

1—2

× 8 × 9.9 × sin 50°

= 47.66 cm2

(b) (i)

7 cm5.2 cm

30°

A

B

A′C

(ii) ∠BA′C = 180° − ∠BAC = 180° − 42°18′ = 137°42′

3. (a) cos ∠ABC = 102 + 122 − 152–––––––––––––

2(10)(12) ∠ABC = 85°28′

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Additional Mathematics SPM Chapter 10

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(b) (i)

13 cm

15 cm

10 cm

12 cm

50°

C

A

D

D ′

B

In ∆ACD,

sin ∠ADC––––––––– 15

= sin 50°––––––– 13

sin ∠ADC = 15 × sin 50°–––––––––––13

∠ADC = 62°7′ ∠AD′C = 180° – 62°7′ = 117°53′

(ii) ∠DAC = 180° − 50° – 62°7′ = 67°53′

DC–––––––––

sin 67°53′ = 13–––––––

sin 50°

DC = 13 × sin 67°53′––––––––––––––sin 50°

= 15.72 cm

(iii) ∠D′AC = 180° − 50° – 117°53′ = 12°7′

Area of ∆ACD′ = 1—

2 × 13 × 15 × sin 12°7′

= 20.47 cm2

4. (a) Area of ∆ABC = 4.41

1—2

× 4.2 × 4.2 × sin ∠ABC = 4.41

sin ∠ABC = 4.41 × 2––––––––4.2 × 4.2

∠ABC = 150°

∠BAC = 1180° – 150°––––––––––2 2

= 15°

(b)

4.2 cm40°

C

V

B

tan 40° = VB–––– 4.2

VB = 4.2 × tan 40° = 3.524 cm

4.2 cm 4.2 cm15 0°15°

CAE

B

Let E be the midpoint of AC, In ∆ABE,

sin 15° = BE–––– 4.2

BE = 4.2 × sin 15° = 1.087 cm

1.087 cm

3.524 cm

EB

V

In ∆VBE,

tan ∠BEV = 3.524––––––1.087

∠BEV = 72°51′

Therefore, the angle between the plane VAC and the plane ABC is 72°51′.

1. (a) ∠BDC = 180° − 64° = 116°

∠CBD = 180° − 116° − 32° = 32°

Therefore, BD = DC = 3 cm

In ∆BCD, BC2 = 32 + 32 − 2(3)(3) cos 116° = 25.89 BC = 5.088 cm

(b) In ∆ABC, AB2 = AC2 + BC2 − 2(AC)(BC) cos ∠C = 102 + 5.0882 − 2(10)(5.088) cos 32° AB = 6.292 cm

(c) Area of ∆ABD = 1—

2 × AD × BD × sin ∠BDA

= 1—2

× 7 × 3 × sin 64°

= 9.437 cm2

Given ∠ABC is an obtuse angle.

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Additional Mathematics SPM Chapter 10

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2. (a) In ∆ABD, BD2 = 122 + 52

= 169 BD = 13 cm

In ∆DBC, ∠DBC = 180° − (70° + 60°) = 50°

DC–––––––

sin 50° = 13–––––––

sin 60°

DC = 13––––––– sin 60°

× sin 50°

= 11.5 cm

(b) Area of ∆BCD = 1—

2 × BD × DC × sin ∠BDC

= 1—2

× 13 × 11.5 × sin 70°

= 70.24 cm2

(c) Let the vertical height from B to DC be h.

1—2

× h × DC = 70.24

1—2

× h × 11.5 = 70.24

h = 70.24 × 2–––––––––11.5

= 12.22 cm

Therefore, the height from B to DC is 12.22 cm.

3. (a) In ∆ABC, AC2 = 42 + 72 − 2(4)(7) cos 120° = 93 AC = 9.644 cm

(b) ∠ADC = 180° − ∠ABC = 180° − 120° = 60° In ∆ADC,

sin ∠DCA––––––––– 10

= sin 60°––––––– 9.644

sin ∠DCA = sin 60°––––––– 9.644

× 10

∠DCA = 63°54′

∠DAC = 180° − 63°54′ − 60° = 56°6′

(c) Area of ∆ABC

= 1—2

× 4 × 7 × sin 120°

= 12.12 cm2

Area of ∆ACD = 1—

2 × 10 × 9.644 × sin 56°6′

= 40.02 cm2

Area of quadrilateral ABCD = 12.12 + 40.02 = 52.14 cm2

4. (a) In ∆PQS,

PS–––––––

sin 80° = 8–––––––

sin 60°

PS = 8 × sin 80°–––––––––– sin 60°

= 9.097 cm

RS = 1—2

PS

= 1—2

× 9.097

= 4.549 cm

(b) ∠QPS = 180° − (80° + 60°) = 40°

PR = PS + SR = 9.097 + 4.549 = 13.65 cm

In ∆PQR, QR2 = 82 + 13.652 − 2(8)(13.65) cos 40° = 83.02 QR = 9.112 cm

(c) Area of ∆PQR = 1—2

× 8 × 13.65 ×sin 40°

= 35.1 cm2

5. (a) In ∆ABC, Area of ∆ABC = AB3 cm2

1—2

× x × y × sin 60° = AB3

1—2

xy1 AB3–––2 2 = AB3

xy = 4 Using cosine rule, 52 = x

2 + y2 − 2xy cos 60°

25 = x2 + y2 − 2xy1 1—

2 2 25 = x

2 + y2 − xy 25 = (x + y)2 − 2xy − xy 25 = (x + y)2 – 3xy 25 = (x + y)2 − 3(4) (x + y)2 = 37 x + y = ABB37

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Additional Mathematics SPM Chapter 10

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(b) Let h be the vertical height from A to BC. Area of ∆ABC = AB3

1—2

× 5 × h = AB3

h = 2AB3––––

5 = 0.6928 cm

6. (a) Given volume of the cone = 100π cm3

1—3

πr2h = 100π

1—3

× r2 × 12 = 100

r2 = 25 r = 5 cm

5 cm

12 cm

AO

V

In ∆VOA, tan ∠VAO = 12–––

5 ∠VAO = tan−1 1 12–––

5 2 = 67°23′

Therefore, the angle between the edge VA and the base is 67°23′.

(b)

5 cm 5 cm30° 30°

A BE

O

Let E be the midpoint of AB. In ∆AOE,

cos 30° = OE–––5

OE = 5 cos 30° = 4.33 cm

4.33 cm

12 cm

O E

V

In ∆VOE,

tan ∠VEO = 12––––4.33

∠VEO = 70°10′

Therefore, the angle between the plane VAB and the base is 70°10′.

7. (a)

5 cm

12 cmA

E

B

In ∆ABE, AE2 = 52 + 122

= 169 AE = 13 cm

50°13 cmA E

V

In ∆AEV, cos 50° = AE––––

AV

= 13––––AV

AV = 13–––––––cos 50°

= 20.22 cm

(b) In ∆AVE, tan 50° = VE–––

AE

= VE–––13

VE = 13 tan 50° = 15.49 cm

15.49 cm

12 cmF

V

E

Let F be the midpoint of AD, In ∆EFV, tan ∠VFE = 15.49–––––

12 ∠VFE = 52°14′

Therefore, the angle between the planes VDA and VCB is 52°14′.

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Additional Mathematics SPM Chapter 10

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(c)

15.49 cm

5 cmB

V

E

In ∆BVE, tan ∠VBE = 15.49–––––

5 ∠VBE = 72°7′

Therefore, the angle between the plane VAB and the base ABCD is 72°7′.

(d) In ∆VBE, VB2 = 15.492 + 52

VB = 16.28 cm

16.28 cm20.22 cm

12 cm B

V

A

cos ∠AVB = 20.222 + 16.282 − 122 ––––––––––––––––––2(20.22)(16.28)

∠AVB = 36°24′

8. (a) In ∆ABC,

sin ∠BCA––––––––– 4

= sin 30°––––––– 5

sin ∠BCA = 4 × sin 30°–––––––––– 5

∠BCA = 23°35′

∠BAC = 180° − 23°35′ − 30° = 126°25′

BC––––––––––

sin 126°25′ = 5–––––––

sin 30°

BC = 5 × sin 126°25′––––––––––––– sin 30°

= 8.047 cm

(b) EB = 1—2

BC

= 1—2

(8.047)

= 4.024 cm

Area of ∆AEC

= 1—2

× 5 × 4.024 × sin 23°35′

= 4.025 cm2

Since ED = 2AE \ Area of ∆CDE = 2 × 4.025 = 8.05 cm2

1. (a) In ∆ABE,

sin ∠ABE–––––––––8 = sin 30°––––––6

sin ∠ABE = 8 sin 30°––––––––6 ∠ABE = 41°49′ \∠CBE = 180° – 41°49′ = 138°11′

In ∆BCE, CE2 = 62 + 102 – 2(6)(10) cos 138°11′ = 225.4 CE = 15.01 cm

(b) Given area of ∆CDE = 50

1—2 × CE × CD × sin ∠DCE = 50

1—2 × 15.01 × 12 × sin ∠DCE = 50

sin ∠DCE = 50 × 2––––––––––15.01 × 12

∠DCE = 33°43′

2. (a) In ∆ABC, 52 = 82 + 102 – 2(8)(10) cos ∠BCA

cos ∠BCA = 82 + 102 – 52–––––––––––

2(8)(10) ∠BCA = 29°41′

(b) ∠DCE = ∠BCA = 29°41′

In ∆CDE,

CE–––––––sin 50° = 14–––––––––

sin 29°41′

CE = 14 sin 50°–––––––––sin 29°41′

= 21.66 cm

3. (a)

9 cm

9 cm

F

BE

In ∆BEF, EF2 = 92 + 92

EF = ABBB162 cm

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Additional Mathematics SPM Chapter 10

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12 cm

12 cm

C

B

O

A

In ∆ABC, AC2 = 122 + 122

= 288 AC = ABBB288 cm

AO = 1—2 AC

= 1—2

ABBB288 cm

8 cm

288– cm2��

V

OA

In ∆AOV,

AV 2 = 82 + 288––––4

= 64 + 72 = 136 AV = ABBB136 cm

��

3 cm3 cm6 cm

136 cm��136 cm

V

R FB C

In ∆BRV, VR2 = (ABBB136)2 – (6)2

VR = 10 cm

In ∆VRF, VF2 = 32 + 102

VF = ABBB109 cm

109 cm��

162 cm��

V

TE F

In ∆VET, cos ∠VET = ET––––

VE

=

ABBB162–––––2––––––––

ABBB109

= ABBB162––––––

2ABBB109 ∠VET = 52°27′ \ ∠VEF = 52°27′

(b) Area of ∆VEF = 2 × Area of ∆ETV

= 2 × 1—2 × ABBB109 ×

ABBB162–––––2 × sin 52°27′

= 52.68 cm2

4. (a)

8 cm

8 cm

C

A B

In ∆ABC, AC = ABBBBB82 + 82

= ABBB128 cm

6 cm

8 cm

F

A B

In ∆ABF, AF = ABBBBB62 + 82

= 10 cm

6 cm

8 cm

F

B C

In ∆BFC, FC = ABBBBB62 + 82

= 10 cm

10

Additional Mathematics SPM Chapter 10

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10 cm 10 cm

F

A C128 cm��

In ∆AFC, AC2 = AF 2 + FC2 – 2(AF)(FC) cos ∠AFC 128 = 100 + 100 – 200 cos ∠AFC

cos ∠AFC = 100 + 100 – 128–––––––––––––––200

∠AFC = 68°54′

(b) The area of ∆AFC = 1—2

× 10 × 10 × sin 68°54′

= 46.65 cm2

(c) Let O be the intersection point of diagonals AC and BD.

6 cm

B

F

O128– cm

2��

In ∆ABC, AC2 = 82 + 82

= 128 AC = ABBB128 cm

\ BO = ABBB128–––––

2 cm

tan ∠FOB = 6––––––ABBB128–––––

2

= 12–––––ABBB128

∠FOB = 46°41′

Therefore, the angle between the plane AFC and the plane ABCD is 46°41′.

5. (a)

3 cm

7 cm P

V

Q

In ∆PQV, tan ∠VQP = 3—

7 ∠VQP = tan–1 1 3—

7 2 = 23°12′

Therefore, the angle between the line VQ and the horizontal plane PQR is 23°12′.

(b) In ∆PRV, PR = ABBBBB52 – 32

= ABB16 = 4 cm

7 cm8 cm

4 cm PR

Q

In ∆PRQ,

cos ∠RPQ = 42 + 72 – 82––––––––––

2(4)(7) ∠RPQ = 88°59′

(c) In ∆PVQ, VQ = ABBBBB32 + 72

= ABB58 cm

5 cm

8 cm

V

R Q

58 cm��

(ABB58)2 = 52 + 82 – 2(5)(8) cos ∠VRQ

cos ∠VRQ = 52 + 82 – 58––––––––––2(5)(8)

∠VRQ = 67°12′

Area of ∆VRQ = 1—2

× 5 × 8 × sin 67°12′

= 18.44 cm2

6. (a) Area of ∆BCD = 40–––2

= 20 cm2

1—2

× 8 × 10 × sin ∠BCD = 20

sin ∠BCD = 1—2

∠BCD = 30°

(b)

10 cm

8 cm

A

C B

In ∆ABC ∠B = 180° – 30° = 150°

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Additional Mathematics SPM Chapter 10

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AC2 = 82 + 102 – 2(8)(10) cos 150° AC = 17.39 cm

17.39 cm

12 cm

AC

V

In ∆VAC, tan ∠C = 12––––––

17.39 ∠C = 34°36′

Therefore, the angle between the line CV and the plane ABCD is 34°36′.

7. (a)

–4

53

x0

y

Given sin ∠BAC = 3—5

and ∠BAC is an obtuse

angle.

\ cos ∠BAC = – 4—5

In ∆ABC, BC 2 = 122 + 72 – 2(12)(7) cos ∠BAC

= 122 + 72 – 2(12)(7)1– 4—5 2

BC = 18.09 cm

(b) 12 cm 7 cm

D

A

CB

Area of ∆ABC = 1—2

× 12 × 7 × sin ∠A

1—2

× AD × BC = 1—2

× 12 × 7 × 3—5

1—2

× AD × 18.09 = 1—2

× 12 × 7 × 3—5

AD = 12 × 7––––––18.09

× 3—5

= 2.786 cm

Hence, the perpendicular distance from A to BC is 2.786 cm.

8. (a) ∠C = 180° – (80° + 40°) = 60° The shortest side is BC since its opposite angle

is smallest.

BC–––––––sin 40°

= 16–––––––sin 80°

BC = 16 sin 40°–––––––––sin 80°

= 10.44 cm

(b) Area of ∆ABC = 1—2

× 16 × 10.44 × sin 60°

= 72.33 cm2

(c)

40°

16 cm

C

B

A

T

Area of ∆ABC = 72.33

1—2

× 16 × BT = 72.33

BT = 72.33 × 2–––––––––16

= 9.041 cm

Hence, the shortest distance from B to AC is 9.041 cm.

9. (a) In ∆ABC,

sin ∠ACB–––––––––10

= sin 30°––––––6

sin ∠ACB = 10 sin 30°–––––––––6

∠ACB = 123°33′ \ ∠ACD = 180° – 123°33′ = 56°27′ In ∆ACD, AD2 = 62 + 62 – 2(6)(6) cos 56°27′ = 32.208 AD = 5.675 cm

(b) ∠ADC = 180° – 56°27′––––––––––––2

= 123°33′–––––––2

= 61°47′ \ ∠EDF = 61°47′

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Additional Mathematics SPM Chapter 10

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Area of ∆DEF = 1—2

× 12 × 5.675 × sin 61°47′

= 30 cm2

10. (a) ∠ADB = 180° – 80° = 100° In ∆ABD,

AB––––––––

sin 100° = 8–––––––

sin 60°

AB = 8 sin 100°–––––––––sin 60°

= 9.097 cm(b) In ∆ACD,

AC–––––––

sin 80° = 8–––––––

sin 70°

AC = 8 sin 80°––––––––sin 70°

= 8.384 cm

(c) ∠DAC = 180° – (80° + 70°) = 30°

Area of ∆ADC

= 1—2

× 8 × 8.384 × sin 30°

= 16.77 cm2

11. (a) ∠ACD = 180° – 60° = 120°

In ∆ACD, AD2 = 82 + 52 – 2(8)(5) cos 120° = 129 AD = 11.36 cm

(b) In ∆ABC,

AB–––––––sin 60°

= 8–––––––

sin 40°

AB = 8 sin 60°––––––––sin 40°

= 10.78 cm

(c) Area of ∆ABD = Area of ∆ABC + Area of ∆ACD

= 1—2

× 8 × 10.78 × sin 80° + 1—2

× 5 × 8 × sin 120°

= 59.79 cm2

12. (a)

6 cm6 cm

5 cm

5 cm

ECB

DF

A

In ∆ABE, AE 2 = 102 + 62

= 136 AE = 11.66 cm

In ∆BCD, CD2 = 52 + 122

= 169 CD = 13 cm

(b) EF = 1—3

AE

= 1—3

× 11.66

= 3.887 cm

CF = 2—3

CD

= 2—3

× 13

= 8.667 cm

6 cm

8.667 cm3.887 cm

E

F

C

In ∆CEF,

cos ∠CFE = 3.8872 + 8.6672 – 62–––––––––––––––––

2(3.887)(8.667) ∠CFE = 36°25′

(c) Area of ∆CEF

= 1—2

× 3.887 × 8.667 × sin 36°25′

= 10 cm2

Area of ∆BDC = 1—2

× 5 × 12

= 30 cm2

Area of BDFE = 30 – 10 = 20 cm2

13. (a) In ∆ADC,

sin ∠DAC–––––––––10

= sin 35°–––––––8

sin ∠DAC = 10 sin 35°–––––––––8

∠DAC = 45°48′

(b) ∠DAB = 35° + 45°48′ = 80°48′ In ∆ABD, BD2 = 82 + 102 – 2(8)(10) cos 80°48′ = 138.42 BD = 11.77 cm

13

Additional Mathematics SPM Chapter 10

© Penerbitan Pelangi Sdn. Bhd.

14. (a)

O5 cm5 cm

5 cm

6 cm

5 cm

CA

B

D

In ∆BCO, cos ∠BCO = 62 + 52 – 52

––––––––––2(6)(5)

∠BCO = 53°8′

(b) ∠BOC = 180° – 2 × 53°8′ = 180° – 106°16′ = 73°44′

Area of ∆AOD = 1—2

× 5 × 5 × sin 73°44′

= 12 cm2

15. (a) ∠BOC = 180° – 70° = 110°

In ∆BOC, BC2 = 82 + 82 – 2(8)(8) cos 110° BC = 13.11 cm

(b) Area of ∆ABC = 21 1—2

× 8 × 8 ×sin 70°2 = 60.14 cm2

16. (a)

O C

A

B

60°8 c

m

8 cm

30°

10 cm8 cm

OA = OB = AB = 8 cm

∠C = 1—2

∠AOB

= 30°

In ∆ABC,

sin ∠BAC–––––––––10

= sin 30°––––––8

sin ∠BAC = 10 sin 30°–––––––––8

∠BAC = 38°41′

(b) ∠ABC = 180° – (30° + 38°41′) = 180° – 68°41′ = 111°19′

In ∆ABC, AC2 = 82 + 102 – 2(8)(10) cos 111°19′ AC = 14.91 cm

(c) ∠BOC = 2 × ∠BAC = 2 × 38°41′ = 77°22′

Minor arc length BC = 77°22′––––––360°

× 2 × π × 8

= 10.8 cm

17. (a) In ∆RNS, NR2 = 122 + 52

NR = 13 cm

RN

M

13 cm

8 cm

In ∆MNR,

tan ∠MRN = 8–––13

∠MRN = tan–1 1 8–––13 2

= 31°36′

Therefore, the angle between the line MR and the plane PQRS is 31°36′.

(b) R

N

S 12 cm

5 cm

In ∆SNR,

tan ∠SNR = 12–––5

∠SNR = tan–1 1 12–––5 2

= 67°23′

Therefore, the angle between the plane MNR and the plane ADSP is 67°23′.

14

Additional Mathematics SPM Chapter 10

© Penerbitan Pelangi Sdn. Bhd.

(c) In ∆SRQ, SQ2 = 102 + 122

SQ = ABBB244 cm

B

S Q

8 cm

��244 cm

In ∆SBQ,

tan ∠BSQ = 8–––––ABBB244

∠BSQ = 27°7′

18. (a)

B CD

A

8 cm 7 cm

35°

In ∆BAD,

sin 35° = AD–––8

AD = 8 sin 35° = 4.589 cm

Therefore, the shortest distance from A to BC is 4.589 cm.

(b)

B CD

A

8 cm7 cm 7 cm

35°

In ∆ABD,

sin ∠ADB–––––––––8

= sin 35°––––––7

sin ∠ADB = 8 sin 35°––––––––7

∠ADB = 40°58′, 139°2′

19. (a)

O

SP

RQ

15 cm20 cm

12 cm

12 cm

20 cm

In ∆ROS,

cos ∠ROS = 122 + 202 – 152–––––––––––––

2(12)(20)

∠ROS = 48°21′

\ ∠QOR = 180° – 48°21′ = 131°39′

(b) In ∆ROQ, QR2 = 202 + 122 – 2(20)(12) cos 131°39′ = 863 QR = 29.38 cm

20. (a) PQ

––––BC

= 1—3

PQ

––––6

= 1—3

PQ = 2 cm

AP

––––PB

= 1—2

AP

––––2

= 1—2

AP = 1 cm

AQ––––QC

= 1—2

AQ

––––4

= 1—2

AQ = 2 cm

B C

QP

A

4 cm

6 cm

2 cm

1 cm 2 cm

In ∆ABC, cos ∠BAC = 32 + 62 – 62

––––––––––2(3)(6)

∠BAC = 75°31′

(b) Area of ∆APQ–––––––––––––Area of ∆ABC

= 1 1—3 2

2

= 1—9

\ Area of ∆APQ : Area of trapezium BCQP = 1 : 8.