19[A Math CD]

1 © Penerbitan Pelangi Sdn. Bhd.

Paper 1

1.

8 mP Q

T

θ

12 m

tan θ = 12–––8

θ = 56.3°The angle of elevation of T from P is 56.3°.

Answer: A

2.

12 m

h m

R

QP41.5°

tan 41.5° = h–––12

h = 12 × tan 41.5° = 10.62 m

Answer: B

3.

9.5 m

H

GF

38.4°

38.4°

tan 38.4° = FH––––9.5

FH = 9.5 × tan 38.4° = 7.53 m

Answer: B

4.

9 m

6 m

x

x

R

Q P

tan x = 6—9

x = 33°41′

The angle of depression of P from R is 33°41′.

Answer: D

5.

20 m26.5°

E

G

F

tan 26.5° = FG––––20

FG = 20 × tan 26.5° = 9.97 m

Answer: B

6.

62 m

48°U

T

V

tan 48° = 62––––UV

UV = 62–––––––tan 48°

= 55.8 m

Answer: B

CHAPTER

19 Angles of Elevation and Depression

CHAPTER

2

Mathematics SPM Chapter 19

© Penerbitan Pelangi Sdn. Bhd.

Paper 1

1. P

RQ 12 m

14 m

tan ∠QRP = 1412

∠QRP = 49°249

The angle of elevation of P from R is 49°249.

Answer: C

2. G

EF

7 m

N

65°

65° x

tan 65° = 7–––FN

FN = 7–––––––tan 65°

= 3.264 m FE = 2 × 3.264 = 6.528 m

tan x = 7––––––6.528

x = 47°

The angle of elevation of G from E is 47°.

Answer: D

3.

U S

TW

V

The angle of elevation of V from T is ∠WTV.

Answer: D

4.

30 m

50 m

P

XY

tan ∠PYX = 3050

∠PYX = 30°589

The angle of depression of point Y from P is 30°589.

Answer: A

5.

32 m

x m42°

tan 42° = 32x

x = 32tan 42°

= 35.54

Answer: C

6.

5°10° 3.5 m

PP

T Y X

Let P = Eye level

tan 5° = 3.5TX

TX = 3.5tan 5°

= 40.01 m

tan 10° = 3.5TY

TY = 3.5tan 10°

= 19.85 m

XY = TX – TY = 40.01 – 19.85 = 20.16 m

Answer: A

3



7.

12 m

24 m

38°

x

X

TV

U

S

x

tan 38° = UV––––12

UV = 12 × tan 38° = 9.375 m SX = 24 − 9.375 = 14.625 m

tan x = SX––––UX

= 14.625––––––12

x = 50°38′The angle of depression of U from S is 50°38′.

Answer: B

8.

F H

E Gθ

Answer: A

9.

30 m28°

h

Q

P

tan 28° = h–––30

h = 30 × tan 28° = 15.95 m

Answer: B

10.

35°

35°

5 m

Nur

Angle of depression

Cat

Answer: B

11.

8 m

26°Q

P

R

tan 26° = 8––––RQ

RQ = 8––––––tan 26°

= 16.4 m

Answer: D

Paper 1

1.

20 m

62°YX

W

sin 62° = YW––––20

YW = 20 × sin 62° = 17.66 m

Answer: C

4



2. V

X

U6 m

30° 42°

Y

tan 42° = UV––––6

UV = 6 × tan 42° = 5.4024 m

tan 30° = UV––––XU

XU = UV–––––––tan 30°

= 5.4024–––––––tan 30°

= 9.36 m

Answer: C

3.

4 mP Q

T

8 m26°

R

tan 26° = QT

––––8

QT = 8 × tan 26° = 3.9019 m

tan ∠TPQ = QT

––––4

= 3.9019–––––––4

∠TPQ = 44°17′

The angle of elevation of T from P is 44°17′.

Answer: B

4. H

EF G13 m

Angle of depression

9 m48°

tan 48° = HF––––9

HF = 9 × tan 48° = 9.9955 m

tan ∠HGF = HF––––13

= 9.9955–––––––13

∠HGF = 37°33′

The angle of depression of G from H is 37°33′.

Answer: C

5. M

P Q

Nx

Angle of depression

9 m

5 m

4 m 4 m

8.5 m

tan x = 5––––8.5

x = 30.47°

The angle of depression of N from M is 30.47°.

Answer: A

6. T

S

R

P Q

The angle of elevation of T from P is ∠TPQ.

Answer: D

7.

R

T

3 m

18.5°5.5 mP Q

tan 18.5° = RP––––5.5

RP = 5.5 × tan 18.5° = 1.84 m PT = 3 + 1.84 = 4.84 = 4.8 m

Answer: D

5



8.

28°

R

U

T

2 m

16 mV

tan 28° = RU––––16

RU = 16 × tan 28° = 8.51 m TR = 8.51 − 2 = 6.51 m

Answer: B

9.

L N

P

Q

The angle of elevation of the window cleaner, P, from the man, L, is ∠NLP.

Answer: A

10. H

Fy

xG

E

6 m9 m

3 m

15 m

tan x = 9–––15

x = 30°58′

tan y = 3–––15

y = 11°19′x − y = 30°58′ − 11°19′ = 19°39′The difference between the angles of elevation of H and G from E is 19°39′.

Answer: B

11.

16°10°P

T

R

Q24 m

tan 10° = RQ

––––24

RQ = 24 × tan 10° = 4.232 m

tan 16° = TQ

––––24

TQ = 24 × tan 16° = 6.882 m TR = TQ − RQ = 6.882 − 4.232 = 2.65 mThe distance between the two birds is 2.65 m.

Answer: C

12.

22°

22°

Q N

TP

M

30 m 30 m

120 m

tan 22° = MT––––PT

MT = PT × tan 22° = 120 × tan 22° = 48.48 m MN = 48.48 + 30 = 78.48 = 78.5 m

Answer: D

13.

R

T

V

U

Q

20.8 m

51.4 m

Angle of depression

350 m

x

6



VT = 51.4 − 20.8 = 30.6 m

tan x = VT––––UT

= 30.6––––350

x = 5°The angle of depression of U from V is 5°.

Answer: B

14. R

T

P Q

S1.8 m1.8 m

40°

6 m

tan 40° = RT–––ST

RT = ST × tan 40° = 6 × tan 40° = 5.03 m RP = 5.03 + 1.8 = 6.83 mThe height of the bird from the ground is 6.83 m.

Answer: D

15. H

E

GF

6.5 m51.6° 72°

72°

tan 51.6° = HF––––6.5

HF = 6.5 × tan 51.6° = 8.2 m

tan 72° = HF––––FG

FG = HF–––––––tan 72°

= 8.2–––––––tan 72°

= 2.66 m

Answer: C

16.

Y X

V

Z

10 m

26 m12.5°

XZ 2 = XY 2 − YZ 2

XZ = 262 − 102

= 24 m

tan 12.5° = ZV––––XZ

ZV = XZ × tan 12.5° = 24 × tan 12.5° = 5.32 m

Answer: C

17.

P Q

H

63°760 m

tan 63° = HQ

––––760

HQ = 760 × tan 63° = 1492 mThe height of the helicopter from the ground is 1492 m.

Answer: C

18.

G

H

E

F

x

32.5°

120 m

80 m

tan 32.5° = HG––––80

HG = 80 × tan 32.5° = 50.966 m

7



tan x = HG––––FG

= 50.966–––––––120

x = 23.01°

The angle of depression of F from H is 23.01°.

Answer: A

19. T

R Q P35°

80 m

55°

35°55°

tan 35° = TR––––80

TR = 80 × tan 35° = 56.02 m

tan 55° = TR––––RQ

RQ = TR–––––––tan 55°

= 56.02–––––––tan 55°

= 39.23 m

PQ = PR − RQ = 80 − 39.23 = 40.77 = 40.8 m

The distance between the two cars is 40.8 m.

Answer: C

20.

P Q R

HJ

The angle of depression of Q from H is ∠JHQ.

Answer: A

21.

Q

T

RP

16 m

34 m28.5°

QR2 = PR2 − PQ2

QR = 342 − 162

= 30 m

tan 28.5° = TQ

––––16

TQ = 16 × tan 28.5° = 8.687 m

tan ∠TRQ = TQ

––––QR

= 8.687–––––30

∠TRQ = 16°9′

The angle of elevation of T from R is 16°9′.

Answer: B

22.

P

R Q

T

25 m

4.5 m

7 m

V

PQ2 = QR2 − RP2

PQ = ABBBBBB252 − 72

= 24 m

PT = 24–––2

= 12 m

tan ∠VPT = VT––––PT

= 4.5–––12

∠VPT = 20°33′

The angle of elevation of V from P is 20°33′.

Answer: C

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