Angles of Elevation and Depression (SPM WITH WORKINGS)

Angles of Elevation Elevation and Depression

19

Paper 1

4.

1.

R

T

x

6m

x Q

12 m

8m

Q

12 8 θ = 56.3° The angle of elevation of T from from P is 56.3°. tan

P

6 tan x = — 9 x = 33°41′

θ

P

9m

θ = –––

The angle of depression of P from R is 33°41′. Answer: D 5.

Answer: A

G

2.

R

h m

26.5°

E

F

20 m

41.5°

P

12 m

FG tan 26.5° = –––– 20 FG = 20 × tan 26.5° = 9.97 m

Q

h tan 41.5° = ––– 12 h = 12 × tan 41.5° = 10.62 m

Answer: B 6.

Answer: B 3.

T

H 38.4°

62 m

48° U

38.4° F

9.5 m

V

G

62 tan 48° = –––– UV 62 UV = = ––––––– tan 48° = 55.8 m

FH tan 38.4° = –––– 9.5 FH = = 9.5 × tan 38.4° = 7.53 m

Answer: B

Answer: B

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Penerbitan Pelangi Sdn. Bhd.

Mathematics

Chapter 19

SPM

4.

P

Paper 1

30 m

1.

Y

P

14 m

tan

Q

R

12 m

X

50 m

30 50

∠PYX

=

∠PYX

= 30°58

�

�

The angle of depression of point Y from from P is 30°58 . tan

∠QRP

=

∠QRP =

Answer: A

14 12 �

49°24

5.

�

The angle of elevation of P from R is 49°24 . Answer: C

32 m 42°

2.

m x m

G 65°

tan 42° = 7m

x = 65° F

x N

32 x 32 tan 42°

= 35.54

E

Answer: C

7 tan 65° = ––– FN 7 FN = ––––––– tan 65°

6. P

10°

= 3.264 m FE = 2 × 3.264 = 6.528 m

3.5 m 5°

T

7 tan x = –––––– 6.528 x = 47°

Y

Let P = Eye level 3.5 TX 3.5 TX = tan 5°

tan 5° =

The angle of elevation of G from E is is 47°. Answer: D

= 40.01 m

3.

3.5 TY 3.5 TY = tan 10°

tan 10° =

V

W

T

= 19.85 m U

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

S

V from The angle of elevation of V from T is is

∠WTV .

Answer: A

Answer: D

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P

2

X

Mathematics

7.

10. x

U

35°

Angle of depression

35°

24 m X

5m

Cat

Answer: B

38° V

Chapter 19

Nur

S

x

SPM

T

12 m

UV tan 38° = –––– 12 UV = 12 × tan 38° = 9.375 m

11. P

8m

SX = 24 − 9.375 = 14.625 m

26°

R

SX tan x = –––– UX 14.625 = –––––– 12 x = 50°38′

Q

8 tan 26° = –––– RQ 8 RQ = –––––– tan 26° = 16.4 m

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

Answer: D

Answer: B

8. E

G θ

Paper 1 1. F

W

H

Answer: A

20 m

9.

Q

62° X

h

YW sin 62° = –––– 20 YW = = 20 × sin 62° = 17.66 m

28°

P

Y

30 m

h tan 28° = ––– 30 h = 30 × tan 28° = 15.95 m

Answer: C

Answer: B

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Mathematics

SPM

Chapter 19

2.

HF tan ∠ HGF = –––– 13 9.9955 = ––––––– 13 ∠ HGF = = 37°33′

V

U

30°

42° 6m

X

Y

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

UV tan 42° = –––– 6 UV = 6 × tan 42° = 5.4024 m

5. M

Angle of depression

UV tan 30° = –––– XU UV XU = ––––––– tan 30° 5.4024 = ––––––– tan 30°

5m x

9m

4m

= 9.36 m

P

Answer: C 3.

N

4m Q

8.5 m

5 tan x = –––– 8.5 x = 30.47°

T

The angle of depression of N from from M is is 30.47°. Answer: A

26°

P

4m

8m

Q

R

6.

QT tan 26° = –––– 8 QT = 8 × tan 26° = 3.9019 m

tan

T R

S

QT = –––– 4 3.9019 = ––––––– 4 ∠TPQ = 44°17′ ∠TPQ

P

Q

The angle of elevation of T from from P is

∠TPQ.

Answer: D

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

7.

T

Answer: B

3m

4. R

H

Angle of depression depression P

E

48° 9m

F

13 m

G

RP tan 18.5° = –––– 5.5 RP = 5.5 × tan 18.5° = 1.84 m

HF tan 48° = –––– 9 HF = 9 × tan 48° = 9.9955 m

PT = = 3 + 1.84 = 4.84 = 4.8 m Answer: D

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18.5° 5.5 m

4

Q

Mathematics

8.

Chapter 19

SPM

11.

T

R

R T 2m

28° V

10° 16°

P

U

16 m

RU tan 28° = –––– 16 RU = 16 × tan 28° = 8.51 m

Q

24 m

RQ tan 10° = –––– 24 RQ = 24 × tan 10° = 4.232 m

TR = 8.51 − 2 = 6.51 m

TQ tan 16° = –––– 24 TQ = 24 × tan 16° = 6.882 m

Answer: B 9.

TR = TQ − RQ = 6.882 − 4.232 = 2.65 m

Q

The distance between the two birds is 2.65 m. P

Answer: C 12.

M

22°

L

N

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

22° 30 m

Answer: A

30 m

Q

H

6m 9m

G E

15 m

N

MN = = 48.48 + 30 = 78.48 = 78.5 m

3m

y

120 m

MT tan 22° = –––– PT MT = PT × tan 22° = 120 × tan 22° = 48.48 m

10.

x

T

P

F

Answer: D

9 tan x = ––– 15 x = 30°58′

13.

Angle of depression

3 tan y = ––– 15 y = 11°19′ x − y = 30°58′ − 11°19′ = 19°39′

U

x

V

T

51.4 m

20.8 m R

The difference between the angles of elevation of H and G from E is 19°39′.

350 m

Q

Answer: B

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Mathematics

SPM

Chapter 19

VT = = 51.4 − 20.8 = 30.6 m

16.

V

VT tan x = –––– UT 30.6 = –––– 350 x = 5°

26 m

Y

X

12.5° 10 m

Z

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

XZ 2 = XY 2 − YZ 2 XZ = 262 − 10 2 = 24 m

Answer: B 14.

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

R

40°

T 1.8 m

Answer: C

S 1.8 m

P

Q

6m

17. H

RT tan 40° = ––– ST RT = ST × tan 40° = 6 × tan 40° = 5.03 m RP = 5.03 + 1.8 = 6.83 m

63°

P

The height of the bird from the ground is 6.83 m.

HQ tan 63° = –––– 760 HQ = 760 × tan 63° = 1492 m

Answer: D 15.

Q

760 m

H

The height of the helicopter from the ground is 1492 m.

72°

Answer: C 51.6°

F

6.5 m

72°

18. H

G

E

HF tan 51.6° = –––– 6.5 HF = 6.5 × tan 51.6° = 8.2 m

E

x

HF tan 72° = –––– FG HF FG = ––––––– tan 72° 8.2 = ––––––– tan 72°

F

HG tan 32.5° = –––– 80 HG = 80 × tan 32.5° = 50.966 m

= 2.66 m Answer: C

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32.5° 8 0 m

6

120 m

G

Mathematics

21.

HG tan x = –––– FG 50.966 = ––––––– 120 x = 23.01°

Chapter 19

T

P

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

34 m

R

28.5° 16 m

Answer: A 19.

SPM

Q

T

55°

QR2 = PR2 − PQ 2

35°

QR = 342 − 16 2 = 30 m 55°

35°

R

Q

TQ tan 28.5° = –––– 16 TQ = 16 × tan 28.5° = 8.687 m

P

80 m

TR tan 35° = –––– 80 TR = 80 × tan 35° = 56.02 m

tan

TR tan 55° = –––– RQ

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

TR RQ = ––––––– tan 55° 56.02 = ––––––– tan 55°

Answer: B V

22.

= 39.23 m PQ = = = =

TQ = –––– QR 8.687 = ––––– 30 ∠TRQ = 16°9′ ∠TRQ

PR − RQ 80 − 39.23 40.77 40.8 m

4.5 m 25 m

R

The distance between the two cars is 40.8 m.

Q

7m

T

Answer: C

P

20.

PQ2 = QR2 − RP2

H

J

PQ = �� 252 − 7 2 = 24 m

24 PT = ––– 2 = 12 m P

Q

R

tan The angle of depression of Q from H is ∠ JHQ. Answer: A

VT PT 4.5 = ––– 12 ∠VPT = 20°33′ ∠VPT = ––––

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

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Angles of Elevation and Depression (SPM WITH WORKINGS)

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