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Mathematics Form 3 Chapter 5 Trigonometric Ratio Mathematics Form 3 Chapter 5 Trigonometric Ratio
4 3
9. (a) cos ∠BAC = PT3 Standard Practice 5 4. sin x = 2. (a) AB
5 5 sin x = AC ✓
AB = 4 cm, AC = 5 cm Section A / Bahagian A 3 = 3
2
AD = AC + CD 3 AC 5 cos x/ kos x = BC ✓
2
1. cos x = AC = 5 cm AC
= 5 + 12 2 5
2
= 13 cm NP 3 tan y = 1.6 AB
= tan x = ✓
12 10 5 CD BC
cos ∠ADC = 3 = 1.6
13 NP = × 10 5
5
(b) (i) tan x = tan ∠DAE NP = 6 cm CD = 8 cm (b) tan x = 0.4 –1
CE = 4 Answer: B x = tan 0.4
BC 3 NK = 2 × 6 cm = 21.8°
4 = DE + 7 = 12 cm
3 12 5. Answer: A
2
2
DE + 7 = 16 NL = 10 – 6 Section C / Bahagian C
DE = 9 cm = 8 cm
NL 6. Angle between the line BK and the Plane 1. (a) (i) ∠PMQ
4 tan y =
(ii) x = tan NK ADK = ∠BKA (ii) ∠PLQ
–1
3 8
= 53.13° = AK = AD + DK
2
2
12 8
2 = 24 + 7 (b) (i) cos x =
2
2
3 = 17
(c) cos x = 3 = 25 cm NL 8
5 =
AB = 3 Answer: C 10 76.5 17
30 5 tan ∠BKA = 25 NL = 36 cm
3 2. Assume the length of sides AB and BC is
AB = × 30 = 18 cm 10
5 2 cm and AD is the height of the equilateral ∠BKA = tan –1   (ii) tan y = 1
25
2
BD = 30 – 18 triangle. ∠BKA = 21.8° NL = 1
2
= 24 cm LM
2
DC AD = 2 – 1 2 Answer: D 36
tan y = = 1
BD =  LM
3
10 LM = 36 cm
= AD
24 sin 60° = Section B 36
10 AB tan z =
y = tan –1 36
3
24  1
= 1. (a) tan 45° z = tan 1
–1
= 22.62° 2  z = 45°
2
Answer: B
(b) sin 60° 1
3. SW = UW = 8 cm 2
UW 8 cos 60°
sin x = = 
3
VW 17 (c) kos 60° 2
8
x = sin –1  
17
(d) sin 45° 1
x = 28.07°
Answer: D

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BOOKLET ANS MATH F3.indd 12 03/01/2020 10:20 AM