Page 23 - Spotlight A+ SPM Additional Mathematics Form 4 & 5

Form
4
Chapter 9 Solution of Triangles Additional Mathematics

5. The diagram below shows the position of 6. The diagram below shows the position of two
rambutan tree (R), durian tree (D), mango tree ports, A and B and a ship sailing in an ocean.
(M) and a hut in an orchard.

17 m
Durian Mango
Rambutan

13 m 9 m
1 ©PAN ASIA PUBLICATIONS
67°
32°
Port 25 km Port
Hut A B

Given that the distance of durian tree and The distance from port A to port B is 25 km.
mango tree from the hut are the same. Find the Find the distance between the ship and
distance between durian tree and rambutan port B. C3
tree. C4

SPM Simulation HOTS Questions

Paper 2
1. The diagram below shows a quadrilateral (b) Length of BD can be find by using
ABCD. cosine rule.
2
BD = CD + BC − 2(BC)(CD) cos ∠BCD
2
2
= 6 + 7 − 2(6)(7) cos 72.25°
2
2
B
6 cm = 59.39
C
53° BD = 59.39
= 7.71 cm
7 cm CHAP
(c) Using sine rule, 9
D sin ∠BAD sin 53°
10 cm =
A
7.71 10
The area of ΔBCD is 20 cm and ∠BCD is an sin ∠BAD = sin 53° × 7.71
2
10
acute angled. Calculate C4 = 0.6157
(a) ∠BCD, ∠BAD = 38°
(b) length BD,
(c) ∠ADB, and Thus,
2
(d) the area, in cm , of quadrilateral ABCD. ∠ADB = 180° − 53° − 38°
= 89°
Examiner's Comment:
1
(a) Given ΔBCD = 20 cm2 (d) Area ΔABD = (10)(7.71) sin 89°
2
(6)(7) sin ∠BCD = 20 = 38.54 cm 2
2 2
sin ∠BCD = 20 × Area of equilateral ABCD
42 = Area ΔABD + Area ΔBCD
= 0.9524 = 38.54 + 20
∠BCD = 72.25° = 58.54 cm 2
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