Page 721 - SET 1 Koleksi Soalan dan Skema Jawapan SPM KBSM

14 (i)Use cos rule
2
2
13 = 12 + 15 2 2(12)(15)COS <DCE K1

0
< PRS = 56.25 N1

(ii)Find
< BCA = 180 56.25 0
0
= 123.75 0 K1
Use sin rule
K1
=

BC = 18.04 cm N1

0
0
0
0
(iii) < ABC = 180 30 ( 180 56.25 )
= 26.25 0

Use formulae area of triangle
K1
Area of ABC = (30)(18.04)

N1
= 119.68 cm 2

(b)

B

18.04 cm 18.04 cm

56.25
N
C
Find NBC
0
0
< NBC = 180 - 2(56.25 ) K1
0
= 67.50
Use sin rule OR cos rule
K1 =

N1 BC = 20.0453 cm 10