Analysis and Approaches for IBDP Maths Ebook 1 | Page 129

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Solution
(a)
(b)
(c)
EAC ˆ
�180 � 128
� 52
The bearing of A from E
�360 � 52
(M1) for finding true bearing
� 308
A1 N2
[2]
ACE ˆ
�180 �52 � 70
(M1) for valid approach
� 58
(A1) for correct value
AE AC
�
sin ACE ˆ sin AEC ˆ
(M1) for sine rule
AE 300
�
sin58 sin 70
(A1) for substitution
AE � 270.7421802
AE � 271 km
A1 N2
[5]
2 2 2
DE � AD � AE � 2(AD)(AE) cos EAC ˆ
(M1) for cosine rule
2 2 2
DE �(400) �270.7421802
�2(400)(270.7421802) cos 52
(A1) for substitution
DE � 316.153292
DE � 316 km
A1 N2
(d) Note that B lies on AC such that BE � AC . (M1) for valid approach
ˆ BE
sin EAC � AE
sin52
BE
�
270.7421802
BE � 213.3477495
(A1) for correct value
Let v km/h be the speed of the ferry.
BE DE
�
v 100
(M1) for valid approach
213.3477495 316.153292
�
v
100
(A1) for substitution
v � 67.4823748
Therefore, the speed of the ferry is 67.5 km/h. A1 N3
[3]
[5]
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