A boat sails 6km on a bearing of 040o and then 8km on a bearing 100o .(1) how far is the boat from its starting points (2) calculate the bearing of the boat from its starting point. (leave your answer to the nearest km and degree.?

Dende1

4 years ago

Statement of the given problem,
A boat sails 6 km on a bearing of 040° and then 8 km on a bearing of 100°, (a) how far is the boat from the starting point, (b) calculate the bearing of the boat from its starting point?
Let
D1 denotes the displacement vector AB of the given boat
of magnitude 6 km with bearing 040°
D2 denotes the displacement vector BC of the given boat
of magnitude 8 km with bearing 100°
î & ĵ denote two unit vectors along East & North directions respectively.
Hence from above data we get following relations,
D1 = 6*(sin 040°)*î + 6*(cos 040°)*ĵ
D2 = 8*(sin 100°)*î + 8*(cos 100°)*ĵ
Therefore the required,
magnitude of resultant displacement vector AC
= | D1 + D2 |
= | {6*(sin 040°) + 8*(sin 100°)}*î + {6*(cos 040°) + 8*(cos 100°)}*ĵ |
= √[{6*(sin 040°) + 8*(sin 100°)}^2 + {6*(cos 040°) + 8*(cos 100°)}^2]
= √148 = 12.166 (km) [Ans]
direction of resultant displacement vector AC
= arctan [{6*(cos 040°) + 8*(cos 100°)}/{6*(sin 040°) + 8*(sin 100°)}]
= 15.285° (North of East)
hence, bearing of the finishing point C of the given boat from its starting point A
= 090° - 15.285° = 074.715° [Ans]

Jerrycurl12

4 years ago

1.using cosine rule
x²=6²+8²- 2(6)(8)cos120
x²= 36+64 -(-48)
x= square root of 148
x= 12km to the nearest km

2. using sune rule

8/siny = 12/sin 120

siny= 0.575
y= 35°

bearing of the joat from its starting point
= 40°+35°

= 75°

Panthraleo

1 year ago

7.21km