Question

A boat sails 4km on a bearing of 038 degree and then 5km on a bearing of 067 degree.(a)how far is the boat from its starting point.(b) calculate the bearing of the boat from its starting point

Answers

Reiny
simplest way is using vectors.
resultant vector
= (4cos38,4sin38) + (5cos67,5sin67)

= (3.152.., 2.4626..) + (1.9536..., 4.6025..)
= (5.1056.., 7.065...)

distance = √(5.1056..^2 + 7.065.^2)
= appr 8.717 correct to 3 decimals

bearing:
tanØ = 7.065../5.1056..
Ø = 54.146°

OR

Using basic geometry:
make a sketch, on mine I have a triangle with sides 4 and 5 with an angle of 151° between them
by the cosine law:
d^2 = 4^2 + 5^2 - 2(4)(5)cos151
= 8.717 , just like above

let the angle opposite the 5 km side be Ø
by the sine law:
sinØ/5 = sin 151/8.717
sinØ = .27808..
Ø = 16.146
bearing = Ø + 38°
= 16.146+38
= 54.146° , again, just as above
Mubarak Muhammed
I don't understand
pretty
which formula am i going to use to solve it
Cosine rule
A ship sail 4km on a bearing of 038 and then 5km on a bearing of 067 calculate the distance between the starting point and final destination and the bearing ship from the starting point.

Jayeola
Pls questions
John
A ship sails 4km on a bearing of 038° and then 5km on a bearing of 0670
Calculate:
A. The distance between the starting point and final destination
B: the bearing of the ship from it starting point
Coded
I want the full full explanation

lord
i love this
Star
pls how did u get 151
Favour
How did u get .27808...
Did u cross multiply?
Anonymous
How did you get38
Anonymous
What about the bearing from d boat to the starting point or whatever
luxury jay
I want the full explanation
Blessing
The basic geometry is very difficult to understand

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