Asked by Bright ulubi

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
Answered by 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
Answered by Mubarak Muhammed
I don't understand
Answered by pretty
which formula am i going to use to solve it
Answered by 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.

Answered by Jayeola
Pls questions
Answered by 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
Answered by Coded
I want the full full explanation

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