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

Answer 1

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

Answer 2

I don't understand

Answer 3

which formula am i going to use to solve it

Answer 4

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.

Answer 5

Pls questions

Answer 6

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

Answer 7

I want the full full explanation

Answer 8

i love this

Answer 9

pls how did u get 151

Answer 10

How did u get .27808...
Did u cross multiply?

Answer 11

How did you get38

Answer 12

What about the bearing from d boat to the starting point or whatever

Answer 13

I want the full explanation

Answer 14

The basic geometry is very difficult to understand