Bearings and distance?

Question

A ship travels 55km on a bearing of 11 degrees, and then travels on a bearing of 101 degrees for 133km. Find the distance of the end of the trip from the starting point to the nearest kilometer.
54km
144km
10km
188km

Answer 1

We assume we start from (0,0)
x1=0+55*sin 11=10,49449
y1=0+55*cos 11=53,98950
so at first we arrive at point (10.49449,53.9895).From that point we go to :
x2=10.49449+133*sin 101 = 141,05091
y2=53.98950+133*cos 101=28,61190

at last the total distance is computed as follows
d^2=(x2-x0)^2+(y2-y0)^2=141.05091^2+28.6119^2 =>
d=143.9235~144km

Answer 2

The turn was of 90 degrees, so you have a right triangle with legs of 55 and 133

The hypotenuse is the the ending distance from the starting point

h^2 = (55)^2 + (133)^2

Solve for h

Answer 3

144 km
The ship turns at a 90 degree angle so use Pythagorean Theorem
Distance = sqrt(55^2+133^2) approx = 144

Answer 4

these bearings are 90 degrees apart. For a right triangle, use a^2 + b^2 = c^2

55*55 + 133*133 = c^2

c= 143.92

Answer 5

U have to draw a sketch.its a triangle of two adjacent sides of length=55 and 133,the included angle between them is 90 and you need to find the hypoteneous,use the formula
c^2=a^2 +b^2
c^2 = 55^2 + 133^2
c = 144 km