A ship is anchored off a long straight shoreline that runs north and south. From two observation points 18 mil

Question

A ship is anchored off a long straight shoreline that runs north and south. From two observation points 18 miles apart on shore, the bearing of the ship are N 31 degrees E and S 53 degrees E. What is the distance from the ship to each point?

Answer 1

The distances are 8.4394... mi from the northern end and 13.3397 mi from the southern end.

This problem is solved by using the sine formula. Call the points A (south), B (north), and S (ship).

The angle at S, opposite the 18 mi length is 180 - (53 + 31) = 96 deg.

Then 18 / sin 96 = BS / sin 31 = AS / sin 53. Therefore :

BS = 8.4394... mi and AS = 13.3397 mi.

Answer 2

let the points be labelled 1 and 2 resp. thus the distance of point 1 from the ship is the hypotenuse of the triangle. Let 1 be a distance x miles from the ship along the bank.
the distance of point 2 will be 18-x miles

since ship is N31degreesE then the ship is 31 degrees North Eastly of point 1 which means the distance of the ship will be x/cos31

while the distance of point 2 will be (18-x)/cos 53 from the ship..

x can be found by relating the 2 cosine descriptions for the distance of the ship from the bank like:

(18-x)tan53=xtan31 where in all angles are in degrees...
it's difficult to represent verbally ...a diagram would be self-explanatory...

Answer 3

if you sketch this you will have a triangle ABC (Cis the ships position and A is the southernmost point on shore).
you are given angle A = 31 deg, angle B = 53 deg and AB = 18 miles. since there are 180 degrees in a triangle you know angle C is 96 deg. use the law of sines:
AC/sinB = AB/sin C and
BC/sinA +AB/sinC

if you p[lug in the know values, each equation has one unknown which can be solved for:

AC/sin(53) = 18/sin(96)
AC = 18sin(53)/sin(96)

similarly BC=18sin(31)/sin96