Question 1135183
the ship and the two shore points form a triangle.


A is the north point on the shore.
B is the ship.
C is the south point on the shore.
angle A is 56 degrees.
angle C is 36 degrees.
draw a line from point B perpendicular to the line AC and connecting to line AC at point D.
line BD is the shortest distance to the shore.


the length of the line AC is 19.
let the length of the line AD be equal to x and the length of the line DC be equal to 19 - x.


let the length of the line BD be equal to h .


two right triangles are formed with a common line of BD.


these are two right triangles.


tan(56) = h / x


tan(36) = h / (19 - x)


solve for h in both equation to get:


h = x * tan(56).
h = (19 - x) * tan(36).


since they're both equal to h, then:


x * tan(56) = (19 - x) * tan(36)


simplify to get x * tan(56) = 19 * tan(36) - x * tan(36).


add x * tan(36) to both sides of this equation to get x * tan(56) + x * tan(36) = 19 * tan(36).


factor out the x to get x * (tan(56) + tan(36)) = 19 * tan(36).


divide both sides of this equation by (tan(56) + tan(36)) to get x = 19 * tan(36) / (tan(56) + tan(36)).


solve for x to get x = 6.248828112.


that makes 19 - x equal to 12.75117189.


in triangle ABD, you get tan(56) = h / x.


solve for h to get h = x * tan(56) which becomes h = 6.248828112 * tan(56) which makes h = 9.264268658.


in triangle CBD, you get tan(36) = h / (19 - x).


solve for h to get h = (19 - x) * tan(36) which becomes h = 12.75117189 * tan(36) which makes h = 9.264268658.


this confirms that h = 9.264268658 miles which is the shortest distance to land.


here's my diagram of the situation as i see it.


<img src = "http://theo.x10hosting.com/2019/022101.jpg" alt="$$$" >