Question 1043572
1. I had to look up the definition of course, heading, and bearing,
and what I found is some confusion.
In most sources They were defined as the angle from North to the direction the ship is aiming for, or pointing to,
measured clockwise from North.
I am sticking to that, but I believe that for the first problem, it does not matter how you measure the angle, as long as it is done consistently.
 
{{{(4hours)*(18knots)=72}}}{{{nautical}}}{{{miles}}}
{{{(6hours)*(16knots)=96}}}{{{nautical}}}{{{miles}}} .
The situation can be sketched like this:
{{{drawing(660,300,-10,155,-50,25,
locate(-9,2,port),circle(0,0,1),
triangle(0,0,70.43,14.97,0,14.97),triangle(0,0,70.43,14.97,70.43,0),
triangle(70.43,14.97,146.08,-44.13,146.08,14.97),
triangle(0,0,146.08,-44.13,146.08,0),
blue(arrow(0,-20,0,22)),blue(arrow(70.42,-19,70.42,25)),
locate(0,25,blue(N)),locate(72,25,blue(N)),
red(line(0,0,70.43,14.97)),green(line(146.08,-44.13,70.43,14.97)),
rectangle(0,14.97,2,12.97),rectangle(70.43,0,68.43,2),
rectangle(146.08,14.97,144.08,12.97),rectangle(146.08,0,144.08,-2),
green(arc(70.43,14.97,10,10,-90,38)),locate(75.43,20.97,green(128^o)),
red(arc(0,0,10,10,-90,-12)),locate(2,9.6,red(78^o)),
locate(35.2,7.48,red(72)),locate(108.25,-17,green(96)),
locate(33,18.5,x[1]),locate(108.2,18.5,x[2]),
locate(142,12,y[1]),locate(150,-10,y[2]),
arrow(149,-14.58,149,14.97),arrow(149,-14.58,149,-44.13)
)}}}
The right triangles with a red hypotenuse of length {{{red(72)}}} have legs measuring
{{{x[1]=72*sin(78^o)=70.42663}}} (rounded to 5 decimal places) and
{{{y[1]=72*cos(78^o)=14.96964}}} (rounded to 5 decimal places) .
At {{{4}}} hours the ship was about 70 nautical miles East, and 15 miles North of the port.
The legs of the right triangles with a green hypotenuse of length {{{green(96)}}} are such that
{{{x[2]=96*sin(128^o)=75.64903}}} (rounded to 5 decimal places) and
{{{-y[2]=96*cos(128^o)=-59.1035}}} (rounded to 5 decimal places) .
After {{{4+6=10}}} hours the ship was about 76 nautical miles further East, and 59 miles South from its position at {{{4}}} hours.
 
Considering only the start and end positions, we have
{{{drawing(660,300,-10,155,-59,16,
locate(-9,2,port),circle(0,0,1),
triangle(0,0,146.08,-44.13,146.08,0),
blue(arrow(0,-27,0,13)),blue(arrow(70.42,-29,70.42,15)),
locate(1,16,blue(N)),locate(72,15,blue(N)),
rectangle(70.43,0,68.43,2),
arrow(146.08,0,146.08,14.97),arrow(146.08,14.97,146.08,0),
rectangle(146.08,0,144.08,-2),circle(146.08,-44.13,1),
locate(147,-44,end),locate(135,-20,y[2]-y[1]),
locate(33,3,x[1]),locate(108.2,3,x[2]),
locate(142,12,y[1]),locate(150,-10,y[2]),
arrow(149,-14.58,149,14.97),arrow(149,-14.58,149,-44.13),
arrow(73,-46,0,-46),arrow(73,-46,146.08,-46),
locate(65,-38,x[1]+x[2]),rectangle(0,0,2,2),
red(arc(0,0,20,20,-90,16.8)),red(arc(146.08,-44.13,20,20,-90,196.8)),
arc(0,0,30,30,0,16.8),locate(15,-1,theta),
locate(10,8,red(90^o+theta)),locate(140,-53,red(270^o+theta))
)}}}
We have one right triangle, and we know the legs, so we can find
the hypotenuse, and
angle {{{theta}}} .
The lengths of the legs are approximately
{{{x[1]+x[2]=70.42663+75.64903=146.07566}}} and {{{y[2]-y[1]=59.10350-14.96964=44.13386}}
so the hypotenuse, which is the distance from the ship to port at 10 hours, is approximately
{{{sqrt(144.07566^2+44.13386^2)=highlight(152.5972)}}} ;
{{{tan(theta)=44.13386/144.07566=0.30201}}}--->{{{theta=163.8^o}}} .
So,
the distance to the ship from the port is {{{highlight(152.6)}}} nautical miles,
the bearing of the ship from the port is {{{90^o+16.8^o=highlight(106.8^o)}}} , and
the bearing from the ship to the port is
{{{270^o+16.8^o=highlight(286.8^o)}}} .
 
2. The ship is going West.
When the ship reaches point {{{A}}} it records the bearing/heading of the light,
and it does it again, {{{2250}}} feet further, at point {{{B}}} .
If the course is continued, {{{x}}} feet further, the ship will reach point {{{C}}} ,
where it is at a distance of {{{d}}} feet from the light,
as close to the light as its westward trajectory would take it.
{{{drawing(650,390,-300,6200,-850,3050,
triangle(0,0,0,2884.1,3215.5,0),rectangle(0,0,100,100),
triangle(0,0,0,2884.1,5462.5,0),arrow(700,0,-300,0),
blue(arrow(5462.5,-250,5462.5,3000)),locate(-100,3000,L),
circle(0,0,30),circle(3212.5,0,30),circle(5462.5,0,30),
locate(5480,0,A),locate(3230,0,B),locate(20,0,C),
locate(4700,250,red(311^o)),locate(5050,170,red("55 '")),
red(arc(5462.5,0,1550,1550,-90,207.8)),red(line(0,2884.1,5462.5,0)),
blue(arrow(5462.5,-250,5462.5,3000)),blue(arrow(3212.5,-250,3212.5,3000)),
locate(2450,250,green(297^o)),locate(2800,170,green("50 '")),
green(arc(3212.5,0,1550,1550,-90,222)),green(line(0,2884.1,3212.5,0)),
locate(20,1550,d),locate(1400,0,x),locate(4000,0,2250),
arrow(5462.5,0,6500,0),locate(3300,3050,blue(N)),
locate(5550,3050,blue(N)),locate(-300,-50,WEST)
)}}}
We have two right triangles: {{{ALC}}} and {{{BLC}}} .
For triangle {{{ALC}}} , a leg measuring {{{d}}} feet is opposed to an angle measuring {{{297^o}}}{{{"50 '"-270^o=27^o}}}{{{"50 '"=approximately27.83333^o}}} ,
while leg measuring {{{x+2250}}} feet is adjacent to the same angle.
So, {{{tan(27.83333^o)=d/(x+2250)}}} .
For triangle {{{BLC}}} , a leg measuring {{{d}}} feet is opposed to an angle measuring {{{311^o}}}{{{"55 '"-270^o=41^o}}}{{{"55 '"=approximately41.91667^o}}} ,
while leg measuring {{{x}}} feet is adjacent to the same angle.
So, {{{tan(41.91667^o)=d/x}}} .
Dividing one equation vby the other, we get
{{{tan(41.91667^o)/tan(27.83333^o)=(x+2250)/x}}} ,which can be solved for {{{x}}} .
Then, we can use {{{tan(41.91667^o)=d/x}}}<--->{{{d=x*tan(41.91667^o)}}} to find {{{d}}} .
Solving:
{{{tan(41.91667^o)/tan(27.83333^o)=(x+2250)/x}}}
{{{x*tan(41.91667^o)/tan(27.83333^o)=x+2250}}}
{{{x*tan(41.91667^o)/tan(27.83333^o)-x=2250}}}
{{{x(tan(41.91667^o)/tan(27.83333^o)-1)=2250}}}
{{{x(tan(41.91667^o)-tan(27.83333^o))/tan(27.83333^o)=2250}}}
{{{x(tan(41.91667^o)-tan(27.83333^o))=2250*tan(27.83333^o)}}}
{{{x=2250*tan(27.83333^o)/(tan(41.91667^o)-tan(27.83333^o))}}} ,
and finally
{{{d=x*tan(41.91667^o)=2250*tan(27.83333^o)*tan(41.91667^o)/(tan(41.91667^o)-tan(27.83333^o))}}}
That calculates as approximately
{{{d=highlight(2884)}}} feet.