# How do I plot the angle

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Umut Ayyildiz on 29 Jan 2019
Answered: Brian Hart on 30 Jan 2019
A boat moves at 20kmh^−1 along a straight path described by
y =11x/15+43/3
starting at x = −10; y = 7. Plot the angle (in degrees) of the line of sight from an observer at the coordinate origin to the boat as a function of time for 3 hours.
This is what I have so far
x=-10:20:50
y=((11*x)/15)+(43/3)
plot(x,y)
hold on
a=atand(y/x)
legend('boat')

Brian Hart on 30 Jan 2019
Hi Umut,
This is a neat problem! Some nice physics with rectangular-to-polar conversions and trig asymptotes. My code to solve the problem is below. There's a fair amount of background knowledge needed to understand the math.
%define constants
vel = 20 / 60; %km/min
slope = 11/15;
yInt = 43/3;
xStart = -10;
yStart = 7;
tMax = 3*60; %max time in minutes
tInc = 5; %compute for five minute increments
%Make a vector of time values for which to compute x, y position:
tVec = 0:tInc:tMax;
%Compute the radial distance from the starting point at each time interval:
distArr = tVec * vel;
% Compute the x, y position at each time interval:
xArr = xStart + cos(lineAngle) * distArr;
yArr = yStart + sin(lineAngle) * distArr;
figure;plot(xArr,yArr,'x');title('Boat position at each time interval')
%Because the trajectory crosses x=0, we get a phase error. Crude
%correction:
origToPtAngle(find(origToPtAngle<0))=origToPtAngle(find(origToPtAngle<0)) + 180;
figure;plot(tVec,origToPtAngle,'.-');title('Line-of-sight from origin to boat position vs time in minutes (x-axis = 0°)')
With this I get:  There's probably a more elegant way to handle the rectangular-to-polar stuff.