Hello. Welcome to this part of the solving of exercises. Step by step we will take up again the calculation of grid azimuth, the station's orientation and the survey by ray and distance. Let's start with the grid azimuth. We'll recall the definition of the grid azimuth, which is the orientation of a direction with respect to the North of the map. We'll draw two points here, a point S, a point M, trace a vector here between these two points, the direction, here, of the map's North, and we recall that the grid azimuth is measured from the map's North to the considered direction. Here, I have a grid azimuth <i>Î¦SM</i> which is equal to 65 gon. If I consider another point, for example the point N here, well, I have <i>Î¦SN,</i> from the North, here, <i>Î¦SN</i>, which is equal to 280 gon. We'll take a numerical example here to calculate the grid azimuth starting with these two points S and M. We'll take the canvas up again here and we'll first graphically measure this grid azimuth value. So I take the station S here and the point M, I trace my vector here between these points, and likewise I give here, from the station S, the direction X of the map's North. In this case, I take my protractor here in gons and I can trace my grid azimuth <i>Î¦SM</i>, which is equal to 65 gon. This is graphically. I take the numerical example now, so the calculation with equations that are given here, I will calculate the <i>Î”Y</i> and <i>Î”X.</i> So I have my YM minus YS, which, with the coordinates that are here, is equal to 33,03 meters. Next, XM minus XS, which is equal to 20,49 meters. My grid azimuth, <i>Î¦</i>, is equal to the arc tangent of <i>Î”Y</i> over <i>Î”X,</i> so 33,03 over 20,49, which here is equal to 64,652 gon. And in this case, I find myself in the first quadrant, I have a positive <i>Î”Y</i> and likewise a positive <i>Î”X</i>. So I'm in the first quadrant. Likewise, I'll verify my orientation, here, with the graphical values measured which was equal to 65 gon, which is perfectly coherent with the result of my calculation. Now we will consider several points distributed in space to illustrate the problematic of quadrants. Here, I have a certain number of points, data and coordinates, with the station S, a point P1, a point P2, a point P3 and a point P4. Now I will trace the vectors that link my station to these different points, S towards P1. S towards P2. S towards P3. and S towards P4. Likewise, I trace the reference direction, namely the North or X axis of the map. And now, I can note the different grid azimuths here that I have towards the point P1, towards the point P2, towards the point P3, and finally, towards the point P4. So I have my four grid azimuths that I will now measure. So I will measure <i>Î¦SP1,</i> <i>Î¦SP2,</i> <i>Î¦SP3,</i> and <i>Î¦SP4.</i> I take my protractor. <i>Î¦SP1</i>, I can read, is approximately 70 gon. <i>Î¦SP2</i>, I have approximately 130 gon. <i>Î¦SP3</i>, I have approximately 270 on. And finally, <i>Î¦SP4</i> gives me 330 gon. So I graphically measured the four quadrants, knowing that we always have values between zero and 400 gon. We'll now move on to calculations. I take my first point P1, I will calculate my <i>Î¦SP1</i> with the <i>Î”Y</i> which is equal to 40 minus 20, according to the coordinates that I have on the left here, which makes 20 meters. The <i>Î”X</i> which is equal to 30 minus 20, which is equal to 10 meters. My n <i>Î¦SP1</i>, that's the arc tangent of 20 over 10, which is equal to 70,48 gon. In this case here, I have my positive <i>Î”Y</i>, my <i>Î”X</i> also positive, I'm in the first quadrant, as can be seen on this graphic. Similarly, I will now calculate the <i>Î¦SP2</i> with a <i>Î”Y</i> which is equal to 40 minus 20, which is equal to 20, a <i>Î”X</i> which is equal to 10 minus 20, which is equal to negative 10. So my <i>Î¦</i>, here, <i>SP2</i> is equal to the arc tangent of minus 20 over 10, namely negative 70,48 gon. In this case ehre, I have a positive <i>Î”Y</i>, a negative <i>Î”X</i>, I'm in the second quadrant, I need to add 200 gon to the result of the arc tangent. So the <i>Î¦SP2</i> is equal to negative 70,48 plus my 200 gon, which gives us 129,52 gon. So I can verify this well on my graphic with the value, here, which is in the second quadrant. We could make a similar calculation for the points P3 and P4 to have values comprised between zero and 400 gon. Here are the numerical results that the calculations give you with their resolutions for the points P1 to P4.