Question 1203473
solutions are in the displays shown abelow:


first disp;lay shows calulations for acute angles of the triangles formed for point 0 to point A, then point A to point B, then point B to point C.


320 is equal to 270 + 50.
350 is equal to 270 + 80.
50 is equal to 90 - 40.


<img src = "http://theo.x10hosting.com/2023/090421.jpg">


calculator is used to find the horizontal and vetical lengths of the triangle.


first one is for point 0 to point A,


<img src = "http://theo.x10hosting.com/2023/090401.jpg">


second one is for point A to point B.


<img src = "http://theo.x10hosting.com/2023/090402.jpg">


third one is for point B to point C.


<img src = "http://theo.x10hosting.com/2023/090403.jpg">


excel was used to find lengths of the vertical and horizontal legs of the triangles formed, as shown below:


<img src = "http://theo.x10hosting.com/2023/090411.jpg">


sketch below shows the total layout and hot it would appear with all locations together on the same sheet in their approximate locations relative to each other.


this also shows the final calculation from pont O to the final point C.


<img src = "http://theo.x10hosting.com/2023/090412.jpg">


you were asked to find the total distance going north.


that would be the sum of the vertical distances from point 0 to point A, point A to point B, and point B to point C.


that would be 684.86 as shown in the spreadsheet column A row 13.


that would also be 212 + 472 = 684 as showon in the final sketch.


the procedure used was to find the triangles involved and then calculate the vertical and hosizontal lets of those triangles.


in the final diagram, the 472 vertical length shown is the sum of 344 and 128.


here's a reference on bearings.


<a href = "https://www.mathsteacher.com.au/year7/ch08_angles/07_bear/bearing.htm" target = "_blank">https://www.mathsteacher.com.au/year7/ch08_angles/07_bear/bearing.htm</a>


let me know if you have any questions.


theo