By using the given distances and directions, we have;
Part A: 1083.5 feet
Part B: 1921.37 feet
Part C: 58.05°
<h3>How can the distances and directions be calculated?</h3>The distance between David and Lacey = 2000 feet
David's angle to the boat = 70°
Lacey's angle to the boat = 32°
Part A
The distance of the boat from David is found as follows;
Imaginary lines drawn from the boat to David and then to Lacey form a triangle.
In the triangle, let <em>A</em><em> </em>= 70°
B = 32°
Therefore by the sum of angles in a triangle, we have;
C = 180° - (70° + 32°) = 78°
By using sine rule we have;
David's distance from the boat, <em>b</em>, is therefore;
The angle subtended by the coastline, <em>C</em>, is therefore;
- David's distance from the boat is 1083.5 feet
Part B
The distance between the boat and Lacey is found as follows;
- Lacey's distance from the boat is 1921.37 feet
Part C
When <em>b </em>= 1200 feet, we have;
Finding the vertical distance of the boat from the coastline, we have;
1083.5 × sin(70) = 1018.17
We have;
The angle between the coastline and his view to the boat, <em>B'</em>, is therefore;
- B' = arcsine (1018.17÷1200) = 58.05°
Learn more about sine rule here:
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