Respuesta :
By using the given distances and directions, we have;
Part A: 1083.5 feet
Part B: 1921.37 feet
Part C: 58.05°
How can the distances and directions be calculated?
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 A = 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;
[tex] \frac{a}{sin(A)} = \frac{b}{sin( B )} = \frac{c}{sin(C)}[/tex]
David's distance from the boat, b, is therefore;
[tex] \frac{b}{sin( 32 )} = \frac{2000}{sin(78)}[/tex]
The angle subtended by the coastline, C, is therefore;
[tex] b = \frac{2000}{sin(78)} \times sin( 32 ) = 1083.5 [/tex]
- David's distance from the boat is 1083.5 feet
Part B
The distance between the boat and Lacey is found as follows;
[tex] \mathbf{ \frac{a}{sin( 70)} }= \frac{2000}{sin(78)}[/tex]
[tex] a = \frac{2000}{sin(78)} \times sin( 70 ) = 1921.37 [/tex]
- Lacey's distance from the boat is 1921.37 feet
Part C
When b = 1200 feet, we have;
Finding the vertical distance of the boat from the coastline, we have;
1083.5 × sin(70) = 1018.17
We have;
[tex] \frac{1200}{sin(90)} = \frac{1018.17}{sin( B' )} [/tex]
[tex] {sin( B' )} = \frac{sin(90)}{1200} \times 1018.17 [/tex]
The angle between the coastline and his view to the boat, B', is therefore;
- B' = arcsine (1018.17÷1200) = 58.05°
Learn more about sine rule here:
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