To find:theship bearing and distance from port at 6 P.M.
Answer to Problem 24CT
The bearing of the ship E
Explanation of Solution
Given:
The ship leaves port at 12 P.M. and the ship changes course to S
Calculation:
The given information to form a triangle as shown in Figure 1,
From Figure 1, it is observed that the point D represents the ship at 6 P.M. and the point A represents the port.
The ship gets on the point B after traveling 2 hours, the ship traveled the distance
In triangle BCD , find the distance CD .
That is, the value of
In triangle BCD , find the distance BC .
That is, the value of
The value of AC is,
That is, the value of
In triangle ACD , find the angle of A .
The angle with east west line is,
Therefore, the bearing of the ship E
In triangle ACD , the distance AD is computed as follows,
Therefore, the distance from port to ship at 6 P.M. is 23.862 nm.
Chapter 4 Solutions
PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)
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