from a ship two lighthouses bear north 40 degree east. after the ship sails at 15knots on a course of 135 degrees for 1 hr, 20 min, the lighthouses bear 10 degrees and 345 degrees. find the distance of the ship in the latter position to the farther lighthouse.
Looks like Steve answered the same question in the first of the "Related Questions" below
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This one takes some drawing. Let's label the near lighthouse A, and the far one B.
The ship began at P, and ended up at Q.
It helps to draw a line due north from Q. A bit of analysis shows that
PQ = 20 mi
∠BPQ = 95°
∠PBQ = 30°
We want the distance QB, from the later ship's position to lighthouse B.
Using the law of sines,
20/sin30° = BQ/sin95°
BQ = 39.85 naut. mi.
Looks like I used some different angles last time. I hope both methods produce the same answer!!
To solve this problem, we'll use trigonometry and the concept of bearing angles.
Let's break down the information given:
1. The initial bearing between the two lighthouses is north 40 degrees east. This means that one lighthouse is 40 degrees east of the North direction.
2. After sailing for 1 hour and 20 minutes (which is equivalent to 1.33 hours) at a speed of 15 knots on a course of 135 degrees, the bearings to the lighthouses change to 10 and 345 degrees.
To find the distance of the ship in the latter position to the farther lighthouse, we'll use the Law of Cosines. The Law of Cosines states:
c^2 = a^2 + b^2 - 2ab * cos(C)
In this case:
- c is the distance we want to find (distance between the ship and the farther lighthouse).
- a and b are the distances between the ship and each lighthouse.
- C is the difference between the two bearing angles.
We'll begin by calculating the difference in bearing angles:
C = 345 - 10 = 335 degrees
Now, we can express the Law of Cosines in terms of the distances:
c^2 = a^2 + b^2 - 2ab * cos(335)
Next, we need to find the values for a and b. We'll start by calculating a, the distance between the ship and the lighthouse bearing 10 degrees.
To do this, we'll use the following trigonometric relationship: tan(angle) = opposite/adjacent. In this case, tan(10) = a/x, where x is the distance between the ship and the first lighthouse. Solving for a:
a = x * tan(10)
Similarly, for b (the distance between the ship and the lighthouse bearing 345 degrees), we have tan(345) = b/y, where y is the distance between the ship and the second lighthouse. Solving for b:
b = y * tan(345)
Now, we substitute the values of a and b into the Law of Cosines equation:
c^2 = (x * tan(10))^2 + (y * tan(345))^2 - 2 * x * tan(10) * y * tan(345) * cos(335)
Finally, we'll simplify the equation and solve for c, the distance between the ship and the farther lighthouse.
Unfortunately, without the specific values of x and y (distances between the ship and the lighthouses), it is not possible to provide an exact numerical solution. Nevertheless, you can use the above equation and input the specific values to calculate the distance c.