A ship leaves port and sails at a bearing of 124degrees. Another ship leaves the same port at the same time sailing at a bearing of 74degrees. When both ships are 80 miles from the port, how far are they from each other?
use the law of cosines. The included angle is 50°
An observer at a point P on a coast sights a ship in a direction N 41°39'
E. The ship is at the same time directly east of a point Q, 18.6 km due north of P. Work to three significant digits.
Find the distance of the ship from point P.
To find the distance between the two ships when both are 80 miles from the port, we can use the law of cosines. The law of cosines relates the sides of a triangle to the cosine of one of its angles. In this case, we have a triangle formed by the two ships and the port.
Let's assume the position of the two ships is represented by points A and B, and the port is represented by point C. The angle between ship A and the port is 124 degrees, and the angle between ship B and the port is 74 degrees. We know that both ships are 80 miles away from the port.
Using the law of cosines, we can relate the distances between the ships (AB), the distance of one ship from the port (AC or BC), and the angle between the ships (angle ACB):
AB^2 = AC^2 + BC^2 - 2 * AC * BC * cos(angle ACB)
In this case, we can set AC = BC = 80 miles (since both ships are equidistant from the port) and angle ACB = 124 - 74 = 50 degrees.
Substituting these values into the equation, we get:
AB^2 = 80^2 + 80^2 - 2 * 80 * 80 * cos(50)
AB^2 = 12800 - 12800 * cos(50)
Simplifying further, we have:
AB^2 = 12800 - 12800 * 0.643
AB^2 = 12800 - 8233.6
AB^2 = 4566.4
Taking the square root of both sides, we find:
AB = sqrt(4566.4)
AB ≈ 67.62 miles
Therefore, when both ships are 80 miles from the port, they are approximately 67.62 miles apart from each other.