A ship leaves port at 6 am and heads due east at 20 knots. At 10 am, to avoid a storm the ship changes course to N 47°. Find the ships distance from port at 4 pm.
Not a right triangle, where in the world did you get that idea?
Draw the figure. Make a triangle. Do you know the law of sine and the law of cosines? If so, figure the distance of the two legs from speed and time, then use the law of sines to figure the other two angles in the triangle. Finally, use the law of cosines to figure the side you are looking for.
can i also have the actual answer and the process
no
To solve this problem, we can break it down into two parts: the distance traveled from 6 am to 10 am, and the distance traveled from 10 am to 4 pm.
First, let's calculate the distance traveled from 6 am to 10 am. The ship is heading due east at 20 knots, which means it is traveling directly to the east. The time elapsed from 6 am to 10 am is 4 hours.
Distance = Speed × Time
Distance = 20 knots × 4 hours
Distance = 80 nautical miles
So, from 6 am to 10 am, the ship has traveled 80 nautical miles to the east of the port.
Next, let's calculate the distance traveled from 10 am to 4 pm. The ship changes course to N 47°, which means it's now heading north at an angle of 47 degrees from the north direction. The time elapsed from 10 am to 4 pm is 6 hours.
To find the distance traveled north, we need to calculate the northward component of the ship's velocity.
Northward Component = Velocity × sine(Angle)
Northward Component = 20 knots × sine(47°)
Northward Component ≈ 20 knots × 0.731
Northward Component ≈ 14.62 knots
In 6 hours, the ship has traveled a distance of:
Distance = Speed × Time
Distance = 14.62 knots × 6 hours
Distance ≈ 87.72 nautical miles
Therefore, from 10 am to 4 pm, the ship has traveled approximately 87.72 nautical miles north of the point it changed course.
Now, to find the ship's distance from the port at 4 pm, we need to calculate the total distance traveled from the port. We can use the Pythagorean theorem, considering the eastward and northward components as the two sides of a right-angled triangle.
Distance from Port = √(Eastward Distance² + Northward Distance²)
Distance from Port = √((80 nautical miles)² + (87.72 nautical miles)²)
Distance from Port ≈ √(6400 + 7695.78)
Distance from Port ≈ √14095.78
Distance from Port ≈ 118.80 nautical miles
Therefore, the ship's distance from the port at 4 pm is approximately 118.80 nautical miles.