A boat sails due north from the port 70 km, then 30 km east and then 60 km north. How far is the boat from port?
130 km north, and 30 km east ... use Pythagoras ... a^2 + b^2 = c^2
All angles are measured CW from +y-axis.
D = 70 + 30i + 60 = 130 + 30i.
D = sqrt(130^2 + 30^2) =
Hi can I ask what letter should I needed to find? If it is a b or c? I really didn't get it. Thank you very much.
Thank you very much!
Well, if the boat sailed due north from the port for 70 km, then 30 km east, and finally 60 km north, I'd say the boat is pretty far from the port. You could look at it this way: the boat clearly didn't want to be near the port anymore, so it decided to take a scenic route. Who needs a straight line when you can zigzag your way to adventure, right?
To find the distance between the boat and the port, we can use the Pythagorean theorem.
First, let's draw a diagram to visualize the boat's movements. Assume the port is at the origin (0,0) on a coordinate plane, and let the distance traveled be represented by vectors.
The boat travels 70 km north, which means it moves in the positive y-direction, giving us a vector (0, 70).
Then, it sails 30 km east, which means it moves in the positive x-direction, giving us another vector (30, 0).
Finally, the boat sails 60 km north, giving us one more vector (0, 60).
Now, we can add these vectors together to find the total displacement of the boat from the port.
Add the north vectors: (0, 70) + (0, 60) = (0, 130).
Add the east vector: (0, 130) + (30, 0) = (30, 130).
Next, we can find the distance between the boat and the port using the Pythagorean theorem, which states that the square of the hypotenuse (distance) of a right triangle is equal to the sum of the squares of the other two sides.
In this case, the two sides of the right triangle are the x-coordinate (30) and the y-coordinate (130).
So, the distance between the boat and the port is calculated as follows:
Distance^2 = (30)^2 + (130)^2
Distance^2 = 900 + 16900
Distance^2 = 17800
Distance = √17800
Distance ≈ 133.41 km
Therefore, the boat is approximately 133.41 km away from the port.