11. Charles leaves the
marina and sails
his boat 10° west
Reasoning and Proving
Representing
Selecting Tools
Problem Solving
of north for 1.5 h at
18 km/h. He then
Connecting
Reflecting
makes a starboard
Communicating
(right) turn to a heading of 60° east of
north, and sails for 1.2 h at 20 km/h.
a) At the end of that time, how far is
Charles from his starting point to the
nearest kilometre?
b) What is the course required for Charles
to return directly to the marina?



Answer :

To find the distance Charles is from his starting point and the course required to return to the marina, we can follow these steps: a) 1. Calculate the distance traveled in the first leg: Distance = speed × time Distance = 18 km/h × 1.5 h = 27 km 2. Find the coordinates of Charles after the first leg: Since Charles went 10° west of north, this creates a right triangle with a 10° angle. Using trigonometry, we can find the northward and westward components of his displacement. Northward displacement = 27 km × sin(10°) Westward displacement = 27 km × cos(10°) 3. Calculate the total displacement after the first leg using Pythagoras theorem: Total displacement = √(Northward displacement² + Westward displacement²) 4. Calculate the distance from the marina after the first leg: Distance = Total displacement b) 1. Determine the angle between Charles' final position and the marina: Since Charles is heading 60° east of north after the second leg, the angle he needs to turn to face the marina is 60°. To return directly to the marina, he would need to turn 60° to the left (port) from his current heading. 2. The course required for Charles to return directly to the marina would be 60° to the left from his current heading, which is 60° east of north. By following these steps, you can find both the distance from the starting point and the course Charles needs to take to return to the marina.

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