a yacht sails 50km north and then another 20km at an unknown angle .Using GPS£¬the skipper finds that bearing from their starting point is N20¡ãW.
(1) What was the direction of the yacht for the 20km of itd journey?
(2) In what direction and how far must the skipper sail to return to the original position?
To answer these questions, we can use basic trigonometry and vector addition. Let's break down the problem step by step.
(1) What was the direction of the yacht for the 20km of its journey?
First, let's determine the direction of the yacht for the 20km part of its journey. We know that the bearing from the starting point is N20°W. The "N" represents north, and the "W" represents west.
To find the angle in degrees, we calculate 180° - 20° = 160°. So, the angle is 160°. However, this angle is measured with respect to true north, while we need the angle relative to the yacht's sailing direction.
Since the yacht sailed 50km north first, the direction of its sailing can be seen as the direction from the starting point to the endpoint of the 50km northward journey. Let's call this point A.
To determine the angle between the yacht's sailing direction and the north, we need to draw a right-angled triangle. One side represents the 20km journey, the other represents the difference between 180° and the given bearing (160°).
Using trigonometry, we can set up the following equation:
tan(angle) = opposite/adjacent
tan(angle) = 20km / (180° - 160°) = 2km/1°
To find the angle, we can take the arctan of 2km/1°:
angle = arctan(2 km/1°) = 63.43° (approximately)
Therefore, the direction of the yacht for the 20km of its journey is N63.43°E.
(2) In what direction and how far must the skipper sail to return to the original position?
To determine the direction and distance, we can use vector addition. The yacht sailed 50km north and then 20km at an angle of 63.43°.
First, let's break the 20km journey into its north and east components. The north component can be calculated using sin(63.43°) * 20km, which gives us approximately 17.89 km. The east component can be calculated using cos(63.43°) * 20km, which gives us approximately 9.75 km.
Next, we can add the north and east components to the initial northward journey of 50km. The resulting vector will be the displacement vector, pointing from the starting point to the final position (which should ideally bring us back to the starting point).
To find the direction of the displacement vector, we can use trigonometry. The angle of the final displacement vector (relative to the north) can be found using the tangent function:
angle = arctan(17.89 km / 9.75 km) = 61.22° (approximately)
Therefore, the direction the skipper must sail to return to the original position is N61.22°E.
To find the total distance the skipper must sail to return to the original position, we can calculate the magnitude of the displacement vector. Using the Pythagorean theorem, the distance (d) can be found:
d = sqrt((50 km + 17.89 km)^2 + (0 km + 9.75 km)^2) = 61.38 km (approximately)
Hence, the skipper must sail approximately 61.38 km in the direction of N61.22°E to return to the original position.