a boat travels 60 km due east. it then adjusts its course by 25 degrees northward and travels another 90 km in the new direction. how far is the boat from its initial position to the nearest kilometer?
d = 60km @ 0 Deg. + 90 km @ 25 Deg.
X = Hor. = 60*cos(0) + 90*cos25 = 98 km.
Y = Ver. = 60*sin(0) + 90*sin25 =38 km.
d = sqrt((98)^2+(38)^2) = 105 km.
To find the distance of the boat from its initial position, we can use the concept of vector addition. Let's break down the boat's journey into two components: the eastward component and the northward component.
1. Eastward Component: The boat travels 60 km due east. In this direction, the boat doesn't deviate north or south, so the northward component is 0 km.
2. Northward Component: After adjusting its course by 25 degrees northward, the boat travels 90 km. In this direction, the boat doesn't deviate east or west, so the eastward component is 0 km.
Now let's find the resultant of these two components using vector addition. We can use the Pythagorean theorem to find the magnitude of the resultant vector.
Magnitude of the resultant vector = √(Eastward Component^2 + Northward Component^2)
Eastward Component = 60 km
Northward Component = 90 km
Magnitude of the resultant vector = √(60^2 + 90^2)
= √(3600 + 8100)
= √11700
≈ 108.11 km (rounded to two decimal places)
Therefore, the boat is approximately 108.11 km away from its initial position.