A plane flies 90KM on a bearing 040 and then flies 90KM due east how far is the plane from the starting point
x^2 = 90^2 + 90^2 - 2*90*90*cos(140°)
To solve this problem, we will use the concept of vector addition.
1. Start by drawing a diagram to visualize the problem. Draw a point to represent the starting point and label it as "A". From point A, draw a line segment 90 km long in the direction of bearing 040°. At the end of this line segment, label the point as "B". Then, draw another line segment 90 km long due east from point B. Label the endpoint of this line segment as "C".
2. We now have a triangle ABC, where AB is the first leg of the flight (90 km on bearing 040°) and BC is the second leg of the flight (90 km due east). We need to find the distance AC, which represents how far the plane is from the starting point.
3. To find AC, we can use the Pythagorean theorem. According to the theorem, in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
4. In triangle ABC, AB and BC are the two sides that form the right angle at point B, so AC is the hypotenuse. Let's call AC as "d" (the distance from the starting point).
5. Apply the Pythagorean theorem as follows:
AC^2 = AB^2 + BC^2
d^2 = 90^2 + 90^2
d^2 = 8100 + 8100
d^2 = 16200
d = √16200
d ≈ 127.28 km
Therefore, the plane is approximately 127.28 km away from the starting point.