Respuesta :
Answer:
To solve this problem, we can use the law of cosines in a triangle formed by points A, B, and C. Let’s denote the distance from A to B as ( x ) km and the distance from B to C as ( 2x ) km. The angle at point B is ( 225° - 135° = 90° ), which makes triangle ABC a right-angled triangle at B.
Using the law of cosines for right-angled triangles, which simplifies to the Pythagorean theorem, we have: [ AC^2 = AB^2 + BC^2 ] [ 350^2 = x^2 + (2x)^2 ] [ 350^2 = x^2 + 4x^2 ] [ 350^2 = 5x^2 ] [ x^2 = \frac{350^2}{5} ] [ x = \sqrt{\frac{350^2}{5}} ]
Calculating the value of ( x ) gives us the distance from A to B.
For the bearing from C to A, since the aircraft initially flies on a bearing of 135° and then on a bearing of 225°, the return bearing from C to A would be the reverse of 225°, which is ( 225° + 180° = 405° ). However, bearings are typically given within a 0° to 360° range, so we subtract 360° to normalize it: [ 405° - 360° = 45° ]
Therefore, the bearing from C to A is 45°.
Let’s calculate the exact value of ( x ):
[ x = \sqrt{\frac{350^2}{5}} ] [ x = \sqrt{\frac{122500}{5}} ] [ x = \sqrt{24500} ] [ x \approx 156.52 ]
So, the distance from A to B, rounded to the nearest whole number, is 157 km, and the bearing from C to A is 45°.
