Question

a plane leaves airport a and travels 550 miles to airport b on a bearing of n35°e. the plane later leaves airport b and travels to airport c 390 miles away on a bearing of s72°e. find the distance from airport a to airport c to the nearest tenth of a mile. the distance from airport a to airport c is about □ miles. (round to the nearest tenth as needed.)

Explanation:

Step1: Find angle at B

Bearing from A to B: N35°E → angle from north to AB is 35°. Bearing from B to C: S72°E → angle from south to BC is 72°. At point B, north-south is straight line, so angle between AB and BC is 35° + 72° = 107°.

Step2: Apply Law of Cosines

In triangle ABC: AB=550, BC=390, angle B=107°.
$AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(107°)$
$\cos(107°) ≈ -0.2924$
$AC^2 = 550^2 + 390^2 - 2 \cdot 550 \cdot 390 \cdot (-0.2924)$
$AC^2 = 302500 + 152100 + 2 \cdot 550 \cdot 390 \cdot 0.2924$
$AC^2 = 454600 + 124532.4 = 579132.4$
$AC ≈ \sqrt{579132.4} ≈ 761.0$

Answer:

761.0