PLZ HELP ON THIS AH its another trigonometry question: An airplane flies 55 degrees east of north from city A to city B, a distance of 470 miles. Another airplane flies 7 degrees north of east from city A to city C, a distance of 890 miles. What is the distance between cities B and C?

523.8 miles

Step-by-step explanation:

The drawing attached represents the question. We create a triangle with the distances from each pair of cities, and we call the distance from B to C by 'd'.

First, we need to find the angle BAC:

55° + BAC + 7° = 90°

BAC = 28°

Then, we can use the law of cosines to find the value of d:

d^2 = 470^2 + 890^2 - 2*470*890*cos(BAC)

d^2 = 470^2 + 890^2 - 2*470*890*0.8829

d^2 = 274365.86

d = 523.8 miles

angle uzt is congruent to angle tzy. angle vzw is congruent to angle yzx.

(credits to leowren)

step-by-step explanation:

goog may be able to you find that answer