Question

At 5:45 p.m., a jet is located 108 mi due east of a city. A second jet is located 214 mi due north of the city. To the nearest tenth of a mile, what is the distance between the two jets? Enter your answer as a decimal in the box.

Answer 1

a^2+b^2=c^2

a^2=108^2 b^2=214^2 c^2=c^2

108^2+214^2=c^2

108^2=11664 214^2=45796 c^2=c^2

11664+45796= 57490

√57490= 239.7

:)

Answer 2

By connecting the distances traveled by the jet and the distance between them, we form a right triangle with hypotenuse equal to the unknown distance.
Using the Pythagorean,
h² = a² + b²
Substituting,
h² = (108 mi)² + (214 mi)²
h = 239.71 miles
Thus, the distance between them is approximately 239.71 miles.