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

239.7 miles is the answer.

Answer 2

Answer: The distance between the two jets is 239.7 miles.

Step-by-step explanation:

Since we have given that

Distance at which first jet is located = 108 miles

Distance at which second jet is located = 214 miles

We need to find the distance between the two jets .

As it is form a "right triangle" so we can apply "pythagorus theorem", which is given as

`H^2=B^2+P^2\\\\H^2=214^2+108^2\\\\H^2=45796+11664\\\\H^2=57460\\\\H=√(57460)\\\\H=239.7\ miles`

Hence, the distance between the two jets is 239.7 miles.