One airplane is 78 miles due south of a control tower. Another airplane is 52 miles from the control tower at a heading of N 38 degrees E. To the nearest tenth of a mile, how far apart are the two airplanes?

123.2 miles

Step-by-step explanation:

Given that One airplane is 78 miles due south of a control tower. Another airplane is 52 miles from the control tower at a heading of N 38 degrees E. To the nearest tenth of a mile, how far apart are the two airplanes?

The angle Ø = 90 + 52 = 142 degree

Or 180 - 28 = 142 degree

Using cosine formula

D^2 = 78^2 + 52^2 - 2(78)(52)Cos142

D^2 = 6084 + 2704 - 8112Cos142

D^2 = 8788 + 6392.34

D^2 = 15180.34

D = 123.2 miles

Therefore, the two planes are 123.2 miles apart.

first, you divide the triangle in two, tô form a retangle triangle. after, using pythagoream theorem:

18^2 = 14^2 + (x/2)^2

(x^2/4) + 196 = 324

x^2/4 = 324 - 196

x^2/4 = 128

x^2 = 512

x = √512

x = √2^9

x = 16√2 cm

i hope i you.

answer: the is put the 450

step-by-step explanation: