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.

-7

step-by-step explanation:

x^2 + 18x + 74 = (x + 9)^2 - 81 + 74

= (x + 9)^2 - 7.

so q = -7.