Question

Two planes left an airport at the same time and in still air. One flew east at 450 mi/h, and the other flew at 380 mi/h. In what direction did the second pilot fly if the two planes were 800 mi apart after 1 hour?

Answer 1

`NW (148.98\°)`

Explanation:

we know that

Applying the law of cosines

`c^(2)=a^(2)+b^(2)-2*a*b*cos(C)`

where

c is the distance between the two planes after 1 hour

a is the distance of the plane 1 after 1 hour from the airport

b is the distance of the plane 2 after 1 hour from the airport

C is the angle between the direction plane 1 and the direction plane 2

step 1

Find the distance a

Multiply the speed by the time

`a=(1\ h)*450(mi)/(h)=450\ mi`

step 2

Find the distance b

Multiply the speed by the time

`b=(1\ h)*380(mi)/(h)=380\ mi`

step 3

Find the measure of angle C

we have

`a=450\ mi`

`b=380\ mi`

`c=800\ mi` -----> given problem

`c^(2)=a^(2)+b^(2)-2*a*b*cos(C)`

solve for angle C

`cos(C)=[a^(2)+b^(2)-c^(2)]/[2*a*b]`

substitute the values

`cos(C)=[450^(2)+380^(2)-800^(2)]/[2(450)(380)]=-0.8570`

`C=arccos(-0.8570)=148.98\°`

The direction plane 2 is equal to

`NW (148.98\°)`

see the attached figure to better understand the problem