Answer:
[tex]NW (148.98\°)[/tex]
Step-by-step explanation:
we know that
Applying the law of cosines
[tex]c^{2}=a^{2}+b^{2}-2*a*b*cos(C)[/tex]
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
[tex]a=(1\ h)*450\frac{mi}{h}=450\ mi[/tex]
step 2
Find the distance b
Multiply the speed by the time
[tex]b=(1\ h)*380\frac{mi}{h}=380\ mi[/tex]
step 3
Find the measure of angle C
we have
[tex]a=450\ mi[/tex]
[tex]b=380\ mi[/tex]
[tex]c=800\ mi[/tex] -----> given problem
[tex]c^{2}=a^{2}+b^{2}-2*a*b*cos(C)[/tex]
solve for angle C
[tex]cos(C)=[a^{2}+b^{2}-c^{2}]/[2*a*b][/tex]
substitute the values
[tex]cos(C)=[450^{2}+380^{2}-800^{2}]/[2(450)(380)]=-0.8570[/tex]
[tex]C=arccos(-0.8570)=148.98\°[/tex]
The direction plane 2 is equal to
[tex]NW (148.98\°)[/tex]
see the attached figure to better understand the problem
