Alright, lets get started.
Please have a look at the diagram attached.
2.5 km at an angle of 35 deg to the ground.
So, AB = 2.5 sin 35 = 1.4 km
BC = 2.5 cos 35 = 2.0 km
Plane changes direction and travels 5.2 km at an angle of 22 deg to the ground.
DE = 5.2 sin 22 = 1.95 km
AE = 5.2 cos 22 = 4.82 km
Total,
X component = [tex]CF = BC + AE = 2 + 4.82 = 6.82 km[/tex]
Y component = [tex]DF = AB + DE = 1.4 + 1.95 = 3.35 km[/tex]
Net displacement d = [tex]\sqrt{6.82^2+3.35^2}= 7.6 km[/tex]
For Angle
tan θ = [tex]\frac{Y}{ X}= \frac{3.35}{6.82} = 0.49[/tex]
So,
θ = [tex]tan^{-1}(0.49) =26[/tex]
Hence magnitude is 7.6 km and angle is 26 degrees. : Answer
Hope it will help.
