Two planes leave Dawson City Airport at the same time. One airplane travels 420 km/hour. The other airplane travels at 375 km/hour. Two hours later they are 1500km apart.
Algebraically determine the angle between their paths, rounded to the nearest degree.
(please solve using trigonometry)
Draw a triangle.
Dawsen city aiport is at one vertice
in 2 hours how far does each plane fly? Those distances are on the two sides radiating from Dawson city
The other side is 1500km
Use cosine rule to find the angle.
In two hours, one airplane has traveled 840 km and the other 750 km.
By the cos rule, we have $1500^2 = 840^2 + 750^2 - 2(750)(840) \cos \gamma$
$981900 = - 2(750)(840) \cos \gamma$
$\cos \gamma = 0.77928571428$
$\gamma = \cos ^{-1} 0.77928571428$
$\gamma = 38.739424597856$
$\gamma = \boxed{39^{\circ}}$