2 planes leave the airport at the same time. their speeds are 140 mph and 170 mph and the angle between their courses is 25 degrees. how far apart are they after 2.5 hours

183.27miles

Step-by-step explanation:

Note that we are considering two planes.

Plane 1

Speed = 140 mph

Time = 2.5 hours

Distance = ?

Speed = distance/time

140 mph = distance/2.5hrs

Distance = 140×2.5

= 350miles

Plane 2

Speed = 170 mph

Time = 2.5 hours

Distance = ?

Speed = distance/time

170 mph = distance/2.5hrs

Distance = 170×2.5

= 425miles

The angle between side is 25 degree

From cosine rule

c^2 = a^2 + b^2 − 2abcos(tetha)

Where a = 350

b = 425miles

c^2 = 350^2+425^2-2(350×425)(cos25)

c^2 = 122500+180625-297500(0.906)

c^2 = 303125-269535

c^2 = 33590

C = √33590

C = 183.27miles

The distance they flew apart after 2.5hrs is 183.27miles

michael scott & dwight schrute

step-by-step explanation:

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δabc ≅ δdcb by sas because

db ≅ ac . . given (side)

bc ≅ bc . . reflexive property (side)

∠dbc ≅ ∠ acb . . alternate interior angles where transversal bc crosses parallel lines bd and ac (included angle)