How far apart are the two planes after 1 hour? When will they be twice as far apart?

Question

two airplanes leave an airport at 11:00 am and the angle between their flight path is 40 degrees. The first plane has a speed of 300 mph and the other plane has a speed of 200 mph

Answer 1

SAS triangle: 200t:40°:300t:d(t)

Similar triangle: 2:40°:3:(d(t)/(100t))

Use Law of Cosines:

(d(t)/(100t))^2 = 2^2 + 3^2 - 2(2)(3)cos(40°)

(d(t)/(100t))^2 = 4 + 9 - 12cos(40°)

(d(t)/(100t))^2 = 13 - 12cos(40°)

d(t)/(100t) = √(13-12cos(40°))

d(t) = (100t)√(13-12cos(40°))

d(1) = 100√(13-12cos(40°)) ≈ 195.127309277104 miles

d(t2) = 2d(1) =

100(t2)√(13-12cos(40°)) = 200√(13-12cos(40°))

(t2)√(13-12cos(40°)) = 2√(13-12cos(40°))

t2 = 2 hours

 

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Answer 2

If you draw the picture, this is an SAS triangle.

 after 1 hour your planes will be 200 & 300 miles out.

 

1. Use The Law of Cosines first to calculate the third side of the triangle

   c² = a² + b² - 2ab cos(C)

 

2. Use The Law of Sines to find the smaller of the other two angles

 

       a    =   b   =    c         

    sin A    sin B    sin C

  

 
3. Use the three angles add to 180° to find the last angle.

 

     A + B + C = 180°