If she maintains a constant speed of 680 miles per hour, how far is she from her starting position?

Question

A pilot flies in a straight path for 1 hour and 30 min. She then makes a course correction, heading 10 degrees to the right of her original course, and flies 2 hours in the new direction. If she maintains a constant speed of 680 miles per hour, how far is she from her starting position?

*answer is in miles 

 

Answer 1

The first leg of the flight is (!.5 hr)(680 mph) = 1020 miles long.  Let's call this side b of our triangle.

The second leg of the flight is (2 hr)(680 mph) = 1360 miles long. Let's call this side c of our triangle.

 

We can find the length of side a of the triangle from the law of cosines.  The angle A between the two sides whose length we know is 170°. (180° - 10° = 170°).

 

a² = b² + c² - (2)*(a)*(b)*cos(A)

 

a² = (1020 miles)² + (1360 miles)² - (2)*(1020 miles)(1360 miles)*cos(170°)

 

a² = 1040400 miles² + 1849600 miles² - (2774400 miles²)*cos(170°)

 

cos(170°) = -0.9848

 

a² = 2890000 miles² + 2732251 miles² = 5622250.6 miles²

 

a = (5622250.6 miles²)^½ = 2371 miles

 

 

Answer 2

Dalia,
 
The above will yield a triangle where leg 2 is the 1.5 hrs, leg 2 = 2 hrs and the hypotenuse is what you are trying to solve for.
 
Step 1: Draw a triangle, draw a line about an inch long then change your direction by the 10° and continue the line for another about inch and a quarter (trying to give some scale so that it all makes sense in the end). Now label the triangle as A (where you originally started the line) B (where you changed direction, & C where you completed the trip. Then label the sides opposite those angles the same letter only lower case a, b, & c.
 
Step 2: find the length of leg 1 (start point to correction point) ... 680 m/h x 1.5 h = 1020 m ... this is the first line.
 
Step 3: find the length of leg 2 (correction point to destination point) ... 680 m/h x 2 h = 1360 m ... this is the second line.
 
Step 4: we know that we changed course by 10°, if we had remained in a straight line then we know that is 180°, our angle will be equal to 180° - 10° = 170°
 
Step 5: lets use the la of cosines to solve this. If you labeled your triangle the same as I did mine then you should be looking at solving for side b using angle B and sides a & c.  b2 = a2 + c2 - 2 x a x c x Cos B = 13602m +10202m - 2 x 1360m x 1020m x Cos (170) = b2 = 5622250.63m ... b = 2371.128556m
 
I would right it out in more steps if I were you so as to understand it better but I simply plugged it into a calculator.
 
Good luck Dalia