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

# 2 Answers

^{2}= b

^{2}+ c

^{2}- (2)*(a)*(b)*cos(A)

^{2}= (1020 miles)

^{2}+ (1360 miles)

^{2}- (2)*(1020 miles)(1360 miles)*cos(170°)

^{2}= 1040400 miles

^{2}+ 1849600 miles

^{2}- (2774400 miles

^{2})*cos(170°)

^{2}= 2890000 miles

^{2}+ 2732251 miles

^{2}= 5622250.6 miles

^{2}

**a**= (5622250.6 miles

^{2})

^{½}=

**2371 miles**

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