SOLUTION: A plane is flying on a bearing of 237 degrees from Town A to Town C which is 364 km away. At Town C, the plane changes its course and travels 527 km on a bearing of 077 degrees to

Click here to see ALL problems on Geometry Word Problems

Question 1102521: A plane is flying on a bearing of 237 degrees from Town A to Town C which is 364 km away. At Town C, the plane changes its course and travels 527 km on a bearing of 077 degrees to Town B. Find the distance from A to B (0dp)
Found 2 solutions by Theo, richwmiller:
Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website!
not sure if there's an easier way to do this but this is the way i did it and it appears i did it right.

here's my diagram.

A to C equals 364 km.
a bearing of 237 degrees from A to C = 180 + 57, therefore angle FAC = 57 degrees.
since triangle AFC is a right triangle, then angle ACF = 33 degrees.
sin(33) = AF/ 364
solve for AF to get AF = 364 * sin(33) = 198.2486087
that's the same length as GC.
cos(33) = CF / 364
solve for CF to get CF = 364 * cos(33) = 305.2760867
that's the same length as GA.

C to B equals 527 km.
a bearing of 77 degrees from C to B makes angle BCE = 13 degrees because 90 - 77 = 13.
sin(13) = BC / 527
solve for BC to get BC = 527 * sin(13) = 118.5492056
cos(13) = FE / 527
solve for FE to get FE = 527 * cos(13) = 208.2169374
that's the same length as FE.

DE is the same length as GC
since DE = BE + DB and BE = 118.5492056, then solve for DB = 198.2486087 - 118.5492056 = 79.6994031.

right right triangle ABD with legs AD and DB.
use pythagorus to solve for AB.
[AB]^2 = [AD]^2 = [DB]^2 = 208.2169374^2 + 79.6994031^2 = 49706.28787
AB = sqrt(49706.28786) = 222.9490701.

that's your solution.

after looking at the diagram, i realized that i could have used the law of cosines to get the same answer, because.

triangle ABC was formed.
AC = 364
CB = 527
angle ACB = 20 degrees.

the law of cosines says that c^2 = a^2 + b^2 - 2ab * cos(C).
AC = b in this formula.
CB = a in this formula
C = 20 degrees in this formula.
BA = c in this formula.

c^2 = a^2 + b^2 - 2ab * cos(C) became:
c^2 = 527^2 + 364^2 - 2 * 527 * 364 * cos(20)
solve for c^2 to get c^2 = 49706.28788
solve for c to get c = sqrt(c^2) = 222.9490701

that's the same answer i got without using the law of cosines.
using the law of cosines would have been more direct.

the law of cosines can be used to find the length of the third side if you know 2 sides of the triangle and the included angle

the law of cosines can also be used to find the included angle if you know 3 sides of the triangle.

here's a reference on law of cosines.

https://www.varsitytutors.com/hotmath/hotmath_help/topics/law-of-cosines

Answer by richwmiller(17219) (Show Source):
You can put this solution on YOUR website!
SAS

third side c = 222.949
Obtuse scalene triangle.
Sides: a = 364 b = 527 c = 222.949
Area: T = 32804.52
Perimeter: p = 1113.949
Semiperimeter: s = 556.975
Angle ∠ A = α = 33.945� = 33�56'43″ = 0.592 rad
Angle ∠ B = β = 126.055� = 126�3'17″ = 2.2 rad
Angle ∠ C = γ = 20� = 0.349 rad
Height: ha = 180.245
Height: hb = 124.495
Height: hc = 294.278
Median: ma = 361.377
Median: mb = 147.204
Median: mc = 438.96
Inradius: r = 58.898
Circumradius: R = 325.93
Vertex coordinates: A[222.949; 0] B[0; 0] C[-214.234; 294.278]
Centroid: CG[2.905; 98.093]
Coordinates of the circumcenter: U[111.475; 306.274]