An aircraft flies from its base 200km on a bearing of 162 degrees, then 350km on a bearing of 260 degrees and then returns directly to base. Calculate the length and bearing of return journey!
HELP PLEASE!
2007-02-18 03:36:13 · 7 answers · asked by Nightmare! 2 in Science & Mathematics ➔ Mathematics
let bearing 00 00 be due east
bearing 162deg= N 72 deg W
bearing 260deg= S10 deg W
{rotating in anticlockwise
direction}
starting point is C
CA=200km,AB=350km
BC=return journey(km)
angleA=72+(270-260)
=82deg
using cosine rule
a^2=b^2+c^2-2bccos82
=200^2+350^2
-2*200*350*cos82
=143,015.7659
a=sqrt(143,015.7659)
=378.1742534km
using sine rule,
sinB=200*sin82/378.1742534
=0.523709952
B=arcsine(0.523709952)
=31.58143726 deg
therefore,bearingBC
=80-31.58143726
=48.41856274 deg
hence, the return journey is
378.1742534km with a
bearing of 48.41856274 deg
{N 41.58143726 deg E}
if you don't follow the working,
contact me
2007-02-19 03:25:13 · answer #1 · answered by Anonymous · 0⤊ 0⤋
Sketch a diagram of the problem.
Deduce an angle within the triangle of the journey.
Apply the cosine rule for the length of the return leg and then the sine rule for the bearing.
The answer is 378 km on a bearing of 048° (nr whole numbers)
2007-02-18 11:36:06 · answer #2 · answered by aepacino 2 · 1⤊ 0⤋
draw it out and you find that you have the distance of two sides
(200 and 250) the angle in between is the change in bearing (260-162=98)
so use the law of cosines to figure out the distance and then the law of sines to find the degree
2007-02-18 03:46:57 · answer #3 · answered by dave 2 · 1⤊ 0⤋
What is the 260° relative to? The ending of the first part of the trip? The starting point?
?
2007-02-18 04:11:47 · answer #4 · answered by Anonymous · 0⤊ 0⤋
sketch a diagram of the subject. Deduce an physique of concepts in the triangle of the journey. save on with the cosine rule for the size of the bypass lower back leg and then the sine rule for the bearing. the respond is 378 km on a bearing of 048° (nr complete numbers)
2016-11-23 16:39:08 · answer #5 · answered by ? 4 · 0⤊ 0⤋
Sketch out your route, take south-north as your Y axis, and east-west as your X axis. give your starting point as 0,0. label your first landing point "A", and your second landing point "B".
Start at the starting point (0,0), and work out the coordinates of "A", then work out the coordinates of "B" using Pythagoras.
Finally use Pythagoras again to find both your return distance, and bearing from the coordinates of "B" to your starting point 0,0.
2007-02-18 22:17:43 · answer #6 · answered by Valmiki 4 · 0⤊ 0⤋
why ?
2007-02-18 03:38:15 · answer #7 · answered by cereal killer 5 · 0⤊ 3⤋