A traveller moves from town p on a HEADING of 055 degree to a town q 200km away.he then moves from q on a HEADING of 155 degree tn a town r 400km from q.find
A.distance between p and r
B.bearing of p from r,correct to the nearest degree
========================= ^&*% landlubber mathematicians
anyway
DRAW IT !!
In triangle PQR angle Q is 180 - (155-55) = 80
so
find length of side q opposite Q from P to R, law of cosines
q^2 = r^2 + p^2 - 2 p r cos 80 (almost a right triangle :)
q^2 = 200^2 + 400^2 - 2(200)(400)(0.174)
q^2 = 40,000 + 160,000 - 27,840 = 172,160
q = 415 km from p to r
now bearing
find angle in triangle at P with law of sines
sin 80/415 = sin P / 400
sin P = (400/415) sin 80 = 0.949
P = 71.1 deg
so bearing of R from P = 55+71.7 = 126.7
we want bearing of P from R so add 180 degrees
= 306.7 degrees so head 306.7 , about 37deg north of west to get home to P
A traveller moves from town p on a bearing of 055 degree to a town q 200km away.he then moves from q on a bearing of 155 degree tn a town r 400km from q.find
A.distance between p and r
B.bearing of p from r,correct to the nearest degree
3 answers
Solution
it is very correct thanks