Question
A girl moves from a town M on a bearing of 065° to a town N, 100km away. She then moves from N on a bearing of 130° to a position P, 200km from N. find the :(a). The distance between M and P. (b). The bearing of M from P correct to the nearest degree
Answers
oobleck
draw a diagram
then use the law of cosines to find MP
use the law of sines to find the angles at M and P
then you calculate (b)
come back if you get stuck, and show your work
then use the law of cosines to find MP
use the law of sines to find the angles at M and P
then you calculate (b)
come back if you get stuck, and show your work
mathhelper
Using the diagram which I assume you made, you should see that, after a bit
of basic geometry, for triangle MNP the angle at N = 115°
And by the cosine law:
MP^2 = 100^2 + 200^2 - 2(100)(200)cos115°
I got MP = appr 258.66 km
by the sine law we find angle M to be ...
sinM/200 = sin115/258.66
angle M = 44.5°
add on you 65° and you have a bearing of 109.5°
or , by vectors
bearing of 65° ---> standard angle of 25°
bearing of 130° ---> standard angle of -40°
resultant = 100(cos25, sin25) + 200(cos-40, sin -40)
= (90.63, 42.26) + (153.21, -128.56)
= (243.84, -86.296)
|resultant| = √(243.84^2 + (-86.296)^2 ) = 258.66 km, as above
direction of resultant:
tan 𝛳 = -86.296/243.84
𝛳 = -19.5°
so bearing = 90+19.5 = 109.5° , as above
of basic geometry, for triangle MNP the angle at N = 115°
And by the cosine law:
MP^2 = 100^2 + 200^2 - 2(100)(200)cos115°
I got MP = appr 258.66 km
by the sine law we find angle M to be ...
sinM/200 = sin115/258.66
angle M = 44.5°
add on you 65° and you have a bearing of 109.5°
or , by vectors
bearing of 65° ---> standard angle of 25°
bearing of 130° ---> standard angle of -40°
resultant = 100(cos25, sin25) + 200(cos-40, sin -40)
= (90.63, 42.26) + (153.21, -128.56)
= (243.84, -86.296)
|resultant| = √(243.84^2 + (-86.296)^2 ) = 258.66 km, as above
direction of resultant:
tan 𝛳 = -86.296/243.84
𝛳 = -19.5°
so bearing = 90+19.5 = 109.5° , as above