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
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
2 answers
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