a man set out to travel from A to C via B.from A,he travels a distance of 8km on a bearing of N30E to B.from B,he goes 60km due east.

Question

Find all it's missing bearing and distance

Answer 1

????????????? "travel" ???????? maebee him wanna walk or drive a kar

or maebee him just wanns GO, GO, GO, GO, GO, GO, GO, GO

???????????????? "a tu c via b"??????????

wi dont yu sae him go a tu b, then on tu c ????????????

??????????????????? "missing beering & distans" ??????????

Maebee yu want a line from start spot tu end spot?????????

????????? "n30e" ?????  me ges yu tri tu sae direkshun be 30 deg rite av north?????

That wood be 60 deg from horizontal line...normal start for referens sistem

First sekshun: north komponent=8km*sine(60 deg)...8*0.8660254038443...=6.928203230275509

Eest part=8km*kosine(60)=8*0.5=4km eest

next step=due eest 60 km

Strate line from start tu end...horizontal part=60+4=64 (eest)

vertikal part=north komponent=6.9282032302 & square=47.9999999

So yu want the long side av a rite triangel...64^2=4,096

add 4096+48=4144 & sqrt(4144)=64

6.928203...^2=47.9999999999

6.9282032302^2=

Answer 2

In the triangle ABC angle B is 90+30=120, AB=8km, BC=60km.

AC^2=AB^2+BC^2-2AB.BCcos120 (Cosine Rule)

AC^2=8^2+60^2+2*8*60cos60=64+3600+960/2=3664+480=4144

AC=√4144=64.37km approx. Angle BAC is found using the Sine Rule:

SinBAC/60=sin120/√4144, sinBAC=60sin120/√4144=0.8072

Therefore the bearing of C from A is 53.82+30=N83.82E