Trigonometry-Bearing (1 Viewer)

[AnandDNA]

Just was stuck on a question and was wondering whether anybody could show me the working out of it. The question is:

The Bearing of B from A is 130(degress). the bearing of C from A is 220(degrees). The distance AB is 40Km. find the distance BC

i am mostly stuck because i dont know which side is the adjacent and opposite.

 

[watatank]

i er don't know how to upload attachments. this would be so much easier if i could.

you draw a picture and try and get as much information as you can. preferably on a plane of some sort. you will see angle BAC is a right angle. you can work out C is 50 degrees. since you have AB as 40 you have all the information you need and you can use the sine rule.

the answer is

sin@ = O/H

sin(ACB) = AB/BC

sin 50 = 40/BC

BC = 40/sin50

sory not the best explanation, but i hope it helps!

 

[AnandDNA]

Thx for the reply but i dont get how u got angle c as 50(degrees)

 

[sle3pe3bumz]

hm i think the problem with that is that you assume CB is parrallel to the x-axis which is false - or i think it is. LOL I uploaded a pic but then I thought it was wrong as I also assumed that. Unless he's right. Well then I'm just totally screwed up in the head. It's late okay people ! ><"

 

[watatank]

yeah thats the assumption i made but i can't see any other way to do it, especially if this is a year 10 level question. if it gave the distance of AC you could use the cosine rule or gave the bearing from B to C or something you could get another angle and use the sine rule. but it doesn't so i can only assume that BC is parallel to the x axis.


p.s. i found out how to upload here's my solution. hope it helps! :wave:

 

[AnandDNA]

Thx for the upload watatank

 

[AnandDNA]

lol now my teacher says that she forgot to put up that the bearing of B from C is 80 degrees. But im still having trouble i think my diagrm is inaccurate

 

[watatank]

ok here's the solution. refer to the diagram.

its simply the sine rule

sin@ = O/H

sin(ACB) = AB/BC

sin 40 = 40/BC

BC = 40/sin40


how i got ACB might be a little confusing though if you cant see it. i used alternate angles and parallel lines to get it. hard to explain though, hope the diagram does it!

if you have any questions just ask. :wave:

 

[AnandDNA]

Thx it makes sense now