Bearings !!!!! Can anybody help?

Question

I have a triangle with the length of:
QS- 98.5m
QR-83.7m
RS-120m

The question is to find the 'BEARING' of R from S.

This is not a right angled triangle, and no angles are given, does anybody know how i might beable to work this out?

PLEASE HELP, DESPERATE

Thanks

Answer 1

Several ways. The first that comes to mind is to figure the area with Heron's formula (area = sqrt(s(s - a)(s - b)(s - c)), where s = (a + b + c)/2). The area can also be figured as 1/2 base times altitude, so you know the altitudes of the triangle. The altitudes give you right triangles where you know the hypotenuse (a side of the triangle), and a leg, so you can compute the trig functions of all the angles. Then the angles can be computed as arcsin or arccos of the angle (arcsin(altitude/side))

Answer 2

You can find all of the angles of the triangle QRS using the cosine formulae, but you cannot determine the bearing of R from S. Bearing are expressed relative to the North direction and this is not given. You can turn QRS in any direction and get a multitude of answers. Check again

Answer 3

You don't have enough information to find a bearing, as it is usually understood. It's fairly standard stuff to find angles in a triangle given the lengths of the sides, using the cosine rule.
Have you missed something out ?

Answer 4

You can't calculate the bearing without more information. Bearing's are calculated from North. With only the lengths of the triangle given, you could rotate it any way you want with respect to North. So you need something defining the position of North relative to the triangle.
If for example you knew Q was North of S, then you could do it using the methods suggested above. I think splitting into two right angle triangles is probably the easiest if that is what you are used to dealing with.

Answer 5

By using the cosine formula you can work out the size of each angle of the triangle.
See the link for some worked examples on this.
But you can't find the bearing of one corner from another unless you're given information on where the true north direction is.

Answer 6

Sketch out triangle QRS and then use cosine rule to solve for the angles within the triangle.
cosine rule: a2 = b2 + c2 - 2bcCosA

However, please note that the directions of the lines are not given in your question. So you can't find the bearing as the North is not stated. I believe that you've got the diagram in your textbook or question sheet, right?

Answer 7

ideally,you need to know the co-ordinates
of points Q,R and S-if you know the
the position of points Q and R say,
there are still two points where S can
be located-you know the length of QS
and RS but there are two points where they
can intersect

in order to find the bearings of R from S,
you would need to know the positions
of R and S-that would allow you to
calculate the bearing of RS
however,if you didn't know the position of Q
you could not find the bearings of QR and
QS
therefore, more info {including the
positions of R and S}is required in order to
determine the bearing of RS

i hope that this helps

if you have any more info,email me
and i will try to put you on the right
track (bearing)

Answer 8

Draw a scaled down version of the triangle then use SOCATOH to work out your answers

Answer 9

Oooo guy, cleansing bearings is a annoying activity. ok 1st, dont pour a answer/water/something liquid on them. it could lead on them to rust up and that wouldnt be relaxing. in case you go with to scrub the exterior of them on an analogous time as their nonetheless in the wheel, what i do is take a toothpick and in basic terms scrape off the airborne dirt and dust and grim. it works rather darn properly.

Answer 10

just cut this triangle to create 2 right angled triangles.

for example :
let h which start from S, ant cut [QR] into A with a right angle
and ø=(SQR) angle.

then we have :

1) QR=QA+AR
2) cos(ø) = QA/QS
3) sin(ø) = h/QS

RS^2 = h^2+AR^2
RS^2 = (sin(ø).QS)^2+(QR-QS.cos(ø))^2
RS^2 = (sin(ø)^2+cos(ø)^2)*QS^2 + QR^2-2.QR.QS.cos(ø)

we know that sin(ø)^2+cos(ø)^2 = 1
then,

RS^2 = QS^2 + QR^2-2.QR.QS.cos(ø)
ie :
cos(ø) = - (RS^2-QS^2-QR^2)/(2.QR.QS)
-> ø = arccos(- (RS^2-QS^2-QR^2)/(2.QR.QS))

sorry for my english, i'm french