#### Triangulation using excel

```Good Morning,

I would like to better understand the method or methods used to calculate and determine bearing and distance from a known point based on observation(s) from two known points.

Let me explain more: Lets use a 60,60,60 triangle so there is no confustion about what side is the base. I have two observation towers, Tower Left, and Tower Right. and the tip or top is the target.  If I observe a bomb/explosion/ect... at a point that is 10 degrees left of the target from the left tower and 10 degrees right of the target from the right tower. I now have a new triangle that is 70,70,40 and I can calculate the leg lengths. where I am lost is trying to determine the bearing & distance from the target to the new spot.  using this example it would be due north of the target i.e. 360 or 12 o'clock, but I am not sure how to find distance at this point.  any assistance in a medthod or solution would be greatly appreciated.  Don

```
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12/12/2011 3:36:58 PM
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```On Dec 13, 4:36=A0am, Donald Ross <donald.ross....@shaw.af.mil> wrote:
> Good Morning,
>
> I would like to better understand the method or methods used to calculate=
and determine bearing and distance from a known point based on observation=
(s) from two known points.
>
> Let me explain more: Lets use a 60,60,60 triangle so there is no confusti=
on about what side is the base. I have two observation towers, Tower Left, =
and Tower Right. and the tip or top is the target. =A0If I observe a bomb/e=
xplosion/ect... at a point that is 10 degrees left of the target from the l=
eft tower and 10 degrees right of the target from the right tower. I now ha=
ve a new triangle that is 70,70,40 and I can calculate the leg lengths. whe=
re I am lost is trying to determine the bearing & distance from the target =
to the new spot. =A0using this example it would be due north of the target =
i.e. 360 or 12 o'clock, but I am not sure how to find distance at this poin=
t. =A0any assistance in a medthod or solution would be greatly appreciated.=
=A0Don

One way is to find the coordinates of the target and the bomb
separately then find the distance and bearing of the bomb from the
target.

Assume that the base of your triangle is your x axis, the left hand
tower is on 0 and the right hand tower is on D.

The target makes a triangle with base angles A in the left and B on
the right.  The top angle is C =3D 180 - A - B.  The left side is L.

The sine rule says L/sinB =3D D/sinC.  You know B, D and C so this gives
L.  This is what you call the leg length.  The coordinates of the
target are x1 =3D LcosA and y1 =3D LsinA.  In your simplified case, A and
B are 60 but the above works for any angles.

Repeat the calculations only using the appropriate angles for the
bomb.  This gives you another set of x2 and y2 as the coordinates of
the target.  We now have the ends of the target/bomb line and can draw
a right angles triangle.

The distance from target to bomb =3D SQRT((y2-y1)^2+(x2-x1)^2).
Pythagoras
The angle from the x direction is ATAN((y2-y1)/(x2-x1)) or from

Snags:
1 Excel works in radians, not degrees so you will need to convert
things at the beginning and the end.  C =3D 180 - A - B becomes C =3D PI()
- A - B.
2 ATAN only give angles from -pi/2 to pi/2 or from -90 to 90.
Sometimes you will

Cheers

xt
```
 0
12/13/2011 8:32:14 PM

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