# Find the third angle.

• Jun 5th 2010, 08:12 PM
geoleo
Find the third angle.
Hello,
I have two triangles ASB and AOB, in different planes. Line AB is
common to triangles ASB and AOB. I have below information:

1)I have Direction cosines of line AB, line AS and line BS.
2)In triangle AOB two sides - OA and OB are of same length. That is AOB is isoceles triangle.
I don't have any angle information of triangle AOB.
3)Vector AS is perpendicular to vector AO. So angle SAO =90 degree.
4)Angle SAB I can find using the dot product of vector AB and vector AS

Can I find angle OAB ? (Worried)
• Jun 6th 2010, 04:28 AM
oldguynewstudent
Quote:

Originally Posted by geoleo
Hello,
I have two triangles ASB and AOB, in different planes. Line AB is
common to triangles ASB and AOB. I have below information:

1)I have Direction cosines of line AB, line AS and line BS.
2)In triangle AOB two sides - OA and OB are of same length. That is AOB is isoceles triangle.
I don't have any angle information of triangle AOB.
3)Vector AS is perpendicular to vector AO. So angle SAO =90 degree.
4)Angle SAB I can find using the dot product of vector AB and vector AS

Can I find angle OAB ? (Worried)

OK since you haven't had another response yet, try this approach.

First figure out angle SAB like you said. Also using the direction cosines, figure out the other angles and sides of ASB. Use the law of cosines and law of sines. You might also be able to use some trig or the Pythagorean theorem if angle SAB is 90 degrees. You really didn't give me the specifics so I can't do any calculations. Try to find other congruencies between the triangles. Since AOB is isoceles, you know that the sum of OAB and AOB is supplementary to OBA. So if you figure out one of the angles, you have everything.

If you're still stuck, send the specifics like the direction cosines or scan in the whole problem.

Good luck.
• Jun 6th 2010, 05:24 AM
geoleo
Thank you.
I will give complete data and problem statement :

I am attaching a figure related to my problem. : Problem_final2.jpg .
In the Figure Triangles ASB and ASC are above the sphere, Triangle AOB is towards inside the sphere, with O(0,0,0) the center of sphere.
So triangles ASB and AOB are in different planes with AB common .
triangles ASC and AOC are in different planes with AC common.
triangles ASC and ASB are in diffrent planes with line AS in common.
Lines SA, SB and SC are tangents to the sphere. so SA,SB,SC make 90 degrees with OA,OB, OC respectively.

1) In triangle AOB I have :
vertex O = (0,0,0)
side OA = 6378.14kms
side OB = 6378.14kms
I don't have any angles in triangles AOB

2) In triangle AOC
vertex O = (0,0,0)
side OA = 6378.14kms
side OC = 6378.14kms
I don't have any angles in triangles AOC

3) In Triangle ASB I have :
dcs of SA=[-0.839790306911 -0.518228990267 -0.161836195347]
dcs of SB=[-0.862922555238 -0.488357439399 0.129891013321]

4) In Triangle ASC
dcs of SA=[-0.839790306911 -0.518228990267 -0.161836195347]
dcs of SC=[-0.898863966995 -0.433861497526 0.061707126937]

Therefore:
5) dcs of AB = dcs of SA- dcs of SB (vector addition theorem)
dcs of AC = dcs of SA- dcs of SC (vector addition theorem)
Note: AB and AC have to be normalized AB = AB/norm(AB) and AC= AC/norm(AC)
where norm(vec) = sqrt( vecx^2 +vecy^2 +vecz^2)

6) So I will get all angles of triangle ASB as below : (output will be in radians, convert to degrees by multiplying R.H.S with 180/pi)
angle ASB = acos(dot(SA,SB))
angle SAB = acos(dot(SA,AB))
angle SBA = acos(dot(SB,AB))

7) Similarly angles of triangle ASC
angle ASC = acos(dot(SA,SC))
angle SAC = acos(dot(SA,AC))
angle SCA = acos(dot(SC,AC))

8) Triangle ASB and AOB are in different planes, with a common side AB.
and Angle SAO =90 degrees . Angle SBO = 90 degrees.

9) Triangle ASC and AOC are in different planes, with a common side AC.
and Angle SAO =90 degrees . Angle SCO = 90 degrees.

10) I thought I can find :
Angle BAO = angle SAO - angle SAB = 90 - angle SAB. - (1)
Angle CAO = angle SAO - angle SAC = 90 - angle SAC . - (2)
Are (1) and (2) valid or not? I think I cannot subtract two angles in different planes.

11) To find Direction cosines of OA(l,m,n) , Using Dot Product of two vectors I can write 3 eqns as below:
l*l1 + m*m1 + n*n1 = 0 ( because RHS=cos(SAO) which is cos(90) =0. ) -(3)
l*l2 + m*m2 +n*n2 = cos(BAO) -(4)
l*l3 + m*m3 +n*n3 = cos(CAO) -(5)

where Direction cosines of SA (l1,m1,n1)
where Direction cosines of AB (l2,m2,n2)
where Direction cosines of AC (l3,m3,n3)

12) Three eqns , three unknowns I can find direction ratios of OA.My aim is to find the coordinates of vertex A.
If I get direction cosines OA (l,m,n) , Then coordinates of A is (l*R, m*R,n*R) where R is length of OA which is 6378.14kms.

But since angles BAO and CAO are not valid because (1) and (2) not valid.. So (4) and (5) also not valid. So I am not getting proper output.

Note : for cross verification:
If I get coordinates of A, I can get coordinates of S. for Simulation purpose I have S coordinates as below:
Coordinates of S=[35244.13549313 23114.23482794 524.26711972]

Let Coordinates of A be (x1,y1,z1). If A is valid, it should satisfy below eqns:

dcs of SA(2)*S(1)-dcs of SA(1)*S(2) = dcs of SA (2)*x1-dcs of SA (1)*y1 ---(6)
dcs of SA(3)*S(2)-dcs of SA(2)*S(3) = dcs of SA (3)*y1-dcs of SA (2)*z1 ---(7)
dcs of SA(3)*S(1)-dcs of SA(1)*S(3) = dcs of SA (3)*x1-dcs of SA (1)*z1 ---(8)

in (6) (7) (8) L.HS need not be exactly same as R.H.S , If approximately equal also I am happy, my method is right
But my A which I found using steps 1 to 12 is not satisfying eqns (6),(7),(8). Its completely off. Please help.
So how can I find Directions cosines of OA or coordinates of point A. is there any other method?
Thanks.

Note: Above equations(6),(7),(8) follows from the fact that
(x-x1)/l1 = (y-y1)/m1 = (z-z1)/n1

where coordinates of S (x,y,z)
coordinates of A(x1,y1,z1)
direction cosines of SA (l1,m1,n1)

Note : S is used only for cross verification ie, simulation purpose. In reality position of S is not known.

(Worried)
• Jun 6th 2010, 06:28 AM
Soroban
Hello, geoleo!

I believe the answer is no.

Quote:

I have two triangles $ASB$ and $AOB$ in different planes.
Side $AB$ is common to both triangles.

1) I have direction cosines of lines $AB, AS$ and $BS.$

2) In $\Delta AOB,\:OA \,=\,OB$

3) $\overrightarrow{AS} \perp \overrightarrow{AO}$

4) I can find $\angle SAB$ using the dot product of $\overrightarrow{AB}$ and $\overrightarrow{AS}.$

Can I find $\angle OAB$ ?

Place $\Delta ASB$ on the "floor".

Code:

            * - - - - - - - - - - - - - - - *           /          o B                  /           /        *    *                /         /        *        *            /         /      *              *        /       /    A o  *  *  *  *  o S  /       /                              /     * - - - - - - - - - - - - - - - *

At $A$, construct $\overrightarrow{AO}$ so that: . $\overrightarrow{AO} \perp \overrightarrow{AS}$

Let $\theta \,=\,\angle OAB$, where . $0 \,<\,\theta\,<\,90^o$

Code:

                O o                       *             * - - - - - - - - - - - - - - - *           /    *    o B                  /           /        *    *                /         /    *  *        *            /         /      *              *        /       /    A o  *  *  *  *  o S  /       /                              /     * - - - - - - - - - - - - - - - *

For any angle $\theta$, we can construct $\Delta AOB$ with $OA = OB.$

Code:

                O o                 .:*                 *::*             * - .::::*- - - - - - - - - - - *           /    *:::θ:o B                  /           /    .::::*    *                /         /    *:θ*        *            /         /    .:*              *        /       /    A o  *  *  *  *  o S  /       /                              /     * - - - - - - - - - - - - - - - *

Since $\theta$ can be any of a brizillion different values . . .

• Jun 6th 2010, 11:04 AM
oldguynewstudent
Quote:

Originally Posted by geoleo
Thank you.
I will give complete data and problem statement :

I am attaching a figure related to my problem. : Problem_final2.jpg .
In the Figure Triangles ASB and ASC are above the sphere, Triangle AOB is towards inside the sphere, with O(0,0,0) the center of sphere.
So triangles ASB and AOB are in different planes with AB common .
triangles ASC and AOC are in different planes with AC common.
triangles ASC and ASB are in diffrent planes with line AS in common.
Lines SA, SB and SC are tangents to the sphere. so SA,SB,SC make 90 degrees with OA,OB, OC respectively.

1) In triangle AOB I have :
vertex O = (0,0,0)
side OA = 6378.14kms
side OB = 6378.14kms
I don't have any angles in triangles AOB

2) In triangle AOC
vertex O = (0,0,0)
side OA = 6378.14kms
side OC = 6378.14kms
I don't have any angles in triangles AOC

3) In Triangle ASB I have :
dcs of SA=[-0.839790306911 -0.518228990267 -0.161836195347]
dcs of SB=[-0.862922555238 -0.488357439399 0.129891013321]

4) In Triangle ASC
dcs of SA=[-0.839790306911 -0.518228990267 -0.161836195347]
dcs of SC=[-0.898863966995 -0.433861497526 0.061707126937]

Therefore:
5) dcs of AB = dcs of SA- dcs of SB (vector addition theorem)
dcs of AC = dcs of SA- dcs of SC (vector addition theorem)
Note: AB and AC have to be normalized AB = AB/norm(AB) and AC= AC/norm(AC)
where norm(vec) = sqrt( vecx^2 +vecy^2 +vecz^2)

6) So I will get all angles of triangle ASB as below : (output will be in radians, convert to degrees by multiplying R.H.S with 180/pi)
angle ASB = acos(dot(SA,SB))
angle SAB = acos(dot(SA,AB))
angle SBA = acos(dot(SB,AB))

7) Similarly angles of triangle ASC
angle ASC = acos(dot(SA,SC))
angle SAC = acos(dot(SA,AC))
angle SCA = acos(dot(SC,AC))

8) Triangle ASB and AOB are in different planes, with a common side AB.
and Angle SAO =90 degrees . Angle SBO = 90 degrees.

9) Triangle ASC and AOC are in different planes, with a common side AC.
and Angle SAO =90 degrees . Angle SCO = 90 degrees.

10) I thought I can find :
Angle BAO = angle SAO - angle SAB = 90 - angle SAB. - (1)
Angle CAO = angle SAO - angle SAC = 90 - angle SAC . - (2)
Are (1) and (2) valid or not? I think I cannot subtract two angles in different planes.

11) To find Direction cosines of OA(l,m,n) , Using Dot Product of two vectors I can write 3 eqns as below:
l*l1 + m*m1 + n*n1 = 0 ( because RHS=cos(SAO) which is cos(90) =0. ) -(3)
l*l2 + m*m2 +n*n2 = cos(BAO) -(4)
l*l3 + m*m3 +n*n3 = cos(CAO) -(5)

where Direction cosines of SA (l1,m1,n1)
where Direction cosines of AB (l2,m2,n2)
where Direction cosines of AC (l3,m3,n3)

12) Three eqns , three unknowns I can find direction ratios of OA.My aim is to find the coordinates of vertex A.
If I get direction cosines OA (l,m,n) , Then coordinates of A is (l*R, m*R,n*R) where R is length of OA which is 6378.14kms.

But since angles BAO and CAO are not valid because (1) and (2) not valid.. So (4) and (5) also not valid. So I am not getting proper output.

Note : for cross verification:
If I get coordinates of A, I can get coordinates of S. for Simulation purpose I have S coordinates as below:
Coordinates of S=[35244.13549313 23114.23482794 524.26711972]

Let Coordinates of A be (x1,y1,z1). If A is valid, it should satisfy below eqns:

dcs of SA(2)*S(1)-dcs of SA(1)*S(2) = dcs of SA (2)*x1-dcs of SA (1)*y1 ---(6)
dcs of SA(3)*S(2)-dcs of SA(2)*S(3) = dcs of SA (3)*y1-dcs of SA (2)*z1 ---(7)
dcs of SA(3)*S(1)-dcs of SA(1)*S(3) = dcs of SA (3)*x1-dcs of SA (1)*z1 ---(8)

in (6) (7) (8) L.HS need not be exactly same as R.H.S , If approximately equal also I am happy, my method is right
But my A which I found using steps 1 to 12 is not satisfying eqns (6),(7),(8). Its completely off. Please help.
So how can I find Directions cosines of OA or coordinates of point A. is there any other method?
Thanks.

Note: Above equations(6),(7),(8) follows from the fact that
(x-x1)/l1 = (y-y1)/m1 = (z-z1)/n1

where coordinates of S (x,y,z)
coordinates of A(x1,y1,z1)
direction cosines of SA (l1,m1,n1)

Note : S is used only for cross verification ie, simulation purpose. In reality position of S is not known.

(Worried)

You have angle ASC right? You know angle OAS is a right angle. So angle AOS is 180 - (angle ASC + OAS). Then angle BOS is congruent to angle AOS so angle AOB = 2* angle AOS.

Or am I missing something? (You already did the hard stuff!)
• Jun 6th 2010, 05:05 PM
geoleo

Actually in this case the Triangles appear to be congruent..

This is a general problem.... I mean...Figure can be even like this....(I have attached below) where no triangles are congruent I believe...Or If they are congruent please let me know.....I am missing something. I would like to know whereever points A , B , C lie on the sphere will I get congruent triangles??

Since In the figure Its a Sphere. Lines SA, SB and SC are tangents to the sphere. There can be many such tangents so many such Points of A,B,C I can get ( I will take 3 such points at a time). But If I can find position of Point A , Then I can find the position of point S.

The Data and Method which I have presented earlier above, If I can get three equations then I can find direction cosines of Radius OA, from which I can find location of Point A.

Yes Soroban, I also think that there is no soln. But I am still hoping that there is some way out....This is part of my project.So I should get some clue, Some Solution.

Is there any method to find Point A? The only initial informations will be

1) Sphere's Center O (0,0,0)
2) Radius R = 6378.14 Kms
3) If u draw lines from Point S above as tangents to sphere for example line AS makes 90 degree with line OA. Like that how much ever lines we can draw ( lower is better Here I took three lines AS, BS and CS)
4) It is assumed that Direction cosines of lines which are tangents to the sphere is known. For eg: lines AS, BS , CS direction cosines are known.

Now any method is there to find Position of Point A, or B or C ? Or Point S directly without finding Point A? (Worried)

I appreciate all your help to me throughtout..(Happy)
• Jun 6th 2010, 05:29 PM
geoleo
Soroban,

In your explanation You have told that we can draw many triangles AOB where OA=OB with different thetas . If point O is fixed (0,0,0) . Still we can draw many triangles???

In triangle AOB , Since Point O (0,0,0) is fixed, and Points A, B are fixed (but we don't know their positions). lines OA and OB are also fixed. I know OA=OB=6378.14kms. Still there is no way to find the angles OAB or OBA?? (Worried)
If I get angle AOB also that can help me.....

oldguynewstudent :

I am trying to see whether any clue is there with the method u have explained.
Thanks to both of you....