I have a Spreadhset with the following representing two sides of a Triangle

Angle Length
Side1, 20 35
Side2, 45 30
Side3,


I need a formula to calculate the Angle (degrees) and Length for the third
Angle

TIA
C

angle3 = 180 - ( angle1 + angle2 )
side3 is a bit trickier, must be a formula out there somewhere on the net
though.
Will have a think about that...

"Cathy" wrote:

I have a Spreadhset with the following representing two sides of a Triangle

Angle Length
Side1, 20 35
Side2, 45 30
Side3,


I need a formula to calculate the Angle (degrees) and Length for the third
Angle

TIA
C

See he
http://www.ajdesigner.com/phptriangl...ion_side_a.php

Hth,
Merjet

Your data doesn't describe a triangle.

Angle Length
Side1 20 35
Side2 45 30
Side3

If you had:

Angle Length
Side1 20 35
Side2 30
Side3

Then the triangle would be:

Angle Length
Side1 20 35
Side2 17 30
Side3 143 61.65

======

If you had:

Angle Length
Side1 20 35
Side2 45
Side3

Then the triangle would be:

Angle Length
Side1 20 35
Side2 45 72.36
Side3 115 92.75


Sometimes, too much information isn't a good thing!


Cathy wrote:

I have a Spreadhset with the following representing two sides of a Triangle

Angle Length
Side1, 20 35
Side2, 45 30
Side3,

I need a formula to calculate the Angle (degrees) and Length for the third
Angle

TIA
C

--

Dave Peterson

Some more info:

Law of Cosines:
http://en.wikipedia.org/wiki/Law_of_cosines

Law of Sines:
http://en.wikipedia.org/wiki/Law_of_sines

and maybe
Law of Tangents:
http://en.wikipedia.org/wiki/Law_of_tangents



merjet wrote:

See he
http://www.ajdesigner.com/phptriangl...ion_side_a.php

Hth,
Merjet

--

Dave Peterson

And sometimes too little information....

The angle provided is not the angle of a corner. It is the angle of the line
(or direction the line is pointing in with 0 degrees being at horizontally
pointing up, 90 degrees pointing vertically to the right etc.)

Regards
C

"Dave Peterson" wrote in message
Your data doesn't describe a triangle.

If you had:

Angle Length
Side1 20 35
Side2 30
Side3

Then the triangle would be:

Angle Length
Side1 20 35
Side2 17 30
Side3 143 61.65

That would contain the formula I would need, as I can calculate Angle by
deducting Angle1 from Angle2

Would be much obliged if you could share the formula used to calculate the
blanks in this example you showed.

Hi Cathy,

If this is a follow up to the post l helped you with earlier in the
week then you need to change these 2 lines:

TiltValue = Range("A1").Value / 6
Theta = (TiltValue / 60 ) * TwoPI

to:

TiltValue = Range("A1").Value
Theta = (TiltValue * (PI/180))

or:

TiltValue = Range("A1").Value / 6
Theta = Radians(TiltValue)

If you look up COS and RADIANS in Excel help you will get explanations
& examples. It saved me having to get out all those geometry books!

Regards

Michael

Thanks again for that Michael

It is not quite a follow on from the previous mail but forms part of the
same spreadsheet.

The previous mail was to enable a visual illustration of what was happening.
I did get your correction on changing the lines to show degrees and this
works a treat. I thank you dearly for this.

This part merely wants to calculate the values using formulaand display the
values in cell.

I will have to look at this over the weekend as I have to get out now before
the weather turns.

Have a nice day
C


"michael.beckinsale" wrote in message
...

TiltValue = Range("A1").Value / 6
Theta = (TiltValue / 60 ) * TwoPI

to:

TiltValue = Range("A1").Value
Theta = (TiltValue * (PI/180))

or:

TiltValue = Range("A1").Value / 6
Theta = Radians(TiltValue)

If you look up COS and RADIANS in Excel help you will get explanations
& examples. It saved me having to get out all those geometry books!

Sorry if I am confusing things by not explaining 100%

This is very usefull to me as my head is hurting going through Trig websites
trying to remember everything again.

Okay let me try again from scratch

# A B C
1 Side Direction Length
2 Line1 20 35
3 Line2 45 30
4 Line3 ? ?

Three lines form a triangle. Lets call it Triangle ABC with these letters on
each of the corners
(it is not a right Triangle)
Line1 = a = AB
Line2 = b = BC
Line3 = c = CA

Line1 starts at position 0.0
Line1 is angled at 20 degrees from 0 (0 being horizontal)
Line1 is 35 meters long

Line2 starts at position 0.0 also
Line2 is angled at 45 degrees from 0
Line2 is 30 meters long

The difference in angle between Line1 and Line2 is therefore 25 degrees

Line3 completes the Triangle
Need to calculate angle from 0
Need to calculate length

Therefore using Daves Example the angle of side1 would be 25

If you had:
# A B C
1 Angle Length
2 Side1 25 35
3 Side2 ? 30
4 Side3 ? ?

And therefore I have to calculate three values
If formulas for these three can be provided, then I will be very happy.

Hope this makes a little more sense now. (Wish I could draw it as I see it
in text)

TIA
C

Line1 starts at position 0.0
Line2 starts at position 0.0

Need to calculate angle from 0

Sure sounds like Vectors:

Length: 14.891395073096243

Angle (Depending on which direction):
141.6356275088955
or: -38.364372491104476

--
Dana DeLouis


"Cathy" wrote in message
...
Sorry if I am confusing things by not explaining 100%

This is very usefull to me as my head is hurting going through Trig
websites trying to remember everything again.

Okay let me try again from scratch

# A B C
1 Side Direction Length
2 Line1 20 35
3 Line2 45 30
4 Line3 ? ?

Three lines form a triangle. Lets call it Triangle ABC with these letters
on each of the corners
(it is not a right Triangle)
Line1 = a = AB
Line2 = b = BC
Line3 = c = CA

Line1 starts at position 0.0
Line1 is angled at 20 degrees from 0 (0 being horizontal)
Line1 is 35 meters long

Line2 starts at position 0.0 also
Line2 is angled at 45 degrees from 0
Line2 is 30 meters long

The difference in angle between Line1 and Line2 is therefore 25 degrees

Line3 completes the Triangle
Need to calculate angle from 0
Need to calculate length

Therefore using Daves Example the angle of side1 would be 25

If you had:
# A B C
1 Angle Length
2 Side1 25 35
3 Side2 ? 30
4 Side3 ? ?

And therefore I have to calculate three values
If formulas for these three can be provided, then I will be very happy.

Hope this makes a little more sense now. (Wish I could draw it as I see it
in text)

TIA
C

OK I follow now.

The internal angle between lines a & b is 45-20=25. This angle is opposite c
and is thus Angle-C

Referring to the formula I posted previously

c = SQRT(a^2 + b^2 - 2*a*b*COS((180-A-B)*PI()/180))

change (180-A-B) to 25 and include the known side lengths a & b

= SQRT(35^2+30^2-2*35*30*COS(25*PI()/180))

= 14.8913950730962

Somehow Dana managed to obtain an extra 2dp of precision beyond the
capabilities of my Excel <g

Regards,
Peter T

"Cathy" wrote in message
...
Sorry if I am confusing things by not explaining 100%

This is very usefull to me as my head is hurting going through Trig
websites
trying to remember everything again.

Okay let me try again from scratch

# A B C
1 Side Direction Length
2 Line1 20 35
3 Line2 45 30
4 Line3 ? ?

Three lines form a triangle. Lets call it Triangle ABC with these letters
on
each of the corners
(it is not a right Triangle)
Line1 = a = AB
Line2 = b = BC
Line3 = c = CA

Line1 starts at position 0.0
Line1 is angled at 20 degrees from 0 (0 being horizontal)
Line1 is 35 meters long

Line2 starts at position 0.0 also
Line2 is angled at 45 degrees from 0
Line2 is 30 meters long

The difference in angle between Line1 and Line2 is therefore 25 degrees

Line3 completes the Triangle
Need to calculate angle from 0
Need to calculate length

Therefore using Daves Example the angle of side1 would be 25

If you had:
# A B C
1 Angle Length
2 Side1 25 35
3 Side2 ? 30
4 Side3 ? ?

And therefore I have to calculate three values
If formulas for these three can be provided, then I will be very happy.

Hope this makes a little more sense now. (Wish I could draw it as I see it
in text)

TIA
C

I forgot about the 'other' angle. First the formula to calculate an internal
angle from three known side lengths -

A=ACOS((b^2+c^2-a^2)/(2*b*c))*180/PI()

or as you want internal Angle-B
B=ACOS((a^2+c^2-a^2)/(2*a*c))*180/PI()

=ACOS((35^2+14.8913950730962^2-30^2)/(2*35*14.8913950730962))*180/PI()

= 58.36437249

Deduct your original 20, elevation of lineA from the horizontal
=38.3643724911045

or depending on direction
= 141.635627508895

Regards,
Peter T



"Peter T" <peter_t@discussions wrote in message
...
OK I follow now.

The internal angle between lines a & b is 45-20=25. This angle is opposite
c
and is thus Angle-C

Referring to the formula I posted previously

c = SQRT(a^2 + b^2 - 2*a*b*COS((180-A-B)*PI()/180))

change (180-A-B) to 25 and include the known side lengths a & b

= SQRT(35^2+30^2-2*35*30*COS(25*PI()/180))

= 14.8913950730962

Somehow Dana managed to obtain an extra 2dp of precision beyond the
capabilities of my Excel <g

Regards,
Peter T

"Cathy" wrote in message
...
Sorry if I am confusing things by not explaining 100%

This is very usefull to me as my head is hurting going through Trig
websites
trying to remember everything again.

Okay let me try again from scratch

# A B C
1 Side Direction Length
2 Line1 20 35
3 Line2 45 30
4 Line3 ? ?

Three lines form a triangle. Lets call it Triangle ABC with these
letters
on
each of the corners
(it is not a right Triangle)
Line1 = a = AB
Line2 = b = BC
Line3 = c = CA

Line1 starts at position 0.0
Line1 is angled at 20 degrees from 0 (0 being horizontal)
Line1 is 35 meters long

Line2 starts at position 0.0 also
Line2 is angled at 45 degrees from 0
Line2 is 30 meters long

The difference in angle between Line1 and Line2 is therefore 25 degrees

Line3 completes the Triangle
Need to calculate angle from 0
Need to calculate length

Therefore using Daves Example the angle of side1 would be 25

If you had:
# A B C
1 Angle Length
2 Side1 25 35
3 Side2 ? 30
4 Side3 ? ?

And therefore I have to calculate three values
If formulas for these three can be provided, then I will be very happy.

Hope this makes a little more sense now. (Wish I could draw it as I see
it
in text)

TIA
C

I thank you both for your all this

Your formula is perfect. I just had difficulty relating your references to
A,a,B,b,C and c.

Again I realise this is probably due to a mistake in my discription. All
this trig stuff is all slowly coming back to me.

What I should have said was
Line1 = b = AC = 20° from 0° = 35 meters
Line2 = c = AB = 45° from 0° = 30 meters
Line3 = a = CB = ?° from 0° = ? Meters
(I simply forgot that side "a" would be opposite to corner "A" :)

Going by the above then
A = BAC = 45° - 20° = 25°

A=ACOS((b^2+c^2-a^2)/(2*b*c))*180/PI()
I already have A. It is 25°

or as you want internal Angle-B
B=ACOS((a^2+c^2-a^2)/(2*a*c))*180/PI()
?Should this not read:
B=ACOS((a^2+c^2-b^2)/(2*a*c))*180/PI()

Is the following therefore correct?
A°=BAC=25°
B°=ABC=ACOS((a^2+c^2-b^2)/(2*a*c))*180/PI() = 96.63893451
C°=ACB=ACOS((b^2+a^2-c^2)/(2*a*b))*180/PI() = 58.36374786

And modelling by previous spreadsheet
# A B C
1 Side Direction Length
2 Line1 20 35
3 Line2 45 30
4 Line3 ? ?

The formula for B4 is:
=180-(ACOS((C4^2+C2^2-C3^2)/(2*C4*C2))*180/PI())+B2
= 141.6362521°

and Formula for C4 is:
=ROUND(SQRT(C2^2+C3^2-2*C2*C3*COS((B3-B2)*PI()/180)),2)
=14.89139507 meters

I have just realised another mistake in my original example, i.e. the angle
of c should have been in the opposite direction. I should be able to work
that out myself and correct what I am trying to achieve.

Thank you so much for all your help.

Kind regards
C

I have just realized another mistake in my original example, i.e. the
angle of c should have been in the opposite direction.

I think you want to describe vector bc in both its Magnitude, and Direction.
You have to keep track of both the internal angles, and a reference angle to
make corrections to get the correct answer.
I was just suggesting an alternate method to avoid keeping track of all
that.
Since this is in a programming group, I was just suggesting subtraction like
this:
v1 = Vector(35, 20)
v2 = Vector(-30, 45)
'...etc

Don't forget an angle below the horizon is negative.

=DEGREES(ATAN((3*SQRT(2)-7*SIN(RADIANS(20)))/(3*SQRT(2)-7*COS(RADIANS(20)))))

-38.3643724911045

(Add 180 to get opposite "direction" of 141.635627508896)


On technique I like to use is to define a Named Constant in Excel to help
convert Degrees to Radians.
For Example:
Deg = Pi()/180

Hence, an alternative for Length might be:

=5*SQRT(85 - 42*SQRT(2)*COS(20*Deg) - 42*SQRT(2)*SIN(20*Deg))

14.8913950730962

Maybe something of interest
http://en.wikipedia.org/wiki/Vector_%28spatial%29

--
HTH
Dana DeLouis



"Cathy" wrote in message
...
I thank you both for your all this

Your formula is perfect. I just had difficulty relating your references to
A,a,B,b,C and c.

Again I realise this is probably due to a mistake in my discription. All
this trig stuff is all slowly coming back to me.

What I should have said was
Line1 = b = AC = 20° from 0° = 35 meters
Line2 = c = AB = 45° from 0° = 30 meters
Line3 = a = CB = ?° from 0° = ? Meters
(I simply forgot that side "a" would be opposite to corner "A" :)

Going by the above then
A = BAC = 45° - 20° = 25°

A=ACOS((b^2+c^2-a^2)/(2*b*c))*180/PI()
I already have A. It is 25°

or as you want internal Angle-B
B=ACOS((a^2+c^2-a^2)/(2*a*c))*180/PI()
?Should this not read:
B=ACOS((a^2+c^2-b^2)/(2*a*c))*180/PI()

Is the following therefore correct?
A°=BAC=25°
B°=ABC=ACOS((a^2+c^2-b^2)/(2*a*c))*180/PI() = 96.63893451
C°=ACB=ACOS((b^2+a^2-c^2)/(2*a*b))*180/PI() = 58.36374786

And modelling by previous spreadsheet
# A B C
1 Side Direction Length
2 Line1 20 35
3 Line2 45 30
4 Line3 ? ?

The formula for B4 is:
=180-(ACOS((C4^2+C2^2-C3^2)/(2*C4*C2))*180/PI())+B2
= 141.6362521°

and Formula for C4 is:
=ROUND(SQRT(C2^2+C3^2-2*C2*C3*COS((B3-B2)*PI()/180)),2)
=14.89139507 meters

I have just realised another mistake in my original example, i.e. the
angle of c should have been in the opposite direction. I should be able to
work that out myself and correct what I am trying to achieve.

Thank you so much for all your help.

Kind regards
C

angle of c should have been in the opposite direction.

As a side note, you gave vectors

a b (45 deg @ 30m)
a c (20 deg @ 35m)

You are asking for b c (sloping down to right)

Hence:
a b + b c = a c
or
b c = a c - a b

That is why I suggested Subtraction.

--
Dana DeLouis

<snip

If interested, since this is a programming group:

Sub Demo()
Dim x, y, Deg
Deg = [PI()/180] 'Technique only, or use Radians()
x = 35 * Cos(20 * Deg) - 30 * Cos(45 * Deg)
y = 35 * Sin(20 * Deg) - 30 * Sin(45 * Deg)
With WorksheetFunction
Debug.Print " Side: " & Sqr(.SumX2PY2(x, y))
Debug.Print "Angle: " & .Degrees(.Atan2(x, y))
Debug.Print "Or:"
Debug.Print "Angle: " & .Degrees(.Atan2(x, y)) + 180
End With
End Sub

Returns:

Side: 14.8913950730962
Angle: -38.3643724911045
Or:
Angle: 141.635627508896

--
HTH :)
Dana DeLouis


"Dana DeLouis" wrote in message
...
angle of c should have been in the opposite direction.

As a side note, you gave vectors

a b (45 deg @ 30m)
a c (20 deg @ 35m)

You are asking for b c (sloping down to right)

Hence:
a b + b c = a c
or b c = a c - a b

That is why I suggested Subtraction.

--
Dana DeLouis

<snip

Looks like you are well on the right track. I haven't looked closely at your
slightly modified formulas but all your results appear same as Dana's &
mine, so I assume all OK.

Regards,
Peter T


"Cathy" wrote in message
...
I thank you both for your all this

Your formula is perfect. I just had difficulty relating your references to
A,a,B,b,C and c.

Again I realise this is probably due to a mistake in my discription. All
this trig stuff is all slowly coming back to me.

What I should have said was
Line1 = b = AC = 20° from 0° = 35 meters
Line2 = c = AB = 45° from 0° = 30 meters
Line3 = a = CB = ?° from 0° = ? Meters
(I simply forgot that side "a" would be opposite to corner "A" :)

Going by the above then
A = BAC = 45° - 20° = 25°

A=ACOS((b^2+c^2-a^2)/(2*b*c))*180/PI()
I already have A. It is 25°

or as you want internal Angle-B
B=ACOS((a^2+c^2-a^2)/(2*a*c))*180/PI()
?Should this not read:
B=ACOS((a^2+c^2-b^2)/(2*a*c))*180/PI()

Is the following therefore correct?
A°=BAC=25°
B°=ABC=ACOS((a^2+c^2-b^2)/(2*a*c))*180/PI() = 96.63893451
C°=ACB=ACOS((b^2+a^2-c^2)/(2*a*b))*180/PI() = 58.36374786

And modelling by previous spreadsheet
# A B C
1 Side Direction Length
2 Line1 20 35
3 Line2 45 30
4 Line3 ? ?

The formula for B4 is:
=180-(ACOS((C4^2+C2^2-C3^2)/(2*C4*C2))*180/PI())+B2
= 141.6362521°

and Formula for C4 is:
=ROUND(SQRT(C2^2+C3^2-2*C2*C3*COS((B3-B2)*PI()/180)),2)
=14.89139507 meters

I have just realised another mistake in my original example, i.e. the
angle
of c should have been in the opposite direction. I should be able to work
that out myself and correct what I am trying to achieve.

Thank you so much for all your help.

Kind regards
C

Thanks for this Dana

At the moment I am just trying to get the math working, however there is
still more I will have to do further down and then this Programming code you
provided will be very usefull.

Regards
C

Hi Cathy
Sent you a mail last week but maybe it didn't get to you. I have a
spreadsheet that calculates lengths and angles in triangles. Separate
sheets for right angles triangles , problems using Cosine Rule,
problems using Sine Rule and problems using dot product. Also contains
various wrong answers students might get if they do the calculations
wrongly.
Send me an email if you would like a copy.
regards
Paul

On Mar 2, 9:09*pm, "Cathy" wrote:
Thanks for this Dana

At the moment I am just trying to get the math working, however there is
still more I will have to do further down and then this Programming code you
provided will be very usefull.

Regards
C

I have done a spreadsheet to calculate if have you have 2 lengths and one
degree also to work out degrees of a triangle if you have the 3 lengths
ALLAN


I have a spreadhset with the following representing two sides of a Triangle

Angle Length
Side1, 20 35
Side2, 45 30
Side3,


I need a formula to calculate the Angle (degrees) and Length for the third
Angle

TIA
C