asked on # Rat's New Year Statics Question

Here is a new Maths question for 2013.

ABC is a triangle of forces.

Prove that the magnitude of the resultant vector which consists of the sum of AB, 2*BC and 3*CA is the square root of b²+c²+2bc.cos(A) and that its direction is that of the diagonal through A of the parallelogram of which AB abd AC are adjacent sides.

Notes:

1) The vectors AB,BC, and CA should have either bars or arrows above the letters to denote their vectorial properties.

2) a,b,c represent the triangles sides opposite the angles at A,B and C as per convention.

And a happy new year to all!

ABC is a triangle of forces.

Prove that the magnitude of the resultant vector which consists of the sum of AB, 2*BC and 3*CA is the square root of b²+c²+2bc.cos(A) and that its direction is that of the diagonal through A of the parallelogram of which AB abd AC are adjacent sides.

Notes:

1) The vectors AB,BC, and CA should have either bars or arrows above the letters to denote their vectorial properties.

2) a,b,c represent the triangles sides opposite the angles at A,B and C as per convention.

And a happy new year to all!

Math / Science

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A+AB+AC+AB+2BC+3CA = A+2AB+2BC+AC+3CA = A+2AB+2BC+2CA = A+2(AB+BC+CA) = A

for proof of the law of cosines, see http://en.wikipedia.org/wiki/Law_of_cosines#Proofs

ozo:

Firstly your statement "The length of the diagonal is b²+c²+2bc.cos(A)" is at odds with the equation in the Wiki article under the heading "Some schools also describe the notation as follows".

Secondly in "and extending it once in the AB direction and twice in the AC direction" why extend AB at all, when in fact it is BC which occurs twice?

Firstly your statement "The length of the diagonal is b²+c²+2bc.cos(A)" is at odds with the equation in the Wiki article under the heading "Some schools also describe the notation as follows".

Secondly in "and extending it once in the AB direction and twice in the AC direction" why extend AB at all, when in fact it is BC which occurs twice?

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James Murphy

I meant to say the square root of b²+c²+2bc.cos(A)

Which becomes the cited notation of some schools:

c^2 = a^2 + b^2 - 2ab cos C,

when we relabel the diagonal in question as "c"

b in your diagram as "a"

c in your diagram as "b"

and the angle in the B (or C) corner of the parallelogram as "A"

and note that cos(B+C) = -cos(A)

I extended my diagram in the way that I did so that when I did the graphical vector addition starting at that corner, the resultant conveniently fell along the diagonal in question

Which becomes the cited notation of some schools:

c^2 = a^2 + b^2 - 2ab cos C,

when we relabel the diagonal in question as "c"

b in your diagram as "a"

c in your diagram as "b"

and the angle in the B (or C) corner of the parallelogram as "A"

and note that cos(B+C) = -cos(A)

I extended my diagram in the way that I did so that when I did the graphical vector addition starting at that corner, the resultant conveniently fell along the diagonal in question

>>and you will find yourself back at A, the other end of the diagonal.

If the triangle is labeled ABC and we start at A, then AB brings us to B. One of the two BCs take us to C. The next BC takes us under the AC (considered as a horizontal line) and then from there we have THREE CA vectors (now considered as horizontal lines in the direction C to A) which certainly does not bring us back to A, in which case the cosine rule for the length of the diagonal does not apply. The answer to prove does not relate diredctly to the cosine rule since there is a plus sign and not a minus sign.

You are also not using an important part of the question, if I may be allowed to say so.

If the triangle is labeled ABC and we start at A, then AB brings us to B. One of the two BCs take us to C. The next BC takes us under the AC (considered as a horizontal line) and then from there we have THREE CA vectors (now considered as horizontal lines in the direction C to A) which certainly does not bring us back to A, in which case the cosine rule for the length of the diagonal does not apply. The answer to prove does not relate diredctly to the cosine rule since there is a plus sign and not a minus sign.

You are also not using an important part of the question, if I may be allowed to say so.

we start at AI started at the corner opposite A, at the other end of the diagonal of the parallelogram, in order to end up at A

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A couple of minor points.

1) the phrase "its direction is that of the diagonal" in the question may be a bit misleading to those whose first langauge is not English. It does not mean "ON the diagonal", but "parallel to the diagonal".

2) The cosine law a²=b+c²-2bc.cos(A) is symetric in the sides and the angles, so that an angle like B+C does not occur. Relabelling will not case the sign to change.

1) the phrase "its direction is that of the diagonal" in the question may be a bit misleading to those whose first langauge is not English. It does not mean "ON the diagonal", but "parallel to the diagonal".

2) The cosine law a²=b+c²-2bc.cos(A) is symetric in the sides and the angles, so that an angle like B+C does not occur. Relabelling will not case the sign to change.

there is a plus sign and not a minus signCosines of supplementary angles have opposite sign

"parallel to the diagonal".The translational position of a vector in a force diagram is irrelevant (unless you're dealing with torques)

A convenient way to prove that vectors in a diagram are parallel is to show that they coincide,

so I constructed my diagram by selecting starting and ending points in what to me was the most convenient way.

Your help has saved me hundreds of hours of internet surfing.

fblack61

>>in order to end up at A

I ask myself why, but in any case the result being that of the length of AD is not the answer. In fact, according to the Wiki page, what would be the result of adding the vectors AB and AC? and how would that square with AB+2*BC+3*CA?

I ask myself why, but in any case the result being that of the length of AD is not the answer. In fact, according to the Wiki page, what would be the result of adding the vectors AB and AC? and how would that square with AB+2*BC+3*CA?

Cosines of supplementary angles have opposite sign.

And adjacent angles of a parallelogram are supplementary.

In a triangle ABC, B+C (which form the angles of the parallelogram adjacent to A) is supplementary to A

And adjacent angles of a parallelogram are supplementary.

In a triangle ABC, B+C (which form the angles of the parallelogram adjacent to A) is supplementary to A

I know all of that and it is not relevant to the question. The magnitude of the result is not the length of the diagonal AD. The length of the diagonal AD is b²+c²-2bc.cos(A) or in terms of the other angles b²+c²-2bc.cos(180-(B+C)) = b²+c²+2bc.cos(B+C).

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length of ADI take it that D is now our label for the corner of the parallelogram opposite A,

which is at A+AB+AC

from

http:#a38733796

A+AB+AC + AB+2*BC+3*CA = A so

AB+2*BC+3*CA = -(AB+AC)

AB+AC + AB+2*BC+3*CA going around the triangle twice

The length |AB+AC| = |AD| is, by the law of cosines, the square root of b²+c²+2bc.cos(A)

The [square of the] length of the diagonal AD is b²+c²-2bc.cos(A)Are we using different D?

I think that's the length of the diagonal BC. (=BA+AC)

Maybe we have a notational confusion.

>>The length |AB+AC| = |AD| is, by the law of cosines, the square root of

a²=b²+c²+2bc.cos(A)

You quoted the Law of Cosines yourself and I pointed out to you that the law is b²+c²-2bc.cos(A). Swapping a,b,c around does not make any difference. The minus sign is always there, the cosine of the angle might make it positive, and since the triangle is arbitary, so is the angle. In fact the article shows via Euclid why the minus sign is relevant.

Since the question asks you to prove a formular with a plus sign, offering an argument that leads to a minus sign is obviously not the correct answer. I can only tell you precisely why if I post the solution.

a²=b²+c²+2bc.cos(A)

You quoted the Law of Cosines yourself and I pointed out to you that the law is b²+c²-2bc.cos(A). Swapping a,b,c around does not make any difference. The minus sign is always there, the cosine of the angle might make it positive, and since the triangle is arbitary, so is the angle. In fact the article shows via Euclid why the minus sign is relevant.

Since the question asks you to prove a formular with a plus sign, offering an argument that leads to a minus sign is obviously not the correct answer. I can only tell you precisely why if I post the solution.

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William Peck

+2bc.cos(A) = -2bc.cos(pi-A) = -2bc.cos( B+C )

B+C is the angle opposite diagonal through A, in the triangle formed by bisecting the parallelogram along the diagonal through A

That puts it in the form of equation in the Wiki article under the heading "Some schools also describe the notation as follows".

B+C is the angle opposite diagonal through A, in the triangle formed by bisecting the parallelogram along the diagonal through A

That puts it in the form of equation in the Wiki article under the heading "Some schools also describe the notation as follows".

Substituting your A = 180-(B+C) in the cosine rule FROM WIKIPEDIA gives b²+c²+2bc.cos(B+C) which still does not correspond to the solution.

You are still making a fundamental error in ignoring information given in the question.

You are still making a fundamental error in ignoring information given in the question.

in the cosine rule FROM WIKIPEDIA *c^2 = a^2 + b^2 - 2ab cos C*,

their*C* corresponds to our B+C

their*c* corresponds to our diagonal through A,

their*a* and *b* correspond to our b and c

their

their

their

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>>their C corresponds to our B+C

No it doesn't. Their C corresponds to the angle at C, not the angle at A. The angle at A may be represented as 180-(B+C) whose cosine may be equal to B+C. But the equation reads the squares of the two sides with the cosine of the INCLUDED angle, which with sides called b and c (opposite angles B and C) is called A. Note 2 in the question made that totally clear and indeed corresponds to the usage in Wikipedia.

No it doesn't. Their C corresponds to the angle at C, not the angle at A. The angle at A may be represented as 180-(B+C) whose cosine may be equal to B+C. But the equation reads the squares of the two sides with the cosine of the INCLUDED angle, which with sides called b and c (opposite angles B and C) is called A. Note 2 in the question made that totally clear and indeed corresponds to the usage in Wikipedia.

Their angle at *C* in their triangle is not the same as our angle at C in our triangle,

Their*C* in their triangle would correspond either to our B, if we take that half of the parallelogram as our law of cosines triangle, or to our C, if we take that half of the parallelogram as our law of cosines triangle, the angle at which is supplementary to our angle A.

We want to the length of the diagonal through A [and D] of our parallelogram, not the diagonal through B and C, which bisects that diagonal. (and for which we would apply a minus sign to cosA)

The included angle we are interested in is the one at ABD or ACD (assuming that D is the point at the opposite end of the diagonal through A of the parallelogram of which AB and AC are adjacent sides.)

In your Note 2,

our |a| = our |BC|

our |b| = our |AC|

our |c| = our |AB|

but when we apply the cosine rule, only one of the sides adjacent to the included angle we want to use come would from our original ABC triangle, the other comes from BD or CD of our parallelogram,

which happen to be parallel and equal in length to AC and AB respectively.

I can't help it if you and Wikipedia happened to choose the same letters to talk about different parts of different diagrams.

(it seemed pretty clear when drawing the diagram, but trying to describe the diagram in English is apparently more confusing than I realized, especially when trying to relate it to a different diagram with homographic labels)

Their

We want to the length of the diagonal through A [and D] of our parallelogram, not the diagonal through B and C, which bisects that diagonal. (and for which we would apply a minus sign to cosA)

The included angle we are interested in is the one at ABD or ACD (assuming that D is the point at the opposite end of the diagonal through A of the parallelogram of which AB and AC are adjacent sides.)

In your Note 2,

our |a| = our |BC|

our |b| = our |AC|

our |c| = our |AB|

but when we apply the cosine rule, only one of the sides adjacent to the included angle we want to use come would from our original ABC triangle, the other comes from BD or CD of our parallelogram,

which happen to be parallel and equal in length to AC and AB respectively.

I can't help it if you and Wikipedia happened to choose the same letters to talk about different parts of different diagrams.

(it seemed pretty clear when drawing the diagram, but trying to describe the diagram in English is apparently more confusing than I realized, especially when trying to relate it to a different diagram with homographic labels)

an angle like B+C does not occurAn angle with magnitude B+C occurs in the parallelogram of which AB abd AC are adjacent sides.

And it is the angle we would want to use in applying the law of cosines to find the length of

of the diagonal through A of that parallelogram.

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rwheeler23

Simply draw a triangle with A to the left, C to the right and B at the top. make AC a horizontal line and the angle at A, ie: BAC acute, say approx 50 degrees. That's the starting diagram. It makes NO difference if you swap B and C.

No disagreement there.

Now draw D, to the right of B, 50 degrees above C, to form the parallelogram of which AB abd AC are adjacent sides.

To find the length of |AD|, the diagonal through A of the parallelogram,

we apply the formula

*c^2 = a^2 + b^2 - 2ab cos C*

substituting

|AD| for*c*

|BD| for*a*

|BA| for*b*

<ABD for*C*

giving

|AD|^2 = |BD|^2 + |BA|^2 - 2|BD||BA| cos(ABD)

<ABD and <BAC are supplementary

so cos(BAC) = - cos(ABD)

and

|BD| = |AC|

|BA| = |AB|

so

|AD|^2 = |AC|^2 + |AB|^2 + 2|AC||AB| cos(BAC)

in your notation,

b²+c²+2bc.cos(A)

To find the length of |AD|, the diagonal through A of the parallelogram,

we apply the formula

substituting

|AD| for

|BD| for

|BA| for

<ABD for

giving

|AD|^2 = |BD|^2 + |BA|^2 - 2|BD||BA| cos(ABD)

<ABD and <BAC are supplementary

so cos(BAC) = - cos(ABD)

and

|BD| = |AC|

|BA| = |AB|

so

|AD|^2 = |AC|^2 + |AB|^2 + 2|AC||AB| cos(BAC)

in your notation,

b²+c²+2bc.cos(A)

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The theme of this question is the "Triangle of forces". If a triangle ABC is a triangle of forces then the sum of the forces AB,BC and CA (taken round the triangle) are equal to zero.

Thus the sum of AB, 2*BC and 3*CA is equal to BC + 2*CA. One can take the triangle again and obtain the vectors CA and BA (which is -AB from BC+CA+AB=0).

However one can also solve the problem by resolving the components of the resultant vector thus:

Horizontally we have from BC b-c*cos(A) and from CA -2b which equals -(b+c*cos(A))

Vertically just c*sin(A) if we take AC as horizontal.

Thus squaring and adding we get b²+c²cos²(A)+2bccos(A)+c²-c²cos²(A) = b²+c²+2bc*cos(A) as required.

Since the tangent of the angle of the resultant vector is y/x we have -c*sin(A)/b+c*cos(A)

But c*sin(A) is the position of B and b+c*cos(A) is the distance from C to a point D and has the same length as AB and the angle is the same as A, and so forms a parallelogram ABCD and the second result follows.

Thus the sum of AB, 2*BC and 3*CA is equal to BC + 2*CA. One can take the triangle again and obtain the vectors CA and BA (which is -AB from BC+CA+AB=0).

However one can also solve the problem by resolving the components of the resultant vector thus:

Horizontally we have from BC b-c*cos(A) and from CA -2b which equals -(b+c*cos(A))

Vertically just c*sin(A) if we take AC as horizontal.

Thus squaring and adding we get b²+c²cos²(A)+2bccos(A)+c²-

Since the tangent of the angle of the resultant vector is y/x we have -c*sin(A)/b+c*cos(A)

But c*sin(A) is the position of B and b+c*cos(A) is the distance from C to a point D and has the same length as AB and the angle is the same as A, and so forms a parallelogram ABCD and the second result follows.

The solution did not use the Triangle of Forces property to solve the problem, but did it by construction, which is unusual but totally valid. My only quibble is the presentation of the solution, not mentioning that the cosine of the angle at B in the parallelogram is the negative value of that at A.

Still, full marks for originality.

Still, full marks for originality.

The Triangle of Forces property was used in http:#a38733796 to cancel 2(AB+BC+CA)

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William Peck

>>The Triangle of Forces property was used in http:#a38733796 to cancel 2(AB+BC+CA)

Should read :-

The Triangle of Forces property was used without mention in http:#a38733796 to cancel 2(AB+BC+CA)

Should read :-

The Triangle of Forces property was used without mention in http:#a38733796 to cancel 2(AB+BC+CA)