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Find Angle Given 3 Sides, NOT Using Law of Cosines

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Date: 12/18/2001 at 14:39:19
From: Diane Pope
Subject: Find angle of triangle given 3 sides and NOT using law of
cosines

I teach a Math Analysis course where we cover trigonometry. A student
3 side lengths and asked them to find an angle. He said he originally
couldn't remember the formula so he added all three sides, took 180
degrees divided by the perimeter to determine the degree per unit,
and then multiplied that result by the length of the side opposite
the angle I asked for. Later he remembered the formula and when he
reworked the problem the answer differed from his first by over
10 degrees. He asked me why it wouldn't work to use proportions.

I can not really come up with a good reason why - his process seemed
to make sense to me, but I know it can't be right or we wouldn't need
the law of cosines to find an angle. What is the fault in his logic?
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Date: 12/18/2001 at 15:32:56
From: Doctor Schwa
Subject: Re: Find angle of triangle given 3 sides and NOT using law of
cosines

The law of sines tells us that the length of each side is proportional
to the SINE of the angle opposite it, and your student assumed that it
was proportional to just the angle.

In other words, an angle twice as big makes the side opposite it less
than twice as big. Your student assumed it would be twice as big.

Another, maybe more intuitive way to say the same thing is: if you
make the angle twice as big, the arc of a circle would get twice as
long. But the length of the side of a triangle, which is like the
chord of the circle, would get less than twice as long. The two arcs
add up end to end but the two chords make a bent line, and the new
(straight) chord is shorter.

I hope that helps clear things up!

- Doctor Schwa, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 12/18/2001 at 15:39:04
From: Doctor Rob
Subject: Re: Find angle of triangle given 3 sides and NOT using law of
cosines

Thanks for writing to Ask Dr. Math, Diane.

The problem is that some degrees in angle A cut off larger portions of
side a than others do. Thus the side lengths are not proportional to
the angles. The side lengths are proportional to the SINES OF THE
ANGLES, and not the ANGLES THEMSELVES. If they were, the formula
corresponding to the Law of Sines would be

a/A = b/B = c/C = (a+b+c)/(A+B+C) = P/180,

and not

a/sin(A) = b/sin(B) = c/sin(C) = 2*R.

The former is what the student used. The latter is correct.

Funny thing, when I saw the subject of your message, I was sure you
were going to ask how to calculate the angles from the side lengths
without using the Law of Cosines. This is possible, but little known.
The trick is to compute the area two ways. First

K = sqrt(s*[s-a]*[s-b]*[s-c]),  where  s = (a+b+c)/2,

(Heron's Formula), and then the area also is given by

K = (1/2)*a*b*sin(C),
sin(C) = 2*K/(a*b).

Similarly

sin(A) = 2*K/(b*c),
sin(B) = 2*K/(c*a).

You may want to add this fact to your store of knowledge.

Feel free to write again if I can help further.

- Doctor Rob, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 12/18/2001 at 15:53:41
From: Doctor Peterson
Subject: Re: Find angle of triangle given 3 sides and NOT using law of
cosines

Hi, Diane. I see you already got two answers, but a third may give a
slightly different perspective.

There's no real logic (in the sense of something provable) in his
reasoning. The foundation of what he did doesn't even exist: it's
meaningless to talk about the "degrees per unit" in a triangle,
because the angles are not proportional to the lengths of the opposite
sides. You can only work with ratios where some sort of proportion
exists.

What he's really finding is something like this. Make a circle whose
circumference is the sum of the three lengths a, b, and c:

*********** c
*****           *****
****                     ****
A  **                             **
+----                             *
** \   ----------                    **
*    \            ---------             *
*      \                    ----------    *
*       \                             ----+ B
*         \                              /  *
*          \          +                /    *
*           \          O             /      *
*           \                     /       *
*            \                  /         *
b *            \                /         *
**           \             /         **
*           \          /          *
**          \       /          **  a
****       \    /        ****
*****   \ /     *****
****+******
C

The central angles subtended by the arcs will be proportional to the
arc lengths, so dividing the 360 degrees by the circumference (the sum
of the lengths) gives the central angle per unit along the
circumference. The central angle subtended by each arc is therefore
the product of this ratio and the length of the arc. The inscribed
angle subtending any arc is half the central angle, so angles A, B,
and C are the angles he calculated:

A = 180a / (a+b+c)
B = 180b / (a+b+c)
C = 180c / (a+b+c)

But this has nothing to do with the triangle with _sides_ a, b, c,
except to the extent that our triangle ABC whose _arcs_ have these
lengths approximates a similar triangle - which, as you can see, is
not a close approximation in general.

So, why doesn't it work to use proportions? Because the sides are not
proportional to the angles. (They are, of course, proportional to the
_sines_ of the angles; but you don't know the sum of the sines as you
do the sum of the angles, so that doesn't help.)

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Trigonometry

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