Clarification/Follow-up by Dathaeus on 03/03/06 4:50 am:
And one more thing, I think i was jus tired or had a brain freeze on this one... in a circle with two line coming out, where they both intersect on the circle and then spread out and exit the circle, is the length of that angle created by the two lines equal to 1/2 of the arc length it cuts as it exits? I cant remember a formula for that, but I just made myself an example to make the lines exit out where the diameter of the circle would intersect both of them, creating an arc length of 180 degrees, but I am pretty sure the angle created by the 2 lines is 90 degrees. IO just need to know if that is a valid general theorem.

Also, a more complicated question... in the topic of the ambiguous case of solving triangles given 2 sides and an angle using the law of sines.... after u find one of two possible angles (B1), how do you start solving for the second angle (B2)? In one case where B1 was acute and B2 was obtuse, the snswer was as simple as B1 = 180 - B2. But that did not work when B1 AND B2 were acute... so what is the general correct way to handle the second step?

Thanks! I'll check out your masterpiece page now..... :-)

Clarification/Follow-up by IQGuru on 03/07/06 7:20 am:
Re your first followup, yes, an inscribed angle (which is what you described) is measured by one-half of its intercepted arc. This is sometimes labeled Theorem 12 of Geometry.

Re your second followup, your premise is invalid. In any triangle, 180=B1+B2+B3. Thus B1=180-B2-B3. Thus it is not possible that B1=180-B2. Instead, for a right triangle only, B1=90-B2, since B3=90.

All non-right triangles have two acute angles. The third angle may be either acute or obtuse. Given 2 sides and an angle, you need to apply the cosine law (not sine law) to solve the remaining angles.

"IQ"

Clarification/Follow-up by Dathaeus on 03/07/06 9:21 am:
Actually, i thisnk i mis-stated the question... i did mean trying to find the second and third angles, in an anbiguous case, where there are 2 possible answers for the problem, right? so my uncertainty comes where

1) How do i know which of the two possible triangles i am solving for and getting values for on the first law of sines solution?

2) After finding the dimensions of the first triangle, do u recommend my drawing a revised and better scaled triangle so i can better project what the second possible solution would look like? Because it is near impossible to draw an accurate to scale triangle before you get a solution...

3) And after that step, is that when I use the law of cosines to find the second possible solution?

Thanks!

Clarification/Follow-up by IQGuru on 03/07/06 6:43 pm:
Please pardon my dullness this week but it took me a long time to figure out what your question was really asking. Sorry about that -- I should have understood sooner...

(1) Given SAS always completely defines a specific triangle.
(2) Given SSA where A is obtuse also always completely defines a specific triangle.
(3) Given SSA where A is 90 degrees also completely defines a specific triangle.
(4) But given SSA where A is acute always allows the construction of two different triangles, and it is this unique case that you are asking about...

However, the answer to your question (as aksed) is that there is no mathematical answer; because no unique triangle can be defined given SSA where A is acute. (Side A and angle a opposite it will be the same in both triangles.) So, instead, you must either DECIDE which triangle to solve, or DECIDE to solve both of them!

Let us call the Sides A,B,C and the respective oppoite angles a,b,c. where you are given A,B,a... So:
If b is acute then C will be longer than B, and c may be obtuse, right, or acute.
If b is obtuse then C will be shorter than B, and c MUST be acute.

So, before you even start, you should sketch all of the above, and then DECIDE which triangle(s) you want to solve.

In either case, having ALREADY DECIDED whether b is to be acute or obtuse, proceed as follows:

(1) Use the sine law to define angle b:
sin(b)=Bsin(a)/A.
B=ArcSin[Bsin(a)/A].

(2) Use sum of angles to define angle c:
c = 180 -a -b.

(3) Use Sine Law again to determine length of side C:
C=Asin(c)/sin(a)
or
C=Bsin(c)/sin(b).

[Alternately, use the Cosine Law (or Pythagorus if a right triangle) to determine any remaining unknowns.]

Well, hope that helps?
Warm regards,
"IQ"