See here.
There is a problem in my book which I just cannot seem to figure out. It seems more like a mind game than any real math, but whatever. It is a short word description of a triangle inside a circle, and then two questions about it. Drawing it out will likely help.
A triangle ABC with sides a, b, and c is circumscribed by a circle with a center S, and radius R.
A) Show that 2R*sinC=c
B) Show that 2R=a/sinA=b/sinB=c/sinC
Ok, so question B obviously has me looking at the law of sines, and based on question A I can see that if 2R*sinC=c then that is the same as saying 2R=c/sinC, and therefore the law of sines answers the rest of that.
My problem is trying to figure out how you come to the conclusion with that first question. It cant be problem with the law of sines, and obviously not with the law of cosines either. So then I figured that if I cut a 90* angle up from the midpoint of side c, I could use the Pythagorean theorem on it. The hypotenuse would be R, and then the short side would be half of c, and since c is a chord, I could use 2R*sinS. I know that sin S must equal 2C, so that leaves me thinking that something like this below would boil down to be the answer. The problem is that to find out what the long leg is, I need to go:
x^{2}=R^{2}+((2R sin(2C))/2)^{2 }
I am a bit unsure how I would solve that out, and even worse, that would just give me the formula for the long leg. I would then need to plug that back into the Pythagorean theorem again, and get:
R^{2}=(2R sin(2C))^{2}+(sqrt(R^{2}+((2R sin(2C))/2)^{2})^{2}
And then my head just starts spinning. I have a feeling that I am way over complicating things, and that there is a faster, simpler route to take, but I just cant seem to see it...
Edit: Made a mistake while explaining it, and had C/2 instead of 2C.
Wow... That was... Really simple, lol. I didnt even think to just take the opposite over the hypotenuse... The last week has been focusing so much on the law of sines and law of cosines, and medians and stuff that I wasnt even thinking about the more basic trig identities...
Thank you very much.