I've got a triangle, and I know two sides and an angle... I need to find the other 2 angles and the other side... I feel stupid for asking, but I'm a bit out of it, and haven't taken a math class in a while... A year ago I did this stuff in my sleep, but now if it doesn't have a right angle, I'm lost (please wait until I'm gone to call me stupid! ) EDIT:: I'm going for any angles/sides as long as they work, I know there is more than one solution EDIT2:: We're trying to disprove "Side Side Angle" I know Side Angle Side proves triangles are congruent, but Side Side Angle doesn't. (hope that helps and doesn't just confuse things!)
c/sin(c)= b/sin(b) 5/sin(93)=3/sin(b) sin(b)*( 5/sin(93))=3 sin(b)=3sin(93)/5 sin(b)=0.5991 b=36.8 degrees 180-a-b-c=0 180-b-c=a 180-93-36.8=a 50.2=a A/sin(a)=B/sin(b) A=Bsin(a)/sin(b) A=3sin(50.2)/sin(36.8) A=3(0.7683)/0.5991 A=3.85 sorry, i'm bored
unfortunately those are the numbers I started with (sorry, I should have mentioned that!) There is a second set of numbers that should work, but I figured it out... I used AutoCAD, zoomed in & eyeballed it, and turned down precision... the math works out close enough to get by tomorrow!
Oh boy, this is going to be a LONG post... Going from Vadim's post: a = 50.2 A = 3.85 b = 36.8 Let's start with angle b. We know that sine is positive in the first and second quadrant, so when we solve for be in this equation we can come up with a second value: B/sin(b) = C/sin(c) 3/sin(b) = 5/sin(93) b = 36.8 or 180 - 38.8 = 143.2 We can prove this works: 3/sin(143.2) = 5/sin(93) 5 = 5 so b = 36.8 && 143.2 From that we know that a = 50.2 && -56.2 So side from there we find A using the second angle: A/sin(-56.2) = 3/sin(143.2) A = -4.2 so A = 3.85 && -4.2 I'm sure something is wrong in that but those are all technically right.
