An obtuse triangle has angle measurements of 100°, 60°, and 20°. If the longest side of this triangle is 20 feet, what is the length of its shortest side, s?

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An obtuse triangle has angle measurements of 100°, 60°, and 20°. If the longest side of this triangle is 20 feet, what is the length of its shortest side, s?

Geometry
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Have you heard about the law of sines?

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Cool because I didnt know how to go about this if you didnt lol Okay so we know \[\large \frac{sin(A)}{a} = \frac{sin(B)}{b} = \frac{sin(C)}{c}\]
So we have the side of 20 ft....and the angle that faces it is 100 degrees right? And we have the angle that faces the 's' side that we want is the 20 degrees right?
So we can set up our law of sines to be \[\large \frac{sin(100)}{20} = \frac{sin(20)}{s}\] and solve for 's'
okay i think i got it, for this problem i got 6.9 ft is that correct? @johnweldon1993
*shouldda calculated it huh* ...yeah 6.9 is correct :)
okay thanks! :-) could you help me with another one? @johnweldone1993 please
Oh god another one????? >.< haha no jk of course :P
haha thanks, okay so ABCD is a parallelogram. Its diagonal, AC, is 18 inches long and forms a 20° angle with the base of the parallelogram. Angle ABC is 130°. What is the length of the parallelogram’s base, AB?
Hint. what do all the angles in any triangle add up to?
Not quite...that is a circle :) the interior angles of a triangle ALWAYS add up to 180 degrees okay? So here...focus on the lower triangle |dw:1432857360434:dw|
So to solve for that angle labeled ?? right now We know those 3 angles will add to make 180...so if we already have 20...and 130....what would the other angle be?
OHHHHH. crap sorry for some reason i was thinking of a square. and yes so 30?? @johnweldon1993
Lol no problem :) and yes 30 good...so we have |dw:1432857553567:dw| now we just have the law of sines again :) we have the angle 30 facing that side we need and we have the angle 130 facing the 18 we have so \[\large \frac{sin(130)}{18} = \frac{sin(30)}{b}\] and solve for 'b'
right. @johnweldon1993 and we cross multiply??
mmhmm just like last time :)
so sin130(b)
So we have yes \[\large sin(130)b = 18sin(30)\] and solve for 'b'
0.7660(b)=9.270? @johnweldon1993
Solving for 'b' would get?
Not quite, lets head back to the original equation after cross multiplying \[\large sin(130)b = 18sin(30)\] If you divide both sides by sin(130) we get \[\large \frac{\cancel{sin(130)}b}{\cancel{sin(130)}} = \frac{18sin(30)}{sin(130)}\] showing that \[\large b = ?\]
Hmm also not what I get....am I doing something wrong? Lol I get b = 11.75
nope i probably did i probably plugged it in wrong in my calculator @johnweldon1993 but thanks so much for your help!
Just wanted to make sure and I got it again so I would say go with it :P not to say yours is wrong of course >.< lol but Anytime :)

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