Write the steps needed to solve a triangle if A=120°,a=28,and b=17.

start with \[\frac{\sin(120)}{28}=\frac{\sin(B)}{17}\] or \[\frac{17\sin(120)}{18}=\sin(B)\] and solve for B

\[\frac{17\sin(120)}{28}=\sin(B)\] typo

Yep, then once you get B, do 180 - A - B = C. Now you have C and can do the same formula to find c, the last side.

http://cyberchalky.files.wordpress.com/2011/08/sine.gif This will explain the rule.

use sine rule

a^2 = b^2 + c^2 -2bc cos(A) as well

the sine rule can produce false results at times; whereas the cosine rule is more exacting

my trig teacher showed us why, but i cant recall the details ...

the "ambiguous case" but i think cosine doesn't help there in any case. plus law of sines is much snappier.

\[B=\arcsin(\frac{17\sin(120)}{28})=31.2\]

C is now obvious because they have to add to 180 and for side c put \[\frac{28}{\sin(120)}=\frac{c}{\sin(31.72)}\]or \[c=\frac{28\sin(31.72)}{\sin(120)}=17\]

a huge help in solving these is to draw a reasonable triangle and make sure that your answers make sense.

for example the longest side is opposite the largest angle

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