How do you solve the triangle given a=39, b=52, <C=122 degrees?

1 Answer
Nov 14, 2015

Answer:

Explanation:

The Law of Cosines states that #c^2=a^2+b^2-2abcosC#. We have #a,b# and #C# and should try to find side #c#.

Plug in values.
#c^2=39^2+52^2-2(39)(52)cos122˚#
#c^2~~6374.3525#
#c~~ 79.8395#

Now, using The Law of Sines, we can figure out the other two angle lengths.

The Law of Sines states that #sinA/a=sinB/b=sinC/c#, so we can say that #sinA/39=(sin122˚)/79.8395# and #sinB/52=(sin122˚)/79.8395#.

Cross multiply and use #arcsin# to figure out angles #A# and #B#.