A triangle has sides A, B, and C. The angle between sides A and B is #(pi)/3#. If side C has a length of #4 # and the angle between sides B and C is #( 3 pi)/8#, what are the lengths of sides A and B?

1 Answer
Mar 8, 2016

#A=4.27#
#B=3.66#

Explanation:

This is a mult-step problem. Now, conventionally, a side length is denoted by a lowercase letter, rather than an uppercase one like in the problem above. Still, I don't want to confuse anyone, so I'll write the problems in the same way as it is given. We do need to label the angles though, so I'm going to say that #(3pi)/8# is signified by #x#, and #pi/3# by #z#, with the remaining angle called #y#.

So, I don't want to deal with the angles given in radians. I want to cahnge the angles from radians to degrees, and to do that it is a simple conversion. If we begin with #(3pi)/8#, and multiply it by #360/(2pi)#, then we have #(1080pi)/(16pi)#. The #pi#s divide out, and that leaves us with #67.5^o#.

So that's one angle converted, one to go.

#pi/3*360/(2pi)# gives us #60^o#.

So now we know that C=4, #z=60#, and #x=67.5#.
Because all the angles in a triangle must add up to #180#, we can take the whole, #180#, and subtract #67.5# and #60# from it. That leaves us with #52.5#, which equals #y#.

I'm going to solve for #A# using the law of sines, like this:
#A/(sinx)=C/(sinz)#. After we fill in the variables with what we know, we have #A/(sin 67.5)=4/(sin 60)#, which simplifies to #A=4.267#.

Now we just need to solve for #B#.
#B/(siny)=C/(sinz)#. This becomes #B/(sin52.5)=4/(sin60)#, and can be simplified to #B=3.664#.

Now we're done. Nice job!