Precalculus, Chapter 6, 6.2, Section 6.2, Problem 24

Given: C=101^@, a=3/8, b=3/4

c^2=a^2+b^2-2ab*cos(C)
c^2=.375^2+.75^2-2(.375)(.75)cos(101)
c^2=.8105
c=.90

cos(B)=[a^2+c^2-b^2]/(2ac)
cos(B)=[(.375)^2+(.90)^2-(.75)^2]/[(2)(.375)(.90)]
cos(B)=.5750
B=arccos(.5750)
B=54.90^@

A=180-101-54.90
A=24.10^@