The law of Cosines states as follows:
[tex] {c}^{2} = {a}^{2} + {b}^{2} - 2ab \cos(C) [/tex]
where a is the side opposite angle A, b is the side opposite angle B, and c is the side opposite angle C
for problem 1), we can switch it to:
[tex] {a}^{2} = {b}^{2} + {c}^{2} - 2bc \cos(A) \\ {a}^{2} = {(23)}^{2} + {(28.5)}^{2} \\ - 2(23)(28.5) \cos(87) \\ {a}^{2} = 1341.25 - (1311)(.052) [/tex]
[tex] {a}^{2} = 1341.25 - 68.61 = 1272.64 \\ a = \sqrt{1272.64} = 35.67[/tex]
Now you can use the law of sines for a 2nd angle:
[tex] \frac{b}{ \sin(B) } = \frac{a}{ \sin(A) } \\ \sin(B) = \frac{b\sin(A)}{a} = \frac{23\sin(87)}{35.67} \\ B = {sin}^{ - 1} ( \frac{22.97}{35.67} ) = 40.08[/tex]
and for the 3rd just use the 180-triangle rule:
C = 180 - 40.08 - 87 = 52.92
Now for problem 2)
again, the Law of Cosines, but now we'll switch it to solve for C:
[tex] {c}^{2} = {a}^{2} + {b}^{2} - 2ab \cos(C) \\ \cos(C) = ( {c}^{2} - {b}^{2} - {a}^{2}) \div - 2ab \\ = ( {19.9}^{2} - {16.7}^{2} - {23.5}^{2}) \div - 784.9 \\ C = {\cos}^{ - 1} ( \frac{-435.13}{-784.9}) = 56.33[/tex]
Now for a 2nd angle use the Law of Sines:
[tex] \frac{c}{ \sin(C) } = \frac{a}{ \sin(A) } \\ \frac{19.9}{ \sin(56.33) } = \frac{23.5}{ \sin(A) } \\ \sin(A) = 23.5 \sin(56.33) \div 19.9 [/tex]
[tex]A = {sin}^{ - 1} ( \frac{19.56}{19.9} ) = 79.36[/tex]
finally, use the 180-triangle rule to find B:
B = 180 - 79.36 - 56.33 = 44.31
