(* 6.5, p. 473, #9, 14, 24, 41 *)

(* #9, ASA *)
AngleA=46*Pi/180;
SideC=65;
AngleB=20*Pi/180;
AngleC=Pi-AngleB-AngleA;
N[%]*180/Pi
SideB=Sin[AngleB]*SideC/Sin[AngleC];
N[%]
SideA=Sin[AngleA]*SideC/Sin[AngleC];
N[%]

(* #14, ASA *)
AngleA=23*Pi/180;
SideC=50;
AngleB=110*Pi/180;
AngleC=Pi-AngleB-AngleA;
AngleC*180/Pi
SideB=Sin[AngleB]*SideC/Sin[AngleC];
N[%]
SideA=Sin[AngleA]*SideC/Sin[AngleC];
N[%]

(* #24, SSA, case of two distinct solutions. If there is one solution, then if it's not a right triangle there's another solution. *)
SideA=75;
AngleA=30*Pi/180;
SideB=100;
AngleB=ArcSin[SideB*Sin[AngleA]/SideA];
N[%]*180/Pi
AngleC=Pi-AngleB-AngleA;
N[%]*180/Pi
SideC=Sin[AngleC]*SideA/Sin[AngleA];
N[%]

(* 
Second solution: 
The second solution is obtained by using the supplementary angle. That is, the sum of the two solution angles is 180. This is because the sine function is symmetric about pi/2. 
*)
AngleB=Pi-AngleB;
N[%]*180/Pi
AngleC=Pi-AngleB-AngleA;
N[%]*180/Pi
SideC=Sin[AngleC]*SideA/Sin[AngleA];
N[%]

(* #41, Story problem:
involves the addition of an extra angle which must be taken into account.
 *)
AngleA=30*Pi/180;
SideC=60;
AngleB=141*Pi/180;
AngleC=Pi-AngleB-AngleA
SideB=Sin[AngleB]*SideC/Sin[AngleC];
N[%]
SideA=Sin[AngleA]*SideC/Sin[AngleC];
N[%]