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restart;

 1.  The Law of Sines

The law of sines is a three-sided equation. It equates the ratio of the sine of each angle to the length of its opposite side for all the angles and sides of a triangle. It is usually used as a set of three equations.

>	sin(A)/a = sin(B)/b : % = sin(C)/c;

>	sin(A)/a = sin(B)/b  ; sin(B)/b = sin(C)/c  ; sin(A)/a = sin(C)/c  ;

 2.  Solving Triangles - Two Sides & Included Angle Known

The law of cosines has four unknowns - the three sides and an angle. If we know the three sides of a triangle, we can then find "the angle" - which is the angle opposite of the "hyptenuse" (although this is not necessarily a right triangle and the side opposite of the angle is not necessarily the longest side.)

         Example 2.1 :    A triangle have sides of length : a = 31,  b = 42, and angle A = 41 degrees. Find the angle B.

>	subs( {a = 31, b = 42, A = 41*(Pi/180) }, sin(A)/a = sin(B)/b  ); Eq := evalf(%);

>	sin(B) = solve( Eq, sin(B) ); B = solve( Eq, B), ` radians`; B = evalf( solve( Eq, B)*180/Pi), ` degrees`;

         Example 2.2 :     A triangle have sides of length : a = 102, b = 79, and A = 22 degrees. Find c.


In this problem, we need to two steps. First, find angle B, and use this to find angle C. Then use angle C to find side c.

>	subs( {A = 22*(Pi/180), a = 102, b = 79 }, sin(A)/a = sin(B)/b  ); evalf(%); angle_B := solve( %, B): %, ` radians`;

>	angle_C := subs( {A = 22*(Pi/180), B = angle_B}, Pi - A - C);

>	subs( {A = 22*(Pi/180), a= 102, C = angle_C }, sin(A)/a = sin(C)/c  ); evalf(%); side_c := solve( %, c);

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 3.  Solving Triangles - Two Angles & Included Side Known

Here is another scenario.

         Example 3.1 :     A = 52 degrees,  B = 65 degrees, side c = 100. Find sides a and b.

First we find angle C, which is easy to find. Then use one of the three equations of the Law of Sines invovling A,a, C, and c. This is the basic method of using the Law of Sines - to find three of the four bits of information about any two pairs, then solve for the fourth.

>	C = 180 - 52 - 65;

>	subs( {A = 52*(Pi/180),  C = 63*(Pi/180), c = 100 }, sin(A)/a = sin(C)/c  ); Eq := evalf(%);

>	solve( %, a);

Now that we know A, B, C, a, and c, we can find b in a number of ways. Let's just find it the same way that we found a. (We could also relate A and B in the Law of Sines, or even use the Law of Cosines.)

>	subs( {B = 65*(Pi/180),  C = 63*(Pi/180), c = 100 }, sin(B)/b = sin(C)/c  ); Eq := evalf(%);

>	solve( %, b);