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Solving of oblique triangles


Case 1. Three sides  a, b, c  are given. Find angles A, B, C. By the law of cosines we find one of the angles:

the second angle we find by the law of sines:

the third angle is found by the formula:  C = 180 – ( A + B ).

E x a m p l e . Three sides of a triangle are given: a = 2,  b = 3,  c = 4. Find angles of this triangle.
S o l u t i o n .

Case 2. Given: two sides a and b and angle C between them. Find a side c and angles A and B. By the law of cosines we find a side c :

c 2   =  a 2 +  b 2 - 2 ab cos C ;

and then by the law of sines an angle A :

here it is necessary to emphasize that A is an acute angle, if b / a > cos C, and an obtuse angle, if  b / a < cos C. The third angle  B = 180 ( A + C ).

Case 3. Any two angles and a side are given. Find the third angle and two other sides. It is obvious, that the third angle is calculated by the formula: A+ B+ C = 180, and then using the law of sines we find two other sides.

Case 4. Given two sides  a and b and angle B, opposite one of them. Find a side c and angles  and  C. At first by the law of sines we find an angle A :

The following cases are possible here:
 1)   a > ba sin B > b    there is no solution here;
 2)   a > ba sin B = b    there is one solution here,  A is a right angle;
 3)   a > ba sin B < b < a    there are two solutions here:  A  may be taken either an acute or an obtuse angle;
 4)   a   b    there is one solution here,  A an acute angle. After determining an angle A, we find the third angle:
       C = 180 ( A+ B ). Ii is obvious that if  A can have two values, then also C can have two values. Now the third
       side can be determined by the law of sines:

      If  we found two values for  C , then also a side  c  has two values, hence, two different triangles satisfy  the given
      conditions.

E x a m p l e . Given:  a = 5, b = 3,  B = 30. Find a side c and angles A and C .

S o l u t i o n . We have here: a > b  and  a sin B < b .  ( Check it please ! ).
Hence, according to the case 3  two solutions are possible here:

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