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  A  and  C. At first by the law of sines we find an angle A : The following cases are possible here:  1)   a > b ;  a · sin B > b  –  there is no solution here;  2)   a > b ;  a · sin B = b  –  there is one solution here,  A is a right angle;  3)   a > b ;  a · 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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