
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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