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 :
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 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 there are two solutions here: A may be
taken either an acute or an obtuse angle;
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
|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 ( Check it please ! ).
Hence, according to the case 3 two solutions are possible here: