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; 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 ( Check it please ! ). Hence, according to the case 3 two solutions are possible here: |