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Do not worry about your difficulties in mathematics, I assure you that mine are greater
". Einstein, Albert (1879-1955)
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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 ).
Example:
Three sides of a triangle are given: a = 2, b = 3, c = 4. Find angles of this triangle.
Solution:
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.
Example:
Given: a = 5, b = 3, B = 30°. Find a side c and angles A and C .
Solution:
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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