Download Using the Law of Sines and the Law of Cosines to Solve an

Using the Law of Sines and the
Law of Cosines to Solve an Oblique Triangle
Oblique Triangle
c

a


Law of Sines
sin  sin  sin 


a
b
c
Law of Cosines
a 2  b 2  c 2  2bc cos 
b 2  a 2  c 2  2ac cos 
c 2  a 2  b 2  2ab cos 
b
An oblique triangle is a triangle in which none of the angles are right angles. An oblique triangle will either
have three acute angles or, two acute angles and one obtuse angle.
Case 1:
SAA or ASA – One side and two angles are known.  One Triangle
Given:
( ,  , and a) OR ( , , and a) OR (, , and b)
SAA:
( ,  , and b) OR ( , , and c) OR (, , and c)
ASA: ( ,  , and c) OR ( , , and b) OR (, , and a)
 First, subtract from 180° to solve for the unknown third angle.
 Then, use the Law of Sines TWICE to solve for the remaining two sides.
Case 2:
SSA 1 – Two sides and the angle opposite one of them are known.
Ambiguous Case  One Triangle, Two Triangles, or No Triangle
Given:
(a, b, and  ) OR (a, c, and  ) OR (b, c, and )
SSA:
(a, b, and  ) OR (a, c, and  ) OR (b, c, and )
 First, use the Law of Sines to solve for one of the unknown angles.
o If sin = 1, then  = 90°, and one right triangle results. (“One Triangle” Case)
o If sin > 1, then no triangle results. (“No Triangle” Case)
o If sin < 1, then there are TWO values of  that satisfy this relationship. The first value of
 1 will be in Quadrant I generated directly by the sin-1 key. The second value of  2 will be
in Quadrant II and is found by subtracting  1 from 180°. Thus, possibly TWO oblique triangles
result. (“One Triangle” or “Two Triangle” Case)2
 Second, use the computed value for and the given angle to subtract from 180° to find the
unknown third angle. (When sin < 1 there will be two computed values of  to consider, and
you will need to subtract from 180° twice, once for each value of .)
o If you can not find the third angle because the sum of the angles would be > 180°, then no
triangle results for that value of .
 Then use the Law of Sines to solve for the remaining third side. (Use the “Given” AS pair and
the “computed” value that you just found for the third angle.)
Case 3:
SAS – Two sides and the included angle are known.  One Triangle
Given: (a, b, and ) OR (a, c, and  ) OR (b, c, and )
 First, use the Law of Cosines to solve for the unknown side.
 Then, use the Law of Cosines to solve for one of the two unknown angles. 3
 Last, subtract from 180° to find the remaining third angle. (Or, use the Law of Cosines again.)
Case 4:
SSS – Three sides are known.  One Triangle
Given: (a, b, and c)
 First, use the Law of Cosines TWICE to solve for TWO of the three unknown angles.
 Then, subtract from 180° to find the remaining third angle. (Or, use the Law of Cosines again.)
Foot Notes:
1)
If the acronym spells out a bad word, but backwards, (SSA  ASS) then that is the ambiguous case that
can result in three possible outcomes: One Triangle, Two Triangles, or No Triangle.
 One Triangle – This can be generated one of two ways:
 When sin = 1, then  = 90°, and one right triangle results
 When sin < 1, then there are two values of  to consider. If one of the two possible values of 
results in an impossible triangle then one value of  remains, resulting in one oblique triangle.
 Two Triangles – When sin < 1, then there are two values of  to consider. If both values of 
result in triangles, then two oblique triangle result.
 No Triangle – When sin > 1, no triangle results since the sine function has range: sin   1 , so that
the equation sin = a, has no solution for a > 1.
2)
If sin > 0, 0° <  < 180°,   90° then  could either lie in Quadrant I or Quadrant II. Thus, there are
TWO angles to consider. Your calculator only gives you the value for  in Quadrant I because of how
the range of the Inverse Sine function (from the sin-1 key) is defined. To find the second angle located
in Quadrant II you must find the angle in Quadrant II that has the same reference angle as the angle in
Quadrant I. To do this subtract the acute angle, , from 180° to obtain the corresponding angle in
Quadrant II..
For example, let’s says that sin = 0.8, then  = sin–1(0.8), so that  = 53.13010235…
Thus  1 = 53.1° and  2 = 180° – 53.1° = 126.9°.
Now, there are two triangle possibilities to consider one corresponding to each value of .
3)
It would have been simpler computationally to use the Law of Sines here, but for sin < 1 there are
TWO possible solutions that need to be investigated, while the Law of Cosines results in ONE unique
solution. The Law of Cosines produces one unique solution for cos > 0, 0° <  < 180°, because only
angles in Quadrant I satisfy these two conditions. This corresponds to the same answer generated by
your calculator when you use the cos-1 key.
Space for additional notes: