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Dr. Neal, WKUMATH 117 The Law of Sines
Besides Side-Angle-Side (SAS) and Side-Side-Side (SSS), there
are two other congruenceforms that completely determine a triangle.
These are Angle-Side-Angle (ASA) andSide-Angle-Angle (SAA).
In each case, the third angle can be determined because all the
angles must sum to
180. So with ASA and SAA, we have all three angles, but we must
find the other twosides. The Law of Cosines will not apply because
it requires two sides to find the thirdside. And we may not have a
right triangle, so we usually cant use right-triangle trig.In this
case, we use the Law of Sines.
We again label the sides of the triangle as a, b, c and the
angles as A, B, and C.As usual, side a is opposite angle A, side b
is opposite angle B, and side c is oppositeangle C. Then,
a
sinA=
b
sinB=
c
sinC
Law of Sines
a
b
c
C A
B
Example 1. Find the remaining sides in the following
triangles:
10
85
55
a
c
100 30
40
w
v (i) (ii)
Solution. (i) First, let A = 85 and C= 55. Then B = 180 85 55 =
40. The givenside is b = 10. To find sides a and c, we have
a
sin85=
10
sin40 ! a =
10 sin85
sin 40!15.498
c
sin55=
10
sin 40 ! c =
10 sin55
sin 40!12.7437
(ii) Let W = 30 and V = 180 100 30 = 50. Then
w
sin30=
40
sin100=
v
sin 50 ! w =
40 sin30
sin100!20.3 and v =
40 sin 50
sin100!31.114
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Dr. Neal, WKU
When we have Side-Angle-Side, then we can use the Law of Cosines
to find thethird side. But it may be easier to use the Law of Sines
to find the other two angles. Butnow we use the reciprocal
form:
sinA
a=
sinB
b=
sinC
c
Example 2. Find the remaining side and angles in the triangle
below:
30
50
40
A
B
Solution. To find side c, we use the Law of Cosines: c2 =402
+502 ! 2(40)(50) cos 30 ;
c = (402+ 50
2! 2"40" 50 " cos30) !25.217
Now let c = 25.217, C= 30, a = 50, and b = 40. Then
sinA
50=
sin 30
25.217=
sin B
40
Then, sin A =50 sin30
25.217and sinB =
40 sin30
25.217. But both of these equations have two
solutions, a first quadrant angle and a second quadrant angle.
However, the twoshortest sides in a triangle must have opposite
angles that are acute (less than 90).Thus, angle B must be less
than 90 .
So B =sin!1 40 sin30
25.217
"
#$
%
&'!52.4776. Finally, A !180 30 52.4776 = 97.5224.
Note: Using the inverse sine to solve for A, we have sin!1 50
sin30
25.217
"#$ %&' !82.478 which is
not the correct value of A . We then need A = 180 82.478 to give
an obtuse (secondquadrant) angle.
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Dr. Neal, WKU
Side-Side-Angle (SSA)
A congruence form that may notgive a unique triangle is
side-side-angle. If the givenangle is small enough, then there
could be two triangles made. But if the given angle isobtuse, then
there will be only one possible triangle.
Example 3. Consider a triangle with A = 20, b = 17, and a = 10.
Give two possibilitiesfor the dimensions of this triangle.
Solution. From the drawing, we see that we have SSA. So we
cannot use the Law of
Cosines and the only relationship we can establish issinA
a=
sinB
b which gives
sin20
10=
sinB
17. Thus, sinB =
17sin20
10which has two solutions:
One Solution:
B = sin!1 17 sin20
10
"#$ %&' !35.55
20
10
B
17
Another Solution:
B = 180 35.55 !144.45
20
10
B
17
20
1010
BB
17
Side-Side-Angle May Give Two Solutions
