Download Law of Sines

Dr. Neal, WKU
MATH 117
The Law of Sines
Besides Side-Angle-Side (SAS) and Side-Side-Side (SSS), there are two other congruence
forms that completely determine a triangle. These are Angle-Side-Angle (ASA) and
Side-Angle-Angle (SAA).
In each case, the third angle can be determined because all the angles must sum to
180º. So with ASA and SAA, we have all three angles, but we must find the other two
sides. The Law of Cosines will not apply because it requires two sides to find the third
side. And we may not have a right triangle, so we usually can’t 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 opposite
angle C . Then,
B
a
a
b
c
=
=
sin A sin B sin C
c
A
C
Law of Sines
b
Example 1. Find the remaining sides in the following triangles:
85º
c
40
10
w
100º
55º
a
(i)
30º
v
(ii)
Solution. (i) First, let A = 85º and C = 55º. Then B = 180º – 85º – 55º = 40º. The given
side is b = 10. To find sides a and c , we have
a
10
=
sin85º sin40º
!
a=
10 sin85º
≈ 15.498
sin 40º
c
10
=
sin55º sin 40º
!
c=
10 sin55º
≈ 12.7437
sin 40º
(ii) Let W = 30º and V = 180º – 100º – 30º = 50º. Then
w
40
v
=
=
sin30º sin100º sin 50º
! w=
40 sin30º
≈ 20.3
sin100º
and v =
40 sin 50º
≈ 31.114
sin100º
Dr. Neal, WKU
When we have Side-Angle-Side, then we can use the Law of Cosines to find the
third side. But it may be easier to use the Law of Sines to find the other two angles. But
now we use the “reciprocal” form:
sin A sin B sin C
=
=
a
b
c
Example 2. Find the remaining side and angles in the triangle below:
B
50
A
30º
40
Solution. To find side c , we use the Law of Cosines: c 2 = 402 + 502 ! 2(40)(50) cos 30º ;
c = (402 + 502 ! 2 " 40 " 50 " cos30º) ≈ 25.217
Now let c = 25.217, C = 30º, a = 50, and b = 40. Then
sin A sin 30º sin B
=
=
50
25.217
40
50 sin30º
40 sin30º
and sin B =
. But both of these equations have two
25.217
25.217
solutions, a first quadrant angle and a second quadrant angle. However, the two
shortest sides in a triangle must have opposite angles that are acute (less than 90º).
Thus, angle B must be less than 90º .
Then, sin A =
!1 " 40 sin30º%
' ≈ 52.4776º. Finally, A ≈ 180º – 30º – 52.4776º = 97.5224º.
So B = sin $
# 25.217 &
" 50 sin30º %
Note: Using the inverse sine to solve for A , we have sin !1 $
' ≈ 82.478º which is
# 25.217 &
not the correct value of A . We then need A = 180º – 82.478º to give an obtuse (second
quadrant) angle.
Dr. Neal, WKU
Side-Side-Angle (SSA)
A congruence form that may not give a unique triangle is side-side-angle. If the given
angle is small enough, then there could be two triangles made. But if the given angle is
obtuse, then there will be only one possible triangle.
Example 3. Consider a triangle with A = 20º, b = 17, and a = 10. Give two possibilities
for the dimensions of this triangle.
Solution. From the drawing, we see that we have SSA. So we cannot use the Law of
sin A sin B
Cosines and the only relationship we can establish is
which gives
=
a
b
sin20º sin B
17 sin 20º
. Thus, sin B =
which has two solutions:
=
10
17
10
One Solution:
17
!1 " 17 sin20º %
B = sin $
#
10
10
' ≈ 35.55º
&
20º
B
Another Solution:
17
10
B = 180º – 35.55º ≈ 144.45º
20º
17
20º
10
B
10
B
Side-Side-Angle May Give Two Solutions
B