College Trigonometry, 2011a

11.2 The Law of Sines 897
minimizes the chances of propagated error. 3 Third, since many of the applications which require
solving triangles ‘in the wild’ rely on degree measure, we shall adopt this convention for the time
being. 4 The Pythagorean Theorem along with Theorems 10.4 and 10.10 allow us to easily handle
any given right triangle problem, but what if the triangle isn’t a right triangle? In certain cases,
we can use the Law of Sines to help.
Theorem 11.2. The Law of Sines: Given a triangle with angle-side opposite pairs (α, a),
(β,b) and(γ,c), the following ratios hold
or, equivalently,
sin(α)
a
= sin(β)
b
= sin(γ)
c
a
sin(α) =
b
sin(β) =
c
sin(γ)
The proof of the Law of Sines can be broken into three cases. For our first case, consider the
triangle △ABC below, all of whose angles are acute, with angle-side opposite pairs (α, a), (β,b)
and (γ,c). If we drop an altitude from vertex B, we divide the triangle into two right triangles:
△ABQ and △BCQ. If we call the length of the altitude h (for height), we get from Theorem 10.4
that sin(α) = h c and sin(γ) = h a
so that h = c sin(α) =a sin(γ). After some rearrangement of the
last equation, we get sin(α)
a
using the triangles △ABQ and △ACQ to get sin(β)
α
c
B
β
γ
a
= sin(γ)
c
. IfwedropanaltitudefromvertexA, we can proceed as above
= sin(γ) , completing the proof for this case.
α
c
b
A
C A
C
b
Q
A
C
b
For our next case consider the triangle △ABC below with obtuse angle α. Extending an altitude
from vertex A gives two right triangles, as in the previous case: △ABQ and △ACQ. Proceeding
as before, we get h = b sin(γ) andh = c sin(β) sothat sin(β)
b
= sin(γ)
c
.
B
h
γ
c
a
c
β
h ′
B
Q
γ
B
c
β
α
a
γ
B
c
β
Q
h
a
γ
A
b
C
A
b
C
3 Your Science teachers should thank us for this.
4 Don’t worry! Radians will be back before you know it!