Begin2.DVI

Example 6-10.

Derive the law of sines for the triangle illustrated in the figure 6-10.

Solution The sides of the given triangle are formed from the vectors A
,B
and C
and

since these vectors are free vectors they can be moved to the positions illustrated

in figure 6-10. Also sketch the vector −C
as illustrated. The new positions for

the vectors A,
B,
C
and −C
are constructed to better visualize certain vector cross

products associated with the law of sines.

Figure 6-10. Triangle for law of sines.

Examine figure 6-10 and note the following cross products

C
×A
= (A
−B
)×A
=A
×A
−B
×A
=−B
×A
=A
×B

and B
×(−C
) = B
×(−A
+B
) = B
×(−A
) + B
×B
=A
×B .

Taking the magnitude of the above cross products gives

|C
×A
|=|A
×B
|=|B
×(−C
)|

or

AC sin θB=AB sin θC=BC sin θA.

Dividing by the product of the vector magnitudes ABC produces the law of sines

sin θA

A

=sin θB

B

=sin θC

C

.