40
Chapter 2
•
General Triangles
§2.1
Example2.1
b
a
=
10
c
A
=
41
◦
B
C
=
75
◦
Case 1: One side and two angles.
Solve the triangle
△
ABC
given
a
=
10,
A
=
41
◦
, and
C
=
75
◦
.
Solution:
We can find the third angle by subtracting the other two angles
from 180
◦
, then use the law of sines to find the two unknown sides. In this
example we need to find
B
,
b
, and
c
. First, we see that
B
=
180
◦
−
A
−
C
=
180
◦
−
41
◦
−
75
◦
⇒
B
=
64
◦
.
So by the Law of Sines we have
b
sin
B
=
a
sin
A
⇒
b
=
a
sin
B
sin
A
=
10 sin 64
◦
sin 41
◦
⇒
b
=
13
.
7
,
and
c
sin
C
=
a
sin
A
⇒
c
=
a
sin
C
sin
A
=
10 sin 75
◦
sin 41
◦
⇒
c
=
14
.
7
.
Example2.2
b
=
30
a
=
18
c
A
=
25
◦
B
C
Case 2: Two sides and one opposite angle.
Solve the triangle
△
ABC
given
a
=
18,
A
=
25
◦
, and
b
=
30.
Solution:
In this example we know the side
a
and its opposite angle
A
,
and we know the side
b
. We can use the Law of Sines to find the other
opposite angle
B
, then find the third angle
C
by subtracting
A
and
B
from
180
◦
, then use the law of sines to find the third side
c
. By the Law of Sines, we have
sin
B
b
=
sin
A
a
⇒
sin
B
=
b
sin
A
a
=
30 sin 25
◦
18
⇒
sin
B
=
0
.
7044
.
Using the
✄
✂
a0
✁
sin
−
1
button on a calculator gives
B
=
44
.
8
◦
.
However, recall from Section 1.5 that
sin (180
◦
−
B
)
=
sin
B
.
So there is a second possible solution for
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 Fall '11
 Dr.Cheun
 Calculus, Angles, Law of sines, Sin, Harshad number, triangle

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