The following chart
gives
a good review of the last two chapters for non-right triangles:

Given
Use

SAS

law of Cosines

SSS

law of Cosines

ASA or AAS

law of Sines

SSA

ambiquous case (0, 1, or 2 triangles)

Sample
Problems

1) Solve the triangle
with
sides a = 3, b = 5, c = 7. Round to the nearest tenth of a degree.

Solution:

Find the largest
angle
of the triangle first. This will be / C because the
longest
side is c.

Use this form:

c^{2 }= a^{2}
+ b^{2} - 2ab Cos C

49 = 9 + 25 -
2(3)(5)
Cos C

49 = 34 - 30 Cos C

15 = -30 Cos C

-1/2 = Cos C

Cos^{-1}
(-1/2) = C

120 = C

/ C = 120^{o}

Now that we have an
angle,
we can switch to the law of sines.(Easier to use)

Find / B

(Sin B)/b = (Sin
C)/c

(Sin B)/5 = (Sin
120)/7

7 Sin B = 5 Sin 120

Sin B = (5 Sin
120)/7

Sin B =
0.6185895741317

B = 38.2

/ B = 38.2^{o}

To find /
A, subtract
from 180

180 - (120 + 38.2)
=
180 - 158.2 = 21.8

/ A = 21.8^{o}

2) Solve the triangle
if
a = 3, b = 7 and / C = 37^{o}

Solution

We are given two
sides
and the included angle. We must find the third side. The
missing
side is c.

Use the form:

c^{2 }= a^{2}
+ b^{2} - 2ab Cos C

c^{2}
= 9 + 49 - 2(3)(7) Cos 37

c^{2}
= 58 - 42 Cos 37

c^{2}
= 24.457308

c = 4.9

Now use the law of
sines
and find the smallest angle. The smallest angle is definitely an
acute angle. The law of sines can not distinquish between acute
and
obtuse because both angles give a positive answer.

The smallest angle
is
opposite side a, the shortest side.

(Sin A)/3 = (Sin
37)/4.9

4.9 Sin A = 3 Sin 37

Sin A = (3 Sin 37)/
4.9

Sin A = .36845817

A = 21.6^{o}

To find /
B, subtract
from 180

180 - (21.6 + 37) =
121.4

/ B = 121.4^{o}

3) A farmer has a
triangular
field with sides 120 yards, 170 yards, and 220 yards. Find the
area
of the field in square yards. Then find the number of acres
if
1 acre = 4840 square yards.

Solution:

We need to find an
angle
so we can use the area formula

let a = 120, b =
170,
c = 220. Find / C

c^{2 }= a^{2}
+ b^{2} - 2ab Cos C

48400 = 14400 +
28900
- 40800 Cos C

5100 = -40800 Cos C

-5100/40800 = Cos C

Cos^{-1}(-5100/40800)
= C

97.2^{o} = C

Now find the area

K = (1/2)(ab Sin C)

K = (1/2)(120)(170)
Sin
97.2

K = 10120 square
yards

The number of acres
is
found by:

10120/4840 = 2.1

2.1 acres

4) Find the area of
the following figure:

Solution:

Divide the
quadrilateral
into two triangles:

You can find the
area
of the triangle on the left:

K =
(1/2)(14)(12)Sin
110

K = 78.9 square
units.

To find the area of
the
triangle on the right, we need to find x, and z.

To find x, we can
use
the law of cosines

x^{2}
= 196 + 144 - 2(14)(12) Cos 110

x^{2}
= 454.918

x = 21.3

To find z, we first
have
to find y. We can use the law of sines for the left triangle:

(Sin y)/14 = (Sin
110)/21.3

21.3 Sin y = 14 Sin
110

Sin y = (14 Sin
110)/21.3

Sin y = .6176383

y = 38.1^{o}

To find z, subtract
130
- 38.1 = 91.9

We can now find the
area
of the right side triangle:

K =
(1/2)(21.3)(8)(Sin
91.9)

K = 85.2

Add the two areas:

85.2 + 78.9 = 164.1

The area is 164.1
square
units

We can now try some
applications
to navigation and surveying!!