We can use trigonometry to work with navigation problems as
well as surveying problems. Let's take a look at
navigation. The course of a plane or ship is measured clockwise
from the north direction. Study the examples below:

Notice that the courses are all three digit numbers.
The first has a course bearing of 110^{o}, the second a course
heading of 240^{o}, and the third a course heading of 050^{o}.

Sample problem

1)
A plane proceeds on a course of 310^{o} for 2 hours at 150
mph. It then changes direction to 200^{o} continuing for
3 more hours at
160 mph. At this time, how far is the ship from its starting
point?

Solution:

draw a diagram:

The 50 on the right side of the picture comes from a
complete circle
minus the 310. The 50 in the middle angle comes from alternate
interior
angles.

The 20 comes from the course heading of 200. A
straight line
is 180 plus 20 more makes the bearing 200.

We can use law of cosines to find the missing side.
The included
angle measures 70^{o}. (Sum of the two angles)

x^{2}
= 300^{2} + 480^{2} - 2(300)(480)Cos 70

x^{2}
= 320400
- 288000Cos 70

x^{2}
= 221898.1987

x = 471

The plane is 471 miles from its starting point.

Surveying

In surveying, a compass reading is usually given as an acute
angle measured from a north- south line toward the east or west.
Study the examples:

Northwest = N 45^{o} W

Northeast = N 45^{o} E

Southeast = S 45^{o} E

Southwest = S 45^{o} W

Sometimes a plot of land is taxed by its area. Work
the following
problem.

1)
A post is driven in a certain spot. Proceed due east for 300 ft,
then proceed S 40^{o} E for another 150 feet. Turn
direction again S 60^{o} W for 400 feet and then back to the
post in a straight line. Find the area.

Solution:

draw a diagram

To find the area, we can divide the area into two triangles
like we did in the last section. One way to divide it is:

We can calculate the top triangle's area right now!

K = (1/2)(300)(150)Sin 130

K = 17236 sq ft.

To find the other area, we need the length of the blue line
and the degree of the bottom angle.

To find the length of the blue line, use law of cosines:

x^{2}
= 300^{2} + 150^{2} - 2(300)(150)Cos 130

x^{2}
= 170350.8849

x = 413

Now find the angle using the law of sines:

(Sin 130)/413 = (Sin y)/300

(300 Sin 130)/413 = Sin y

.556448748 = Sin y

33.8^{o} = y

The other angle is found by 80 - 33.8 = 46.2^{o}

We can now find the area of the bottom triangle:

K = (1/2)(413)(400)Sin 46.2

K = 59617 sq ft.

Add the two areas together"

59617 + 17236 = 76853 sq ft.

That's
it for chapter 9. On to the Sample Test!!

Current quizaroo # 9

1) In the triangle
XYZ, angle
X is 90^{o}, side x is 9 and side y is 5. Find angle Y to
tenths
place.

a)
56.3^{o}

b)
33.7^{o}

c)
29.1^{o}

d)
61.9^{o}

e)
45.3^{o}

2) Find the area of
the triangle
ABC if a = 5, b = 7 and angle C = 42^{o}. Round to tenths.

a) 11.7 square
units

b) 23.4 square units

c)
13.0 square units

d)
17.5 square units

e)
35.0 square units

3) In triangle XYZ, angle X = 135^{o}, angle Y
= 11^{o} and z = 4. Find side x and round to three
significant digits.

a)
14.8

b)
12.1

c)
5.06

d)
4.35

e)
9.38

4) In triangle
XYZ, the sides are x = 3, y = 5 and z = 7. Find the measure of
the biggest angle and round to the nearest tenth of a degree.

a) 60^{o}

b) 30^{o}

c) 150^{o}

d) 100^{o}

e) 120^{o}

5) Ship A sights
ship B on a compass bearing of 115^{o}. Give the compass
bearing of ship A from Ship B.