9 - 5 Applications to Navigation and Surveying
 
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 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 110o, the second a course heading of 240o, and the third a course heading of 050o.

Sample problem
1)  A plane proceeds on a course of 310o for 2 hours at 150 mph.  It then changes direction to 200o 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 70o. (Sum of the two angles)
x2 = 3002 + 4802 - 2(300)(480)Cos 70
x2 = 320400 - 288000Cos 70
x2 = 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 45o W
Northeast = N 45o E
Southeast = S 45o E
Southwest = S 45o 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 40o E for another 150 feet.  Turn direction again S 60o 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:
x2 = 3002 + 1502 - 2(300)(150)Cos 130
x2 = 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
33.8o = y
The other angle is found by 80 - 33.8 = 46.2o
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 90o, side x is 9 and side y is 5.  Find angle Y to tenths place.  
 
a)  56.3o
b)  33.7o
c)  29.1o
d)  61.9o
e)  45.3o
 
 
 
2)  Find the area of the triangle ABC if a = 5, b = 7 and angle C = 42o.  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 = 135o, angle Y = 11o 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)  60o
b)  30o
c)  150o
d)  100o
e)  120o
 
 
  5)  Ship A sights ship B on a compass bearing of 115o.  Give the compass bearing of ship A from Ship B.   
 
a)  245o
b)  295o
c)  65o
d)  115o
e)  25o
 
 
 click here for answers!!