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9.5 APPLICATION OF TRIG

9.5 APPLICATION OF TRIG. NAVIGATION. The course of a plane or ship is measured clockwise from the north direction. 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.

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9.5 APPLICATION OF TRIG

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  1. 9.5 APPLICATION OF TRIG NAVIGATION

  2. The course of a plane or ship is measured clockwise from the north direction. 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.

  3. 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 plane from its starting point? draw a diagram

  4. 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.

  5. A ship proceeds on a course of 300° for 2 hours at a speed of 15 knots ( 1 knot = 1 nautical mile per hour). Then it changes course to 230°, continuing at 15 knots for 3 more hours. At that time, how far is the ship from its starting point? ≈ 62 miles

  6. Town T is 8 km northeast of village V. City C is 4 km from T on a bearing of 150° from T. What is the bearing and distance of C from V? ≈ 7.96 km Bearing of 74º

  7. A plane flies 600 km on a course of 300°. It then flies south for a while and finally flies on a course of 40° to return to its starting point. Find the total distance traveled. ≈ 2327 km

  8. 9.5 APPLICATION OF TRIG SURVEYING

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

  10. Northwest = N 45o W Northeast = N 45o E Southeast = S 45o E Southwest = S 45o W

  11. Sometimes a plot of land is taxed by its area.  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. draw a diagram

  12. To find the area, we can divide the area into two triangles   One way to divide it is:

  13. 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

  14. 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.

  15. Sketch the plot of land described. Then find its area. From a granite post, proceed 195 ft east along Tasker Hill Road, then along a bearing of S32°E for 260 ft, then along a bearing of S68°W for 385 ft, and finally along a line back to the granite post. ≈ 84,797 ft2

  16. From the north corner of Park Avenue, a surveyor walked N38°E for 128 m and then S48°E. Find the surveyor’s distance and compass reading from his starting point. ≈ 190 miles due east

  17. Assignment • Page 362 1-18 • Page 357 #29 a-e

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