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Warm-up:

Warm-up:. 1. Solve the right triangle if a=25 and c=35 B C A 2. Find the angle of depression from the top of a lighthouse 250 feet above water level to the water line of a ship 2 miles off shore. Answers. Precalculus Lesson 6.1.

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Warm-up:

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  1. Warm-up: 1. Solve the right triangle if a=25 and c=35 B C A 2. Find the angle of depression from the top of a lighthouse 250 feet above water level to the water line of a ship 2 miles off shore.
  2. Answers
  3. PrecalculusLesson 6.1 Check: NONE
  4. Objective Use the Law of Sines to solve oblique triangles. Read p. 353 “Bearings”
  5. The Law Of Sines Notes 6.1
  6. An oblique triangle is a triangle that does not contain a right angle. To solve an oblique triangle, we need to be given at least one side and two other parts.
  7. Four Cases AAS and ASA SSA (ambiguous case) SSS SAS Law of Sines Law of Cosines
  8. The Law of Sines
  9. Applying the Law of Sines 600 c 1800- (750 + 450) = 600 c = 12.247
  10. Use Law of Sines to answer the following EX 1 Given A = 123o, B = 41o, and a = 10 inches, find c. Answers C = 16° c =3.287
  11. Use Law of Sines EX 2 A triangular plot of land has interior angles A=95o and C= 68o. If the side between the angles is115 yards, Find the length of the other two sides? Answers B = 17° a = 391.838 yds. c = 364.694
  12. EX 3 Solve ambiguous Case Triangles (SSA) (Read p. 430) Given A = 26o, b = 5 ft., and a = 21 ft., solve the triangle. **Compare the side opposite the given angle to the other given side. (is a> b). IF YES, then one triangle. IF NO, then two or none. Answers B = 5.991° C = 148.009° c = 25.379 ONE TRIANGLE
  13. Ex 3 (continued) Given B=78o, c=12, and b=5, find angle C. c) Given A=29o, a = 6, and b = 10. Solve the triangle. Answer No Triangle
  14. Two Triangles. A=29°, a=6, b=10same for both cases Case 1 Case 2 Answers B2is supplementary to B1 B2 = 126.097° C2 = 24.903° c2 = 5.211 B1 = 53.903° C1 = 97.097° c1 = 12.281
  15. EX 4Find the area of an oblique triangle using the law of Sines. C = 143o, a = 7 meters, and b = 18 meters. Area = ½ absinC Answer Area = ½ acsinB 37.914 m2 Area = ½ bcsinA
  16. Ex 4 (b) Find the area of the triangle ABC. B 880 22 cm 680 240 A 54.06 cm C b Area = ½ bcsinA Area = 551.318 cm2 b = 54.056 cm
  17. Applying the Law of Sines Stan observes that the angle of elevation of a plane is 510. At the same time, Paul observes it to be 340. They are 340 m apart. How far is each person from the plane? How high is the plane? 340 510 340 m
  18. How high is the plane? The Solution A 950 b c b = 265.24 m Paul is 265.24 m away. D 340 510 B 340 m C AD = 190.85 sin 51 = 148.32 m The plane is 148.32 m high. c = 190.85 m Stan is 190.85 m away.
  19. EX 5 Application problem. Two fire ranger towers lie on the east-west line and are 5 miles apart. There is a fire with a bearing of N27oE from tower 1 and N32oW from tower 2. How far is the fire from tower 1? Answer 4.947 miles
  20. Assignments Classwork: p. 434 #5, 29, 41, 51 Homework: p. 434 #6-10 even, 30-34 even, 40, 42, 45, 48, 52, 54 (12 pts)
  21. Closure When given SSA (ambiguous case) how will you determine the number of solutions produced?
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