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Page 276 – The Law of Sines The Law of Sines a sin A = b sin B = c sin C Also remember that there may be no, one or two triangles depending upon the relationship.

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Presentation on theme: "Page 276 – The Law of Sines The Law of Sines a sin A = b sin B = c sin C Also remember that there may be no, one or two triangles depending upon the relationship."— Presentation transcript:

1 Page 276 – The Law of Sines The Law of Sines a sin A = b sin B = c sin C Also remember that there may be no, one or two triangles depending upon the relationship between a and b sin A

2 Solve triangle ABC if A = 29º10’, B = 62°20’ and c = Round angle measures to the nearest minute and side measures to the nearest tenth. First, find the measure of angle C C = 180° - (29°10’ + 62°20’)= 180° - 91°30’= 88°30’ Use the Law of Sines to find a and b a sin A = c sin C a sin 29°10’ = 11.5 sin 88°30’ a = 11.5 sin 29°10’ sin 88°30’ ≈ 11.5(0.4878) a ≈ 5.6 b. sin B = c. sin C b. sin 62°20’ = 11.5 sin 88°30’ b = 11.5 sin 62°20’ sin 88°30’ ≈ 11.5(0.8857) b ≈ 10.2

3 Solve triangle ABC if A = 63°10’, b = 18, and a = 17. Round angle measures to the nearest minute and side measures to the nearest tenth. To determine the number of solutions, compare a with b sin A b sin A = 18 sin 63°10’ = 18(0.8923) = 16.1 Since 63°10’ b sin A and a < b, two triangles exist. Use the Law of Sines to find B 17 sin 63°10’ = 18 sin B sin B = 18 sin 63°10’ 17 sin B ≈ 18(0.8923) 17 sin B ≈ so B ≈ 70°52’ We know there are two triangles, one measurement for B is 70°52’. Since sin is positive in both Q 1 and 2, the other angle is 180° - 70°52’, or 109°8’. Now, we must solve each triangle

4 1 st triangle: A = 63°10’, B = 70°52’, b = 18, and a = 17 C = 180° - (63°10’ + 70°52’)= 180° - 134°2’= 45°58’ To find c, use the law of sines 17. sin 63°10’ = c. sin 45°58’ c ≈ 17 sin 45°58’ sin 63°10’ c ≈ 17(0.7189) c ≈13.7 So, one solution is B ≈ 79°52’, C ≈ 45°58’, and c ≈ 13.7

5 2 nd Solution: A = 63°10’, B = 109.8’, b = 18 and a = 17 C = 180° - (63°10’ + 109°8’)= 180° - 172°18’= 7°42’ To find c, use the law of sines 17. sin 63°10’ = c. sin 7°42’ c ≈ 17 sin 7°42’ sin 63°10’ ≈ 17(0.1340) ≈ 2.6 So, the second solution is B ≈ 109°8’, C ≈ 7°42’, and c≈ 2.6

6 First Triangle AB C 1817 c ≈ °10’70°52’ 45°58’ Second Triangle A B C °10’ 109°8’ 7°42’

7 Solve triangle ABC if A = 43°, b = 20 and a = 11. b sin A = 20 sin 43° ≈ 20(0.6820) ≈ = Since 43° < 90° and a (11) < b sin A (13.64), there is no solution, so there is no triangle

8 Assignment Page # 14 – 24, 31,


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