Download presentation

Presentation is loading. Please wait.

Published bySam Monk Modified over 4 years ago

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 = 11.5. 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) 0.997 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) 0.9997 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 ≈ 0.9448 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) 0.8923 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) 0.8923 ≈ 2.6 So, the second solution is B ≈ 109°8’, C ≈ 7°42’, and c≈ 2.6

6
First Triangle AB C 1817 c ≈ 13.7 63°10’70°52’ 45°58’ Second Triangle A B C 18 17 2.6 63°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) ≈ = 13.64 Since 43° < 90° and a (11) < b sin A (13.64), there is no solution, so there is no triangle

8
Assignment Page 280 - 281 # 14 – 24, 31, 32 - 35

Similar presentations

Presentation is loading. Please wait....

OK

Time and Labor Processing Day 1

Time and Labor Processing Day 1

© 2018 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google