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PRE-CALCULUS LESSON 6.1 LAW OF SINES Spring 2011 Question! How to measure the depth? http://www.mcs.uvawise.edu/dbl5h/resources/latex_examples/files/triglawsines.pdf At the end of this lesson you should be able to Use the Law of Sines to solve oblique triangles (AAS or ASA). Use the Law of Sines to solve oblique triangles (SSA). Find areas of oblique triangles. Use the Law of Sines to model and solve real-life problems. Introduction 1. To solve an oblique triangle, we need to be given at least one side and then any other two parts of the triangle. 3. The sum of the interior angles is 180°. Why can’t we use the Pythagorean Theorem? 4. State the Law of Sines. 2. C a b 5. 6. Area = ½ base*height Area= ½*a*b*sineC A c B Introduction (continued) 7. Draw a diagram to represent the information. ( Do not solve this problem.) A triangular plot of land has interior angles A=95° and C=68°. If the side between these angles is 115 yards long, what are the lengths of the other two sides? Playing with a the triangle If we think of h as being opposite to both A and B, then C b A a h c Let’s drop an altitude and call it h. sin A h and sin B b B h a Let’s solve both for h. h b sin A and h a sin B This means b sin A a sin B and dividing by ab . sinA a sin B b C If I were to drop an altitude to side a, I could come up with a b A c sin B B sin C b c Putting it all together gives us the Law of Sines. sin A Taking reciprocals, we have sin B a b a b sin A sin B sin C c c sin C What good is it? The Law of Sines can be used to solve the following types of oblique triangles •Triangles with 2 known angles and 1 side (AAS or ASA) •Triangles with 2 known sides and 1 angle opposite one of the sides (SSA) With these types of triangles, you will almost always have enough information/data to fill out one of the fractions. sin A a sin B b sin C c Example 1 (AAS) C Let A 45 , B 50 , a 30 85° b 50° 45° A a =30 c B I’m given both pieces for sinA/a and part of sinB/b, so we start there. sin 45 30 Once I have 2 angles, I can find the missing angle by subtracting from 180. C=180 – 45 – 50 = 85° sin 50 b Cross multiply and divide to get b sin 45 30 sin 50 b 30 sin 50 sin 45 C 85° b 3 2 .5 sin A a =30 50° 45° c 42.3 A b sin C a c sin 45 B 30 sin 50 sin 45 sin 85 30 c c sin 4 5 3 0 sin 8 5 c 3 0 sin 8 5 sin 4 5 U sing a calculator, b 32.5 We’ll repeat the process to find side c. Remember to avoid rounded values when computing. U sing a calculator, c 42.3 We’re done when we know all 3 sides and all 3 angles. Example 2 ASA A triangular plot of land has interior angles A=95° and C=68°. If the side between these angles is 115 yards long, what are the lengths of the other two sides? The Ambiguous Case (SSA) Three possible situations 1. No such triangle exists. 2. Only one such triangle exists. 3. Two distinct triangles can satisfy the conditions. Example 3(SSA) Use the Law of Sines to solve the triangle. A = 26, a = 21 inches, b = 5 inches C a sin A 21 sin 26 sin B a = 21 in b sin B 5 26 B b = 5 in A a > b : One Triangle sin B 1 5 sin 26 B sin ( 21 B 5.99 5 sin 26 21 ) Example 3(SSA) Use the Law of Sines to solve the triangle. A = 26, a = 21 inches, b = 5 inches C C 180 ( 26 5 . 99 ) 148 . 01 a = 21 in B 21 sin 26 c 5.99° 26 b = 5 in A c sin 148 . 01 21 sin 148 . 01 sin 26 25 . 38 a > b : One Triangle Example 4(SSA) Use the Law of Sines to solve the triangle. A = 76, a = 18 inches, b = 20 inches a b sin A sin B a = 18 in 18 20 sin 7 6 sin B sin B 1 .0 7 8 There is no angle whose sine is 1.078. B b = 20 in h 76 a < h:None h b sin A h 20 sin 76 h 19 . 4 There is no triangle satisfying the given conditions. A Example 5(SSA) Let A = 40°, b = 10, and a = 9. C We have enough information for a=9 b = 10 sin A sin B a sin 40 9 h=6.4 b A sin B c B h < a < b: Two 10 Cross multiply and divide 40° 9 sin B 10 sin 40 sin B 10 sin 40 9 h b sin A h 10 sin 40 h 6 .4 C To get to angle B, you must unlock sin using the inverse. sin 1 sin B sin B sin 1 1 10 sin 40 9 10 sin 40 45.6 9 Once you know 2 angles, find the third by subtracting from 180. b = 10 A a=9 45.6° 40° B c =14.0 sin A sin C a c sin 40 C = 180 – (40 + 45.6) = 94.4° We’re ready to look for side c. 94.4° sin 94.4 9 c c sin 4 0 9 sin 9 4 .4 c 9 sin 9 4 .4 sin 4 0 1 4 .0 Example 5 Finding the Second Triangle Let A = 40°, b = 10, and a = 9. Start by finding B’ = 180 - B Now solve this triangle. b= 10 C’ 40° a=9 B’ B = 45.6° A b= 10 a=9 A 40° 134.4° B’ c’ B’ = 180 – 45.6 = 134.4° C’ 5.6° Next, find C’ = 180 – (40 + 134.4) C’ = 5.6° b= 10 a=9 134.4° 40° A c =1.4 B’ sin A sin C a c sin 40 sin 5.6 9 c c sin 40 9 sin 5.6 c 9 sin 5.6 sin 40 1.4 A New Way to Find Area C We all know that A = ½ bh. And a few slides back we found this. h b sin A and h a sin B b A a h c Area 1 bc sin A 1 ab sin C 1 ac sin B 2 2 2 Area = ½*product of two given sides * sine of the included angle B Example 6 Finding the area of the triangle Find the area of a triangle with side a = 10, side b = 12, and angle C = 40°. Example 7 Application Two fire ranger towers lie on the east-west line and are 5 miles apart. There is a fire with a bearing of N27°E from tower 1 and N32°W from tower 2. How far is the fire from tower 1? The angle at the fire is 180° - (63° + 58°) = 59°. N 1 S N x 63° 58° 2 5 mi S 22 Example 8 Application http://www.mcs.uvawise.edu/dbl5h/resources/latex_examples/files/triglawsines.pdf Practice p.398-400 #s 2-38, even 24