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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).

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## Pre-calculus lesson 6.1 Law of Sines

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**Pre-calculus lesson 6.1Law 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.**C**a b A B c Introduction • To solve an oblique triangle, we need to be given at least one side and then any other two parts of the triangle. • The sum of the interior angles is 180°. • Why can’t we use the Pythagorean Theorem? • State the Law of Sines. • Area = ½ base*height • Area= ½*a*b*sineC**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?**C**a b B A c Playing with a the triangle If we think of h as being opposite to both A and B, then h Let’s solve both for h. Let’s drop an altitude and call it h. This means**C**a b B A c If I were to drop an altitude to side a, I could come up with Putting it all together gives us the Law of Sines. Taking reciprocals, we have**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.**C**85° a b =30 B 50° 45° A c Example 1 (AAS) I’m given both pieces for sinA/a and part of sinB/b, so we start there. Cross multiply and divide to get Once I have 2 angles, I can find the missing angle by subtracting from 180. C=180 – 45 – 50 = 85°**C**85° a b =30 B 50° 45° A c We’ll repeat the process to find side c. Remember to avoid rounded values when computing. 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 No such triangle exists. Only one such triangle exists. Two distinct triangles can satisfy the conditions.**a = 21 in**b = 5 in 26 A Example 3(SSA) Use the Law of Sines to solve the triangle. A = 26, a = 21 inches, b = 5 inches C B a > b : One Triangle**a = 21 in**b = 5 in 26 A Example 3(SSA) Use the Law of Sines to solve the triangle. A = 26, a = 21 inches, b = 5 inches C B 5.99° a > b : One Triangle**b = 20 in**a = 18 in 76 B A Example 4(SSA) Use the Law of Sines to solve the triangle. A = 76, a = 18 inches, b = 20 inches h a < h:None There is no angle whose sine is 1.078. There is no triangle satisfying the given conditions.**b = 10**40° Example 5(SSA) Let A = 40°, b = 10, and a = 9. C We have enough information for a = 9 h=6.4 A c B h < a < b: Two Cross multiply and divide**b = 10**40° C To get to angle B, you must unlock sin using the inverse. 94.4° a = 9 45.6° A B c =14.0 Once you know 2 angles, find the third by subtracting from 180. C = 180 – (40 + 45.6) = 94.4° We’re ready to look for side c.**C’**b= 10 a = 9 134.4° 40° A B’ c’ Example 5Finding the Second Triangle Let A = 40°, b = 10, and a = 9. Start by finding B’ = 180 - B Now solve this triangle. b= 10 a = 9 B = 45.6° 40° B’ A B’ = 180 – 45.6 = 134.4°**C’**b= 10 a = 9 134.4° 40° A B’ c Next, find C’ = 180 – (40 + 134.4) C’ = 5.6° 5.6° =1.4**C**a b B A c A New Way to Find Area We all know that A = ½ bh. And a few slides back we found this. h Area = ½*product of two given sides * sine of the included angle**Example 6Finding the area of the triangle**Find the area of a triangle with side a = 10, side b = 12, and angle C = 40°.**Application**• 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 N x 63° 58° 1 2 5 mi S S**Example 8 Application**http://www.mcs.uvawise.edu/dbl5h/resources/latex_examples/files/triglawsines.pdf**Application**Practice p.398-400 #s 2-38, even

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