3 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.
4 IntroductionTo 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*heightArea= ½*a*b*sineCABCabcMake sure students understand the opposite side and the 2 adjacent sides (vs. adjacent and hypotenuse).
5 Introduction (continued) 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?Make sure students understand the opposite side and the 2 adjacent sides (vs. adjacent and hypotenuse).
6 Playing with a the triangle If we think of h as being opposite to both A and B, thenABCabchLet’s solve both for h.Let’s drop an altitude and call it h.This means
7 ABCabcIf I were to drop an altitude to side a, I could come up withPutting it all together gives us the Law of Sines.Taking reciprocals, we have
8 What good is it?The Law of Sines can be used to solve the following types of oblique trianglesTriangles 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.
9 Example 1 (AAS)ABCabcI’m given both pieces for sinA/a and part of sinB/b, so we start there.45°50°=3085°Cross multiply and divide to getOnce I have 2 angles, I can find the missing angle by subtracting from C=180 – 45 – 50 = 85°
10 We’ll repeat the process to find side c. 45°50°=3085°ABCabcMake sure students can get the right answer with their calculator.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.
11 Example 2 ASAA 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?Check for calculator ability
12 The Ambiguous Case (SSA) Three possible situationsNo such triangle exists.Only one such triangle exists.Two distinct triangles can satisfy the conditions.
13 Example 3(SSA)Use the Law of Sines to solve the triangle. A = 26, a = 21 inches, b = 5 inchesCAb = 5 ina = 21 in26Ba > b : One Triangle
14 Example 3(SSA)Use the Law of Sines to solve the triangle. A = 26, a = 21 inches, b = 5 inchesCAb = 5 ina = 21 in26B5.99°a > b : One Triangle
15 Example 4(SSA)Use the Law of Sines to solve the triangle. A = 76, a = 18 inches, b = 20 inchesABb = 20 ina = 18 in76ha < h:NoneThere is no angle whose sine isThere is no triangle satisfying the given conditions.
16 Example 5(SSA) Let A = 40°, b = 10, and a = 9. C We have enough information for40°b = 10a = 9h=6.4AcBh < a < b: TwoCross multiply and divide
17 To get to angle B, you must unlock sin using the inverse. 40°b = 10To get to angle B, you must unlock sin using the inverse.94.4°a = 945.6°ABc=14.0Once you know 2 angles, find the third by subtracting from 180.C = 180 – ( ) = 94.4°We’re ready to look for side c.
18 Example 5 Finding the Second Triangle Let A = 40°, b = 10, and a = 9.Start by finding B’ = BNow solve this triangle.b= 10a = 9AB’C’b= 10a = 940°134.4°c’40°B’B = 45.6°AB’ = 180 – 45.6 = 134.4°
20 A New Way to Find Area A B C a b c We all know that A = ½ bh. And a few slides back we found this.hArea = ½*product of two given sides * sine of the included angle
21 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°.
22 Example 7 ApplicationTwo fire ranger towers lie on the east-west line and are 5miles 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°.NNx63°158°25 miSSApplication
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