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PRE-CALCULUS LESSON 6.1 LAW OF SINES Spring 2011.

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Presentation on theme: "PRE-CALCULUS LESSON 6.1 LAW OF SINES Spring 2011."— Presentation transcript:

1 PRE-CALCULUS LESSON 6.1 LAW OF SINES Spring 2011

2 Question! How to measure the depth?

3 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. At the end of this lesson you should be able to

4 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. 2. The sum of the interior angles is 180°. 3. Why cant we use the Pythagorean Theorem? 4. State the Law of Sines. 5. Area = ½ base*height 6. Area= ½*a*b*sineC A B C a b c

5 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?

6 Playing with a the triangle A B C a b c Lets drop an altitude and call it h. h If we think of h as being opposite to both A and B, then Lets solve both for h. This means

7 A B C a b 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

8 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.

9 Example 1 (AAS) A B C a b c Once I have 2 angles, I can find the missing angle by subtracting from 180. C=180 – 45 – 50 = 85° 45° 50° =30 85° Im given both pieces for sinA/a and part of sinB/b, so we start there. Cross multiply and divide to get

10 45° 50° =30 85° A B C a b c Well repeat the process to find side c. Remember to avoid rounded values when computing. Were done when we know all 3 sides and all 3 angles.

11 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?

12 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.

13 Example 3(SSA) Use the Law of Sines to solve the triangle. A = 26, a = 21 inches, b = 5 inches A b = 5 in a = 21 in 26 a > b : One Triangle B C

14 Example 3(SSA) Use the Law of Sines to solve the triangle. A = 26, a = 21 inches, b = 5 inches A b = 5 in a = 21 in 26 a > b : One Triangle B C 5.99°

15 Example 4(SSA) Use the Law of Sines to solve the triangle. A = 76, a = 18 inches, b = 20 inches There is no angle whose sine is There is no triangle satisfying the given conditions. AB b = 20 in a = 18 in 76 a < h:None h

16 Example 5(SSA) Let A = 40°, b = 10, and a = 9. A B C 40° b = 10 a = 9 We have enough information for Cross multiply and divide c h < a < b: Two h=6.4

17 To get to angle B, you must unlock sin using the inverse. 40° b = 10 A B C a = ° Once you know 2 angles, find the third by subtracting from 180. C = 180 – ( ) = 94.4° 94.4° c Were ready to look for side c. =14.0

18 Example 5 Finding the Second Triangle A 40° b= 10 a = 9 B = 45.6° B Start by finding B = B B = 180 – 45.6 = 134.4° Now solve this triangle. AB C b= 10 a = 9 40° 134.4° c Let A = 40°, b = 10, and a = 9.

19 AB C b= 10 a = 9 40° 134.4° c Next, find C = 180 – ( ) C = 5.6° 5.6° =1.4

20 A New Way to Find Area We all know that A = ½ bh. And a few slides back we found this. A B C a b c h Area = 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 22 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° mi 63° 58° x N S N S Example 7 Application

23 Example 8 Application

24 24 Application p #s 2-38, even Practice


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