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There is no doubt that the 3 PTRs are extremely useful when solving problems modeled on a right triangle. Unfortunately, the world does not consist.

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Presentation on theme: "There is no doubt that the 3 PTRs are extremely useful when solving problems modeled on a right triangle. Unfortunately, the world does not consist."— Presentation transcript:

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3 There is no doubt that the 3 PTRs are extremely useful when solving problems modeled on a right triangle. Unfortunately, the world does not consist only of right triangles…

4 As a matter of fact, right triangles end up being more of a rarity than commonplace. Does that mean when we come across a situation that can only be modeled with a non-right triangle that we abandon our pursuit?….

5 No Way!!!! There exists 2 Laws of Trigonometry that allow one to solve problems that involve non-right Triangles:

6 A triangle is uniquely determined by two angles and a particular side A C B c b a O1O1 O2O2

7 If a corresponding angle and side are known, they form an “opposing pair” A C B c b a O1O1 O2O2

8 The Sine Law can be used to determine an unknown side or angle given an “opposing pair” A C B c b a O1O1 O2O2

9 Find the length of b A C B c b 5 30 o 65 o

10 Construct CN with height h A C B c b 5 30 o 65 o N h

11 By the right triangle SIN ratio A C B c b 5 30 o 65 o N h Sin 30 o = h b Sin 65 o = h 5 h 30 o 65 o

12 Solve both equations for h X b Sin 30 o = h b Sin 65 o = h 5 X 5 bSin30 o = h h = 5Sin65 o Because the equations are equal bSin30 o = 5Sin65 o

13 b = 5Sin65 o Sin30 o b = 9.1 Consider the general case:

14 A C c b a N h B Sin A = h b Sin B = h a bSinA = aSinB

15 a a bSinA = SinB a b ab b SinA = SinB a

16 Extend this to all 3 sides of a triangle, and the Sine Law is generated! b SinA = SinB a c = SinC

17 Find the length of a a A C c o 57 o N a Sin73 o =24 Sin57 o a = 27.4

18 5.9 O 10.3 O 2.9 km Find h h

19 5.9 O 10.3 O 2.9 km Find h 1. Find O O O = 180 O – 5.9 O – 10.3 O = O

20 5.9 O 10.3 O 2.9 km O Find X X X SIN 10.3 O 2.9 SIN163.8 O = X = 1.86km

21 5.9 O 10.3 O 2.9 km Find h h 1.86 km SIN 5.9 O = h 1.86 km h = m

22 The Ambiguous Case

23 Find A o A 11 SinA = 9 Sin48 o A=65.3 o Does that make sense? No Way!!!

24 Side 9 can also be drawn as: o A Could A be 65 o in this case?

25 This type of discrepancy is called the “Ambiguous Case” Be sure to check the diagram to see which answer fits: O, or 180 o - O

26 Page 366


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