1. Trigonometry Angles & Degree Measure Trigonometric Ratios Right Triangle Problems Law of Sines Law of Cosines Problem Solving

2. Angles and Degree Measure Angles are always measured from the right x- axis to the terminal side of the angle.   

3. Angles and Degree Measure Angles are often expressed in units of degrees, but can be into smaller increments called minutes and seconds. These units are used to rename the decimal part of a degree measure. 1 o = 60’ and 1’ = 60 “ 30.25 o = 30 o + (.25 o )(60’/1 o ) = 30 o 15’ 0” 30.29 o = 30 o + (.29 o )(60’/1 o ) = 30 o 17.4’ the 0.4’ is then multiplied again to get seconds; (.4’)(60”/1 ’ ) = 24”. This gives a final answer of 30 o 17’ 24”

4. Angles and Degree Measure Solving for degrees from minutes and seconds is also possible. Example: 30 o 27’ 45” = the sum of each of the values expressed in degree units. 30 + (27’)(1 o /60’) + (45”)(1 o /3600”) = 30 + 0.45 +.0125 = 30.4625 0 = 30.46 o

5. Angles and Degree Measure Coterminal angles are angles that have the same starting and ending point but are expressed in different values based on the direction of the rotation or the number of rotations.  Going around CCW and returning to the terminal side results in (360 + ). The rotations can be any integer value. Even with increasing angle measure, each angle is still equal to the original angle. Going around CW is (-) and results in (-360 + ).

6. Angles and Degree Measure Reference angles, denoted by , are always measured from the x-axis of the quadrant that the terminal sides is located back to the terminal side. Reference angles are positive.  

7. Trigonometric Ratios Sine Function – sin  Sine ratio is the value of the length of the opposite side divided by the hypotenuse of a right triangle. sin  = a/c Values for sin  oscillate between 0 to 1 to 0 to –1 to 0 b a c  Right Triangle a – opposite side b – adjacent side c – hypotenuse  - theta (reference )

8. Trigonometric Ratios Sine Function – sin  Sine ratio works for angle values between 0 and 360 o and then repeats. sin  = a/c is also expressed as follows: sin  = opposite side hypotenuse b a c  Right Triangle a – opposite side b – adjacent side c – hypotenuse  - theta (reference )

9. Trigonometric Ratios Cosine Function – cos  Cosine ratio is the value of the length of the adjacent side divided by the hypotenuse of a right triangle. cos  = b/c Values for cos  oscillate between 1 to 0 to -1 to 0 to 1 b a c  Right Triangle a – opposite side b – adjacent side c – hypotenuse  - theta (reference )

10. Trigonometric Ratios Cosine Function – cos  Cosine ratio works for angle values between 0 and 360 o and then repeats. cos  = b/c is also expressed as follows: cos  = adjacent side hypotenuse b a c  Right Triangle a – opposite side b – adjacent side c – hypotenuse  - theta (reference )

11. Trigonometric Ratios Tangent Function – tan  Tangent ratio is the value of the length of the opposite side divided by the adjacent side of a right triangle. tan  = a/b Values for tan  oscillate between  and -  within each 180 o increment b a c  Right Triangle a – opposite side b – adjacent side c – hypotenuse  - theta (reference )

12. Trigonometric Ratios Tangent Function – tan  Tangent ratio works for angle values between 0 and 360 o and then repeats. tan  = a/b is also expressed as follows: tan  = opposite side adjacent side b a c  Right Triangle a – opposite side b – adjacent side c – hypotenuse  - theta (reference )

13. Trigonometric Ratios Sine Function change in values as  changes: Sin 0 = 0.0000 Sin 30 = 0.5000 Sin 45 = 0.7071 Sin 60 = 0.8660 Sin 75 = 0.9659 Sin 90 = 1.0000 Sin 120 = 0.8660 Sin 135 = 0.7071 Sin 150 = 0.5000 Sin 180 = 0.0000 Afterwards, sin repeats with negative values b a c  Right Triangle a – opposite side b – adjacent side c – hypotenuse  - theta (reference )

14. Trigonometric Ratios Cosine Function change in values as  changes: Cos 0 = 1.0000 Cos 30 = 0.8660 Cos 45 = 0.7071 Cos 60 = 0.5000 Cos 75 = 0.2588 Cos 90 = 0.000 Cos 120 = -0.5000 Cos 135 = -0.7071 Cos 150 = -0.8660 Cos 180 = -1.0000 Afterwards, cos  repeats values b a c  Right Triangle a – opposite side b – adjacent side c – hypotenuse  - theta (reference )

15. Trigonometric Ratios Tangent Function change in values as  changes: Tan 0 = 0.0000 Tan 30 = 0.5773 Tan 45 = 1.0000 Tan 60 = 1.7321 Tan 75 = 3.7321 Tan 90 = undefined Tan 120 = -1.7321 Tan 135 = -1.0000 Tan 150 = - 0.5773 Tan 180 = 0.0000 Afterwards, tan  repeats values b a c  Right Triangle a – opposite side b – adjacent side c – hypotenuse  - theta (reference )

16. Trigonometric Ratios Note that with each function there is a relation between an angle measure and a ratio of sides of the right triangle. The function of an angle is equal to the decimal value of the specific ratio of sides. b a c  Right Triangle a – opposite side b – adjacent side c – hypotenuse  - theta (reference )

17. Trigonometric Ratios Likewise, the specific ratio is equal to the function of an angle. Ex. Let sin  = 6/10 =.6000 Then  = Sin –1.6000 = 8 6 10  Right Triangle a – opposite side b – adjacent side c – hypotenuse  - theta (reference ) 36.9 o

18. Trigonometric Ratios Likewise, the specific ratio is equal to the function of an angle. Ex. Let sin  = 1/2 =.5000 Then  = Sin –1.5000 = b = 1.732 a = 1 C = 2  Right Triangle a – opposite side b – adjacent side c – hypotenuse  - theta (reference ) 30.0 o

19. Trigonometric Ratios Likewise, the specific ratio is equal to the function of an angle. Ex. Let cos  = 8/10 =.8000 Then  = cos –1.8000 = 8 6 10  Right Triangle a – opposite side b – adjacent side c – hypotenuse  - theta (reference ) 36.9 o

20. Trigonometric Ratios Likewise, the specific ratio is equal to the function of an angle. Ex. Let cos  = 1.732/2 =.8660 Then  = cos –1.8660 = 1.732 1 2  Right Triangle a – opposite side b – adjacent side c – hypotenuse  - theta (reference ) 30.0 o

21. Trigonometric Ratios We can also solve for missing values if we know two of the three variables in a ration. Ex. sin 30 o = 20/r Then r = 20/sin30 o = 20/0.5 = x 20 r 30 o Tan 30 o = 20/x Then x = 20/tan30 o =   is the complementary angle of  and so  = 90 – 30 or 60 o 40 34.6

22. Right Triangle Problems Any situation that includes a right triangle, becomes solvable with trigonometry.  Angle of Elevation  Angle of Depression

23. Right Triangle Problems  This diagram has an inscribed triangle whose hypotenuse is also the diameter of the circle. A B C Given:  = 40 o, AB = 6 cm Determine the area of the shaded region. Solution: Sin 40 o = 6/AC; AC = 6/ sin40 o = 9.33 cm AC/2 = radius of the circle = 4.67 cm; BC = 6/tan 40 o =7.15 cm Ao Ao = r2 r2 = 68.4 cm 2, A A = ½bh = 21.5 cm 2 ; A = 46.9 cm 2

24. Right Triangle Problems  This diagram has a right triangle within a cone that can be used to solve for the surface area and volume of the cone. h r l Given:  = 55 o, l = 6 cm Determine the area and volume of the cone. Solution: Sin 55 o = h/l; h = 6 x sin55 o = 4.91 cm tan 55 o = h/r; r = h/ tan 55 o = 4.91/ tan 55 o = 3.44 cm A cone = r(r + l) =  (3.44)(3.44 + 6)cm 2 ; A cone = 102 cm 2 V cone = ⅓ r 2 h = ⅓  (3.44) 2 (4.91)cm 3 ; V cone = 60.8 cm 3

25. Right Triangle Problems The angle of elevation of a ship at sea level to a neighboring lighthouse is 2 o. The captain knows that the top of lighthouse is 165 ft above sea level. How far is the boat from the lighthouse? h x Solution: tan 2 o = h/x; x = h/tan 2o2o x = 165/ tan 2 o = 4725 ft; 1 mile = 5280 ft, so the ship is.89 mile away from the lighthouse. 

26. Function Values – Unit Circle The unit circle is a circle with a radius of 1 unit.  x y r = 1 Sin  = y; Cos  = x; Tan  = y/x The coordinate of P will lead to the value of x and y which in turn leads to the values for sine, cosine, and tangent. Use the reference angle in each quadrant and the coordinates to solve for the function value.

27. Function Values – Unit Circle    Quadrant I – Angle =  Quadrant II – Angle = (180 - ) Quadrant III – Angle = (180 + ) Quadrant IV – Angle = (360 - ) The reference angle is always measured in its quadrant from the x – axis.

28. Function Values – Unit Circle    Quadrant I – P (x,y) Quadrant II – P (-x, y) Quadrant III – P (-x, -y) Quadrant IV – P (x, -y) With the values of changing from (+) to (-) in each quadrant, and with the functions of sine, cosine, and tangent valued with ratios of x and y, the functions will also have the sign values of the variables in the quadrants.

29. Function Values – Unit Circle Degrees Sine Cosine Tangent 030456090120135150 0½ 2/23/2 1 2/2 ½ 180210225240270300315330 0-½ -2/2-3/2 -3/2-2/2 -½ Degrees Sine Cosine Tangent 1 3/22/2 ½0-½ -2/2-3/2 0 3/3 1 33 - -3-3 -3/3 -3/2-2/2 -½0½ 2/23/2 0 3/3 1 33 - -3-3 -3/3

30. Function Values – Unit Circle Example: Cos 240 = _______ 60 P (-½, -3/2) - ½ Example: Sin 240 = _______ Example: Tan 240 = _______ 33 -3/2 240

31. Law of Sines Law of Sines – For a triangle with angles A, B, C and sides of lengths of a, b, c the ratio of the sine of each angle and its opposite side will be equal. Sin A = B = C a b c A B C c b a h Proof: Sin A = h/b; Sin B = h/a h = b Sin A, h = a Sin B b Sin A = a Sin B; Sin A = Sin B ab

32. Law of Sines Proof, continued: A B C c b a k k is the height of the same triangle from vertex A Sin C = k/b and Sin B = k/c k = b Sin C and k = c Sin B b Sin C = c Sin B; Sin B = Sin C b b c Conclusion: Sin A = Sin B = Sin C a b c

33. Law of Sines Let’s take a closer look: Suppose A < 90 o A c a If a = c Sin A, one solution exists a A c c Sin A If a < c Sin A, no solution exists

34. Law of Sines Let’s take a closer look: Suppose A < 90 o A a c If a > c Sin A and a  c, one solution exists c Sin A a A c If c Sin A < a < c, two solution exists a

35. Law of Sines Let’s take a closer look: Suppose A  90 o A a c If a < c, no solution exists A a If a > c, one solution exists c

36. Law of Cosines A BC D b a c a -xx h b 2 = h 2 + x 2 ; h 2 = b 2 - x 2 Cos C = x/b ; x = b Cos C c 2 = h 2 + (a-x) 2 ; c 2 = h 2 + a 2 –2ax + x 2 Substitute and we get: c 2 = (b 2 - x 2 )+ a 2 –2a(bCos C) + x 2 c 2 = b 2 + a 2 – 2abCos C For any triangle given two sides and an included angle, a 2 = b 2 + c 2 – 2bcCos A b 2 = a 2 + c 2 – 2acCos B c 2 = b 2 + a 2 – 2abCos C PROOF:

37. Law of Cosines - example Given: A = 50 o ; AB = 8 cm, AC = 14 cm x 2 = 8 2 + 14 2 – 2(8)(14)Cos50 Find x. A B C x x 2 = 260 – 144; x 2 = 116 x = 10.8 cm

38. Law of Cosines – Hero’s Formula For any triangle, the area of the triangle can be determined with Hero’s Formula. s = ½(a + b + c) where s = semi-perimeter of the triangle. A =  s(s – a)(s – b)(s – c)

39. Law of Cosines – Hero’s Formula Given: ABC; a = 7, b = 24, c = 25 A =  28(28 – 7)(28 – 24)(28 – 25) = 84 cm 2 7cm 25 cm 24 cm S = ½ (7 + 24 + 25) = 28 Check: A = ½(b)(h) = ½(7)(24) = 84 cm 2