Trigonometry deals with relationships between sides and angles of triangles, especially right triangles. It has been used for thousands of years in fields like astronomy, navigation, architecture, engineering, and more modern fields like digital imaging and computer graphics. Trigonometric functions define ratios between sides of a right triangle and are used to solve for unknown sides and angles. Common applications include calculating distances, heights, satellite positioning, and modeling waves and vibrations.

1. By Nittaya Noinan Kanchanapisekwittayalai phetchabun school Grade 10 Trigonometry

6. Trigonometry ( 三角幾何 ) means “Triangle” and “Measurement” Introduction Trigonometric Ratios In F.2 we concentrated on right angle triangles .

7. Unit Circle A Unit Circle Is a Circle With Radius Equals to 1 Unit.(We Always Choose Origin As Its centre) 1 units x Y

8. Trigonometry Trigonometry is the branch of mathematics that looks at the relationship between the length of the sides and the angles in a right angled triangle. It helps us calculate the length of unknown sides, without drawing the triangle, and it also helps us calculate the size of an unknown angle without having to measure it. Let us look at a right angle triangle.

9. Trigonometry O F C Hypotenuse x 0 Opposite Adjacent O F C Hypotenuse y 0 Opposite Adjacent The ‘Hypotenuse’ is always opposite the right angle The ‘Opposite’ is always opposite the angle under investigation. The ‘Adjacent’ is always alongside the angle under investigation.

10. Trigonometry 6 3 O F C Let us look at the ratio or fraction of the opposite over the adjacent. Let us now split this triangle up. 27 0

11. Trigonometry 4 2 2 3 O E F B C Let us look at the ratio or fraction of the opposite over the adjacent Let us split it again. 27 0

12. Adjacent , Opposite Side and Hypotenuse of a Right Angle Triangle .

13. Adjacent side Opposite side hypotenuse 

14. hypotenuse Adjacent side Opposite side 

19. Exercise 1 In the figure, find sin  Sin  = Opposite Side hypotenuses = 4 7  = 34.85  (corr to 2 d.p.)  4 7

20. Exercise 2 11 In the figure, find y Sin35  = Opposite Side hypotenuses y 11 y = 6.31 (corr to 2.d.p.) 35 ° y Sin35  = y = 11 sin35 

24. Exercise 3  3 8 In the figure, find cos  cos  = adjacent Side hypotenuses = 3 8  = 67.98  (corr to 2 d.p.)

25. Exercise 4 6 In the figure, find x Cos 42  = Adjacent Side hypotenuses 6 x x = 8.07 (corr to 2.d.p.) 42 ° x Cos 42  = x = 6 Cos 42 

28. Exercise 5  3 5 In the figure, find tan  tan  = adjacent Side Opposite side = 3 5  = 78.69  (corr to 2 d.p.)

29. Exercise 6 z 5 In the figure, find z tan 22  = adjacent Side Opposite side 5 z z = 12.38 (corr to 2 d.p.) 22  tan 22  = 5 tan 22  z =

30. Conclusion Make Sure that the triangle is right-angled

33. Values of Trigonometric function 0 30 45 60 90 Sine 0 0.5 1/  2  3/2 1 Cosine 1  3/2 1/  2 0.5 0 Tangent 0 1/  3 1  3 Not defined Cosecant Not defined 2  2 2/  3 1 Secant 1 2/  3  2 2 Not defined Cotangent Not defined  3 1 1/  3 0

38. Problem solved using trigonometric ratios CLICK HERE!

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