# Trigonometric Ratios Contents IIntroduction to Trigonometric Ratios UUnit Circle AAdjacent, opposite side and hypotenuse of a right angle.

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Trigonometric Ratios Contents IIntroduction to Trigonometric Ratios UUnit Circle AAdjacent, opposite side and hypotenuse of a right angle triangle. TThree types trigonometric ratios CConclusion

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

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

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

Adjacent side Opposite side  hypotenuse

There are 3 kinds of trigonometric ratios we will learn. sine ratio cosine ratio tangent ratio Three Types Trigonometric Ratios

Sine Ratios  Definition of Sine Ratio.  Application of Sine Ratio.

Definition of Sine Ratio.Sine Ratio  1 If the hypotenuse equals to 1 Sin  = Opposite sides

Definition of Sine Ratio.Sine Ratio  For any right-angled triangle Sin  = Opposite side hypotenuses

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

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 

Cosine Ratios  Definition of Cosine.  Relation of Cosine to the sides of right angle triangle.

Definition of Cosine Ratio.Cosine Ratio  1 If the hypotenuse equals to 1 Cos  = Adjacent Side

Definition of Cosine Ratio.Cosine Ratio  For any right-angled triangle Cos  = hypotenuses Adjacent Side

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

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 

Tangent Ratios  Definition of Tangent.  Relation of Tangent to the sides of right angle triangle.

Definition of Tangent Ratio.  For any right-angled triangle tan  = Adjacent Side Opposite Side

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

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 =

Conclusion Make Sure that the triangle is right-angled

END

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