sin(90°-α)=cosα cos(90°-α)=sinα tan(90°-α)=cotα cot(90°-α)=tanα sinα /cosα =tanα sin 2 α+cos 2 α=1 sinα =tanα ·cosα cosα =cotα ·sinα cotα =cosα ·cscα tanα ·cotα =1 Connections
Trigonometric Ratios of Acute Angles
Rt △ ABC ∠ ACB=90° To ∠ BAC : Opposite: a=BC Hypotenuse: h=AB Adjacent: b=AC NameAbbreviationExpressionSinesina/h Cosinecosb/h Tangenttana/b Cotangentcotb/a Secantsech/b Cosecantcsch/a
Example 1. Rt △ ABC ∠ ACB=90° BC=6 AB=10 sin ∠ B= cos ∠ B= tan ∠ B=
AC= ( )=8 AC= ( )=8 sin ∠ B= = = sin ∠ B= = = cos ∠ B= = = cos ∠ B= = = tan ∠ B= = = tan ∠ B= = = 6 10
Example 2. 0°< α <90° sin α = 0°< α <90° sin α = cos α = cos α =
Fold the △ CDE along CE , point D is just on AB. Calculate the value of tan ∠ AFE..
∵ AB = 10, rectangle ABCD ∴ DC=10 ∴ FC=10 ∵ FC=10,BC=8,Rt △ FCB ∴ FB=6 ∴ AF=4 If AE=x ∵ AE+ED=8, ED=EF ∴ AE+EF=8 ∴ EF=8-x ∴ x =(8-x) 2 ∴ x=3 ∴ AE=3 ∴ tan ∠ AFE=AE/AF=3/4
Square sin 2 α +cos 2 α =1 cos2 α =cos 2 α -sin 2 α =1-2sin 2 α =2cos 2 α- 1 sin2 α =2sin α cos α tan 2 α +1=1 / cos 2 α 2sin 2 α =1-cos2 α cot 2 α +1=1 / sin 2 α =1-2sin 2 α =2cos 2 α- 1 sin2 α =2sin α cos α tan 2 α +1=1 / cos 2 α 2sin 2 α =1-cos2 α cot 2 α +1=1 / sin 2 α Product sin α =tan α× cos α cos α =cot α× sin α tan α =sin α× sec α cot α =cos α× csc α sec α =tan α× csc α csc α =sec α× cot α Reciprocal tan α× cot α =1 sin α× csc α =1 cos α× sec α =1 Quotient sin α/ cos α =tan α =sec α/ csc α cos α/ sin α =cot α =csc α/ sec α
Rt △ ABC ,∠ C=90°, cosA=1/2, ∠ B=. 3. Rt △ ABC ,∠ C=90°, BC=a, c=___. (A)c=a sinA (C)c=b tanA (B)c=a/sinA (D)c=a/cosA A 30 B.
Solved 54 CHAPTER 2 Acute Angles and Right Triangles Suppose
How do you find the values of all six trigonometric functions of a right triangle ABC where C is the right angle, given a=20, b=21, c=29?
SOLVED: Two students describe the sides of right triangle ABC in relation to angle B. Tomas says that AB is the hypotenuse, AC is the opposite side, and BC is the adjacent
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