Trigonometric Ratios of Acute Angles. Rt △ ABC ∠ ACB=90° To
By A Mystery Man Writer
Description
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.
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.
Geometry - Flip eBook Pages 1-50
Triangle calculator
Answered: Are these triangles similar? What are…
Trigonometric Ratios of Acute Angles. Rt △ ABC ∠ ACB=90° To ∠ BAC : Opposite: a=BC Hypotenuse: h=AB Adjacent: b=AC NameAbbreviationExpressionSinesina/h. - ppt download
In right angled triangle, ABC, angle ABC=90. AM and CN are medians drawn from A and C respectively to BC and AB. How can we show that 4 (CM^2+CN^3) =5AC^2? - Quora
Right Angled Triangle: Explanation, Formula and solved examples
ABC is an acute-angled triangle. The external bisector of ∠BAC meets the line BC at point N. M is the midpoint of BC. P and Q are two points on AN such
Refer to the right triangle ABC with C = 90 degrees. Use the given information to find the six trigonometric functions of A. b = 3, c = 5
Draw acute triangle ABC with measure angle A = 30 degrees. Draw altitude CD from C to AB. If CD=5, find AC. (a) 5 sqrt(3) (b) 5 sqrt(3)/3 (c) 10 sqrt(3)/3 (d)
TRIGONOMETRY TEST N SOLUTION (O'LEVEL MATHEMATICS) Share, PDF, Trigonometry
Trigonometric Ratios of Acute Angles. Rt △ ABC ∠ ACB=90° To ∠ BAC : Opposite: a=BC Hypotenuse: h=AB Adjacent: b=AC NameAbbreviationExpressionSinesina/h. - ppt download
The vertices of triangle ABC lie on the sides of equilateral triangle DEF, with A on EF, B on FD, and C on DE. If CD = 3, CE = BD =
Unlocking the Secrets of Trigonometric Ratios of Acute Angles
Exploring Trigonometric Ratios for Angles Greater than 90° (MCR3U)
Given that triangle ABC=triangle PQR such that angle B= (2x+30) degree, angle Q=55 degree. What is the value of x? - Quora
from
per adult (price varies by group size)