RATIOS IN RIGHT TRIANGLES OPPOSITE SIDE? ADJACENT SIDE? HYPOTENUSE? SINE? COSINE? TANGENT? INVERSE OF TRIGONOMETRIC RATIOS PROBLEM 1a PROBLEM 1b PROBLEM 3c PROBLEM 2c PROBLEM 1c PROBLEM 2b PROBLEM 3a PROBLEM 3b PROBLEM 2a Standards 15, 18, 19 PRESENTATION CREATED BY SIMON PEREZ RHS. All rights reserved

Standard 15: Students use the pythagoream theorem to determine distance and find missing lengths of sides of right triangles. Los estudiantes usan el teorema de Pitágoras para determinar distancia y encontrar las longitudes de los lados de teoremas rectángulos. Standard 18: Students know the definitions of the basic trigonometric functions defined by the angles of a right triangle. They also know and are able to use elementary relationships between them, (e.g., tan(x)=sin(x)/cos(x), etc.) Los estudiantes conocen las definiciones de las funciones básicas trigonométricas definidas para los ángulos de triángulos rectángulos. Ellos también conocen y son capaces de usar relaciones básicas entre ellos. (ej., tan(x)=sin(x)/cos(x), etc.) Standard 19: Students use trigonometric functions to solve for an unknown length of a side of a right triangle, given an angle and a length of a side. Los estudiantes usan funciones trigonométricas para resolver para una longitud desconocida de un triángulo rectángulo, dado un ángulo y la longitud de un lado.

ADJACENT SIDE HYPOTENUSE Standard 18

OPPOSITE SIDE Standard 18

ADJACENT SIDE HYPOTENUSE Standard 18

OPPOSITE SIDE Standard 18

ADJACENT SIDE HYPOTENUSE Standard 18

OPPOSITE SIDE Standard 18

ADJACENT SIDE HYPOTENUSE Standard 18

OPPOSITE SIDE Standard 18

ADJACENT SIDE HYPOTENUSE Standard 18

OPPOSITE SIDE Standard 18

ADJACENT SIDE HYPOTENUSE Standard 18

OPPOSITE SIDE Standard 18

b a c C B A The ADJACENT side to A is side “b” or CA Standard 18

b a c C B A The OPPOSITE side to A is side “a” or BC Standard 18

k m l L K M What is the adjacent side to M? The ADJACENT side to M is side “k” or LM Standard 18

What is the opposite side to M? The OPPOSITE side to M is side “ m ” or KL k m l L K M Standard 18

r t s S R T What is the adjacent side to R? The ADJACENT side to R is side “t” or RS Standard 18

What is the opposite side to R? The OPPOSITE side to R is side “r” or ST r t s S R T Standard 18

z x y Y Z X What is the adjacent side to X? The ADJACENT side to X is side “y” or XZ Standard 18

What is the opposite side to X? The OPPOSITE side to X is side “x” or ZY z x y Y Z X Standard 18

b a c R Q S What is the adjacent side to S? The ADJACENT side to S is side “b” or RS Standard 18

b a c What is the opposite side to S? The OPPOSITE side to S is side “a” or QR R Q S Standard 18

o u i C B A What is the adjacent side to C? The ADJACENT side to C is side “o” or CA Standard 18

C B A What is the opposite side to C? The OPPOSITE side to C is side “i” or AB o u i Standard 18

e i u Q R S What is the adjacent side to Q? The ADJACENT side to Q is side “e” or QS Standard 18

What is the opposite side to Q? The OPPOSITE side to Q is side “u” or RS e i u Q R S Standard 18

o u i C B A Sin C= Hypotenuse Opposite side SINE u i Standard 18

o u i C B A Cos C= Hypotenuse Adjacent side COSINE u o Standard 18

o u i C B A Tan C= Adjacent side Opposite side TANGENT i o Standard 18

o u i C B A Tan C= i o Adjacent side Opposite side TANGENT Sin C= i u Hypotenuse Opposite side SINE Cos C= o u Hypotenuse COSINE Adjacent side Standard 18

b a c C B A Sin A= a Opposite side c Hypotenuse Standard 18

b a c C B A Cos A= c Hypotenuse b Adjacent side Standard 18

b a c C B A Tan A= a Opposite side Adjacent s. b Standard 18

b a c C B A Sin B= c Hypotenuse b Opposite side Standard 18

b a c C B A Cos B= c Hypotenuse a Adjacent side Standard 18

b a c C B A b Opposite side Adjacent c Tan B= Standard 18

k m l L K M Sin M= k Hypotenuse m Opposite side Standard 18

k m l L K M Cos M= k Hypotenuse l Adjacent side Standard 18

k m l L K M Tan M= m Opposite side Adjacent l Standard 18

k m l L K M Sin L= k Hypotenuse l Opposite side Standard 18

k m l L K M Cos L= k Hypotenuse m Adjacent side Standard 18

k m l L K M Tan L= Adjacent m l Opposite side Standard 18

r t s S R T Sin T= s Hypotenuse t Opposite side Standard 18

r t s S R T Cos T= s Hypotenuse r Adjacent side Standard 18

r t s S R T Tan T= t Opposite side Adjacent r Standard 18

r t s S R T Sin R= s Hypotenuse r Opposite side Standard 18

r t s S R T Cos R= s Hypotenuse t Adjacent side Standard 18

r t s S R T Tan R= Adjacent t r Opposite side Standard 18

4 3 5 Y Z X Sin Z= 5 Hypotenuse 4 Opposite side Sin Z= 0.8 Standard 18

Y Z X Cos Z= 5 Hypotenuse 3 Adjacent side Cos Z= 0.6 Standard 18

Y Z X Tan Z= 4 Opposite side Adjacent Tan Z= 1.3 Standard 18

Y Z X Sin X= 5 Hypotenuse Opposite side Sin X=0.6 Sin ( ) = 36.86° m X = Sin ( ) m X = 36.86° ° 0.6 Standard 18

Y Z X Cos X= 5 Hypotenuse 4 Adjacent side Cos X= 0.8 Cos ( ) = 36.86° m X = Cos ( ) m X = 36.86° ° 0.8 Standard 18

Y Z X Tan X= 4 Adjacent 3 Opposite Side Tan X=.75 Tan ( )= 53.14° 36.86° m X = Tan( ) m X = 36.86° 90°- = m Z= ° ° 53.14° Standard 18

Q R S Sin Q= 17 Hypotenuse 15 Opposite side Sin Q= ° 28.07° m Q = Sin( ) m Q = 90°- = m R= ° 28.07° Standard 18

Q R S Cos Q= 41 Hypotenuse 9 Adjacent side Cos Q= ° 12.69° m Q = Cos( ) m Q = 90°- = m R= ° 12.69° Standard 18

Q R S Tan R= 12 Adjacent 9 Opposite Side Tan R= ° 36.86° m R = Tan ( ) m R = 90°- = m Q= ° 53.14° 36.86° Standard 18

Q R S Sin Q= 51 Hypotenuse 45 Opposite side Sin Q= ° 28.07° m Q = Sin( ) m Q = 90°- = m R= ° 28.07° 61.93° Standard 18

Q R S Cos R= 75 Hypotenuse 72 Adjacent side Cos R= ° 16.26° m R = Cos( ) m R = 90°- = m Q= ° 73.31° 16.26° Standard 18

36 i 48 Q R S Tan R= 48 Adjacent 36 Opposite Side Tan R= ° 36.86° m R = Tan ( ) m R = = i 22 2 i = = i 2 |i|=60 i=60 and i=-60 90°- = m Q= ° 53.14° 36.86° SOLVE QRS: Standards 15, 18, 19

Tan F= Tan F=.7446 m F = Tan ( ) m F = u = = u 2 |u|=58.6 u=58.6 and u= °- = m H= ° SOLVE FGH: u G F H = u Adjacent 35 Opposite Side 36.67° 53.32° Standards 15, 18, 19

30 i u Q R S 35° SOLVE SRQ: Sin ( )= u 30 Sin S= Opposite side Sin 35°= u 30 (30) u=30 Sin 35° u=30( ) u= ° Cos 35°= i Adjacent side Cos 35°= i 30 (30) i=30 Cos 35° i=30( ) i= °- = m Q= Hypotenuse 30 Hypotenuse 35° ° 55°.8192 Standards 15, 18, 19

45 i u K L M 35° SOLVE LMK: Sin( )= i 45 Sin K= Opposite side Sin 55°= i 45 (45) i=45 Sin 55° i=45( ) i= ° Cos 55°= u Adjacent side Cos 55°= u 45 (45) u=45 Cos 55° u=45( ) u= °- = m M= Hypotenuse 45 Hypotenuse 55° ° 35°.5735 Standards 15, 18, 19

e i 9 Q R S 30° Tan ( ) = 9 i 30° = Tan ( 30° ) 9 i 1 i = 9 Tan ( 30° ) i = i =15.58 e = = e 2 |e|=18 e=18 and e= = e °- = m Q= 30° 60° SOLVE QRS: i Tan(30°) = 9 Tan(30°) Standards 15, 18, 19

Y Z X i 6 a 36° Tan ( ) = 36° = Tan ( 36° ) 6 i 1 i = 6 Tan ( 36° ) i = i = °- = m Z= 36° 54° a = = a 2 |a|=10.2 a=10.2 and a=-10.2 i = a ° i Tan(36°) = 6 Tan(36°) Standards 15, 18, 19

Tan F= Tan F=.7391 m F = Tan ( ) m F = u = = u 2 |u|=28.6 u=28.6 and u= °- = m H= ° SOLVE FGH: u G F H = u Adjacent 17 Opposite Side 36.46° 53.53° Standards 15, 18, 19

25 i u K L M 33° SOLVE LMK: Sin( )= i 25 Sin K= Opposite side Sin 57°= i 25 (25) i=25 Sin 57° i=25( ) i= ° Cos 57°= u Adjacent side Cos 57°= u 25 (25) u=25 Cos 57° u=25( ) u= °- = m M= Hypotenuse 25 Hypotenuse 57° ° 33°.5446 Standards 15, 18, 19

Y Z X i 7 a 29° Tan ( ) = 29° = Tan ( 29° ) 7 i 1 i = 7 Tan ( 29° ) i = i = °- = m Z= 29° 61° a = = a 2 |a|=14.4 a=14.4 and a=-14.4 i = a ° i Tan(29°) = 7 Tan(29°) Standards 15, 18, 19