RATIOS IN RIGHT TRIANGLES Standards 15, 18, 19 INVERSE OF TRIGONOMETRIC RATIOS END SHOW PROBLEM 1a PROBLEM 2a PROBLEM 3a PROBLEM 1b PROBLEM 2b PROBLEM 3b PROBLEM 1c PROBLEM 2c PROBLEM 3c USING TABLES AND GRAPHS ELEVATION VS DEPRESSION PROBLEM 4 PROBLEM 5 PROBLEM 6 PROBLEM 7 PROBLEM 8 PROBLEM 9 PRESENTATION CREATED BY SIMON PEREZ. 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. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

SINE B Opposite side Sin C= Hypotenuse i Sin C= u u i COSINE Standard 18 Sin C= Hypotenuse i Sin C= u u i COSINE Adjacent side Cos C= Hypotenuse o Cos C= C A o u TANGENT Opposite side Tan C= Adjacent side i Tan C= o PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

Can we get the ratio again? X m X = ? Standard 18 5 Hypotenuse 3 Opposite side Sin X= 5 36.86° 4 Sin X=0.6 m X = Sin ( ) -1 0.6 Z Y 3 m X = 36.86° Can we get the ratio again? Sin ( ) = 36.86° 0.6 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

Can we get the ratio again? X m X = ? Standard 18 5 Hypotenuse 4 Adjacent side Cos X= 5 36.86° 4 Cos X= 0.8 m X = Cos ( ) -1 0.8 m X = 36.86° Z Y 3 Can we get the ratio again? Cos ( ) = 36.86° 0.8 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

Can we get the ratio again? X m X = ? Standard 18 4 Adjacent 3 Opposite Side Tan X= 5 36.86° 4 Tan X= .75 m X = Tan ( ) -1 .75 53.14° m X = 36.86° Z Y 3 Can we get the ratio again? What is the value for the remaining angle? Tan ( )= 36.86° .75 90°- = m Z= 36.86° 53.14° PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

17 Hypotenuse m R = ?, m Q = ? 17 Q 15 Opposite side R Sin Q= 28.07° Standard 18 17 Q 15 Opposite side R Sin Q= 28.07° 61.93° 8 15 Sin Q= .8824 S m Q = Sin ( ) -1 .8824 90°- = m R= 61.93° 28.07° m Q = 61.93° PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

41 Hypotenuse m R = ?, m Q = ? 41 Q R 9 Adjacent side Cos Q= 12.69° Standard 18 41 Q R 9 Adjacent side Cos Q= 12.69° 77.31° 9 40 Cos Q= .2195 S m Q = Cos ( ) -1 .2195 m Q = 77.31° 90° - = m R= 77.31° 12.69° PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

m R = ?, m Q = ? 15 Q 9 Opposite Side R Tan R= 12 Adjacent 53.14° Standard 18 15 Q 9 Opposite Side R Tan R= 12 Adjacent 53.14° 36.86° 9 12 Tan R= .75 S m R = Tan ( ) -1 .75 m R = 36.86° 90° - = m Q = 36.86° 53.14° PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

51 Hypotenuse m R = ?, m Q = ? 51 Q 45 Opposite side R Sin Q= 28.07° Standard 18 51 Q 45 Opposite side R Sin Q= 28.07° 61.93° 24 45 Sin Q = .8824 S m Q = Sin ( ) -1 .8824 90° - = m R = 61.93° 28.07° m Q = 61.93° PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

m R = ?, m Q = ? 75 Hypotenuse 75 Q 72 Adjacent side R Cos R= 16.26° Standard 18 75 Q 72 Adjacent side R Cos R= 16.26° 73.31° 21 72 Cos R= .96 S m R = Cos ( ) -1 .96 m R = 16.26° 90° - = m Q = 16.26° 73.31° PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

1296 + 2304 = i i = 3600 i = 3600 |i|=60 i=60 and i=-60 36 + 48 = i SOLVE QRS: i = 60 Standards 15, 18, 19 Q R 36 Opposite Side Tan R= 53.14° 36.86° 36 48 Adjacent 48 S 1296 + 2304 = i 2 i = 3600 2 Tan R= .75 i = 3600 2 m R = Tan ( ) -1 .75 |i|=60 m R = 36.86° i=60 and i=-60 90° - = m Q = 36.86° 53.14° PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

SOLVE FGH: F Standards 15, 18, 19 36.67° 35 + 47 = u 2 47 Adjacent 47 u G 35 Opposite Side 35 53.32° Tan F= H 1225 + 2209 = u 2 Tan F= .7446 u = 3434 2 m F = Tan ( ) -1 .7446 u = 3434 2 m F = |u|=58.6 90° - = m H = 36.67° u=58.6 and u=-58.6 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

SOLVE SRQ: R i Adjacent side Opposite side Sin S= Hypotenuse i u 30 Standards 15, 18, 19 R i Adjacent side Opposite side Sin S= Hypotenuse i u 30 Hypotenuse u Sin ( )= 35° 35° 55° 30 S Q Sin 35° = u 30 30 (30) Cos 35° = Cos S = u=30 Sin 35° u=30( ) .5736 u=17.2 Cos 35° = i 30 (30) 90°- = m Q= 35° 55° i=30 Cos 35° i=30( ) .8192 i=24.57 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

SOLVE LMK: L Opposite side Sin K= Hypotenuse i u i Sin( )= 55° 35° 55° Standards 15, 18, 19 L Opposite side Sin K= Hypotenuse i u u Adjacent side i 45 Hypotenuse Sin( )= 55° 35° 55° 45 M K Sin 55° = i 45 45 (45) Cos 55° = Cos K = i=45 Sin 55° i=45( ) .8191 i=36.86 Cos 55° = u 45 (45) 90°- = m M= 55° 35° u=45 Cos 55° u=45( ) .5735 u=25.81 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

243 + 81 = e e = 324 e = 324 |e|=18 e=18 and e=-18 15.58 + 9 = e SOLVE QRS: Standards 15, 18, 19 R 9 i Tan ( ) = 30° 15.58= i 9 = Tan ( 30° ) 9 i 1 60° 30° S Q e i Tan(30°) = 9 15.58 + 9 = e 2 Tan(30°) Tan(30°) i = 9 Tan ( 30° ) 243 + 81 = e 2 90° - = m Q = 30° 60° i = 9 e = 324 2 .5774 i =15.58 e = 324 2 |e|=18 e=18 and e=-18 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

X SOLVE XYZ: Standards 15, 18, 19 i 6 Tan ( ) = 36° 8.26 + 6 = a 2 =8.26 36° a = Tan ( 36° ) 6 i 1 i 68.2 + 36 = a 2 a = 104.2 2 54° i Tan(36°) = 6 Z Y Tan(36°) Tan(36°) a = 104.2 2 6 i = 6 Tan ( 36° ) |a|=10.2 i = 6 a=10.2 and a=-10.2 .7265 90° - = m Z = 36° 54° i = 8.26 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

F SOLVE FGH: 36.46° 17 Opposite Side Tan F= 23 Adjacent 23 u G Tan F= .7391 17 53.53° H m F = Tan ( ) -1 .7391 m F = 36.46° |u|= 28.6 17 + 23 = u 2 u=28.6 and u= -28.6 289 + 529 = u 2 u = 818 2 90° - = m H= 36.46° 53.53° u = 818 2 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved Standards 15, 18, 19

SOLVE LMK: L i Opposite side Sin 57° = Sin = 57° 25 Hypotenuse i u u Standards 15, 18, 19 L i Opposite side Sin 57° = Sin = 57° 25 Hypotenuse i u u Adjacent side 25 Hypotenuse 33° 57° M K Sin 57°= i 25 25 (25) Cos 57°= i=25 Sin 57° i=25( ) .8386 i = 20.96 Cos 57°= u 25 (25) u=25 Cos 57° u=25( ) .5446 90° - = m M= 57° 33° u = 13.61 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

SOLVE XYZ: X Standards 15, 18, 19 7 Tan ( ) = 29° i = 14.4 29° a = Tan ( 29° ) 7 i 1 i 12.6 + 7 = a 2 159.5 + 49 = a 2 i Tan(29°) = 7 61° Tan(29°) Tan(29°) Z Y a = 208.5 2 7 i = 7 Tan ( 29° ) a = 208.5 2 i = 7 |a|=14.4 a=14.4 and a=-14.4 .5543 i =12.62 90° - = m Z = 29° 61° PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

Angle to Ratio: cos 37°= 0.7986 sin 19°= 0.3256 tan 67°= 2.3559

Ratio to Angle: acos( 0.9205) = 23° asin (0.0872) = 5° atan (9.5144) = 84°

TRIGONOMETRIC CIRCLE: First Quadrant x y -1 1 y Sin = (x,y) 1 1 y Sin = y =Sin x =Cos x Cos = 1 Cos = x Now, we take the first quadrant and we use it to find Trigonometric ratios from angles and angles from trigonometric ratios. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

Angle to Ratio: Sin Cos cos 70°= .34 sin 50°= .76 0.76 0.34

Sin Cos 52° Ratio to Angle: acos( 0.6) = 52° asin (0.5) = 30°

ANGLE OF DEPRESSION AND ANGLE OF ELEVATION Angles of Depression and Elevation are Alternate Interior Angles and they are congruent. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

45 Tan X= 435 Tan X= .103 m X = Tan ( ) .103 m X = 5.9° Standards 15, 18, 19 A car road rises vertically 45 feet, over a horizontal distance of 435 feet. What is the angle of elevation of the road? 45 Tan X= 435 45 feet X=? Tan X= .103 5.9° 435 feet m X = Tan ( ) -1 .103 m X = 5.9° The angle of elevation of the road is about 5.9°. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

Standards 15, 18, 19 The slope of a rollercoaster track span has 600 meters in length, with a vertical drop of 210 meters. What is the angle of depression for the slope? PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

Standards 15, 18, 19 The slope of a rollercoaster track span has 600 meters in length, with a vertical drop of 210 meters. What is the angle of depression for the slope? PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

Standards 15, 18, 19 The slope of a rollercoaster track span has 600 meters in length, with a vertical drop of 210 meters. What is the angle of depression for the slope? X=? 600 m 210 m PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

210 Cos Y= 600 Cos Y= .35 m Y = Cos ( ) .35 m Y = 69.5° m X= 90° - Standards 15, 18, 19 The slope of a rollercoaster track span has 600 meters in length, with a vertical drop of 210 meters. What is the angle of depression for the slope? 20.5° X=? 600 m 210 m Y=? 210 Cos Y= 600 Cos Y= .35 600 m Y=? 210 m m Y = Cos ( ) -1 .35 m Y = 69.5° We have a right angle, so the angles together are 90°: 90° - m X= 69.5° = 20.5° The angle of depression of the slope is about 20.5°.

15 Tan X= 10 Tan X= 1.5 m X = Tan ( ) 1.5 m X = 56.3° Standards 15, 18, 19 A glider is flying at an altitude of 15 yards, and starts descending when the distance from the expected landing spot is 10 yards away from one person standing below the glider on the ground. Find the angle of depression for the remaining part of the flying distance. 56.3° X 15 yd We have alternate interior angles formed by a transversal crossing parallel lines: X=? Landind spot 10 yd 15 Tan X= 10 15 yd Tan X= 1.5 X=? m X = Tan ( ) -1 1.5 10 yd m X = 56.3° The angle of depression of the glider is about 56.3°.

So, the Pedro’s eye-sight is at about 6 meters above Javier’s. Pedro and Javier are looking at each other. Pedro is on top of a brick wall and javier is on the ground at 20 meters from the wall. If the angle of elevation from javier’s eye is 17°, what is the vertical distance from the horizontal passing through his eye and the horizontal passing through Pedro’s eye. x =? ANGLE OF ELEVATION 17° 20 m 17° 20 m x =? x Tan 17° = x = 20Tan 17° 20 x = 20( ) .306 Tan 17° = x 20 (20) (20) x = 6.11 m So, the Pedro’s eye-sight is at about 6 meters above Javier’s.

Rachel is flying a kite, and she has let out 35 yards of string Rachel is flying a kite, and she has let out 35 yards of string. The angle of elevation for the string and the horizontal is 24°. What is the height of the kite from Rachel’s hand? x =? 35 yards ANGLE OF ELEVATION 24° 35 yards x x =? Sin 24° = x = 35Sin 24° 35 24° x = 35( ) .407 Sin 24° = x 35 (35) (35) x = 14.24 yd. So, Rachel was flying her kite at a height of around 14.24 yards from her hand.

They are approximately 175 feet apart. Mrs. Xiong is on the top of a building. From that height, she is able to see two pedestrians that walk in opposite directions, going away from the building by the sidewalk. She is standing at the midpoint of the roof and the angles of depression for the pedestrians from that point in the roof are 56° and 43° respectively. The building is 100 feet tall. What is the distance between the pedestrians when they are at those angles of depression. (round final answer to the closest unit). x y 56° 43° 100 ft Total distance = x + y 100 = 67.45 + 107.2 = 174.7 They are approximately 175 feet apart. Finding x: Finding y: 100 100 Tan 56° = Tan 43° = x y Tan 56° = 100 x Tan 43° = 100 y (x) (x) (y) (y) (x)Tan 56° =100 (y)Tan 43° =100 x= 100 1.48 y= 100 0.93 Tan 56° Tan 43° x= 100 Tan 56° x = 67.45 ft y= 100 Tan 43° y = 107.2 ft