Trigonometri for Microteaching
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This document provides information about trigonometry including definitions of trigonometric ratios, working with right triangles, and trigonometric ratios for special angles. It defines the sine, cosine, and tangent ratios for an angle in a right triangle. It also defines cosecant, secant, and cotangent as the reciprocals of sine, cosine, and tangent respectively. Examples are provided to demonstrate calculating trigonometric ratios in right triangles. The document also provides trigonometric ratio values for specific angles including 0°, 30°, 45°, 60°, and 90° to help determine an angle value given a ratio.
Trigonometri for Microteaching
- 1. BASIC of TRIGONOMETRY For X grade Senior High School By. Alfiramita Hertanti 1111040151_ICP MATH 2011
- 3. SIMILAR TRIANGLE A BC 10 8 6 PQ R 18 30 Show that triangle ABC and PQR are similar triangles? Mention the ratio of the corresponding sides on both the triangles.
- 4. Measurement of Angle 1 360 round 1 4 round1 2 round1 round 360 𝑜 = 2𝜋 𝑟𝑎𝑑 𝑜𝑟 1 𝑜 = 𝜋 180 𝑟𝑎𝑑 𝑜𝑟 1 𝑟𝑎𝑑 = 57,3 𝑜 DEFIITION 8.2 Degree :“O” Radian: “Rad”
- 5. DEFINITON 8.3 • Round to degree 1. 1 4 𝑟𝑜𝑢𝑛𝑑 = 1 4 𝑥 360 𝑜 = 90 𝑜 • Degree to radian ⇔ 90 𝑜 = 90 𝑜 𝑥 𝜋 180 𝑟𝑎𝑑 = 1 2 𝜋 𝑟𝑎𝑑. 2. 1 3 𝑟𝑜𝑢𝑛𝑑 = 1 3 𝑥 360 𝑜 = 120 𝑜 ⇔ 120 𝑜 = 120 𝑥 𝜋 180 𝑟𝑎𝑑 = 2 3 𝜋 𝑟𝑎𝑑 3. 1 2 𝑟𝑜𝑢𝑛𝑑 = 1 2 𝑥 360 𝑜 =180 𝑜 ⇔ 180 𝑜 = 180 𝑥 𝜋 180 𝑟𝑎𝑑 = 𝜋 𝑟𝑎𝑑 4. 3 4 𝑟𝑜𝑢𝑛𝑑 = 4𝑥 360 𝑜 = 270 𝑜 ⇔ 270 𝑜 = 270 𝑥 𝜋 180 𝑟𝑎𝑑 = 3 2 𝜋 𝑟𝑎𝑑
- 7. 180o 270o 0o,360o 90o Quadrant II Quadrant I Quadrant III Quadrant IV 0o - 90o 90o - 180o 180o - 270o 270o-- 360o
- 8. BASIC CONSEPT OF ANGLE Mr. Yahya was a guard of the school. The Height of Mr. Yahya is 1,6 m. He has a son, his name is Dani. Dani still class II elementary school. His body height is 1, 2 m. Dani is a good boy and likes to ask. He once asked his father about the height of the flagpole on the field. His father replied with a smile, 8 m. One afternoon, when he accompanied his father cleared the weeds in the field, Dani see shadows any objects on the ground. He takes the gauge and measure the length of his father shadow and the length of flagpole’s shadow are 6,4 m and 32 m. But he couldn’t measure the length of his own because his shadow follow ing his progression. PROBLEM
- 9. A B E G C F D XO Where : AB = The height of flagpole (8 m) BC = The lenght of the pole’s shadow DE = The height of Mr. Yahya EC = The length of Mr. Yahya’s Shadow FG = The height of Dani GC = The Lenght of Dani’s shadow 6,4 8 1,6 1,2 32 f
- 10. CE D A B C CG F g8 32 1,6 6,4 1,2 1088 43,52 f 𝐹𝐺 𝐷𝐸 = 𝐺𝐶 𝐸𝐶 = 1,2 1,6 = 𝑓 6,4 . f = 4,8 𝐹𝐶 = 𝑔 = 24,48 a. ____ = ____ = ____ = ________ = ________ = ______ = ____________________ = 24,48 43,52 1088 Opposite side the angleFG GC DE EC EC AB 1,2 1,6 8 Hytenuse of triangles 0,24 the sine of the angle C, written sin x0 = 0.24 b. ____ = ____ = ____ = ________ = ________ = ______ = _______________________ = 24,48 43,52 1088 adjacentGC FC EC DC AC BC 4,8 6,4 32 Hypotenuse of triangle 0,97 the cosine of the angle C, written cos x0 = 0.97 c. ____ = ____ = ____ = ________ = ________ = ______ = _______________________ = 4,8 6,4 32 Opposite side the angleFG GC DE EC BC AB 1,2 1,6 8 adjacent 0,25 the tangent of the angle C, written tan x0 = 0.25
- 11. PROBLEM 1,5 m 8 m 9,5m 𝛼 Undu standing 8 m in front of the pine tree with height of 9.5 m. If the height of Undu is 1,5 m. Determine the trigonometric ratio of Angle 𝛼.
- 12. Where : AC = The height Of Pine Tree ED = The height of Undu DC = The distance between Tree and Undu 1,5 m 8 m A B CD E 𝜶 9,5 m SOLUTION 𝑠𝑖𝑛 𝛼? 𝑐𝑜𝑠 𝛼? 𝑡𝑎𝑛 𝛼? Find EA! 8 2 𝐸𝐴 = 𝐸𝐵2 + 𝐴𝐵2 = 82 + 9,5 − 1,5 2 = 64 + 64 = 128 = 8 2 𝑐𝑜𝑠 𝛼 = 8 8 2 = 1 2 2 𝑡𝑎𝑛 𝛼 = 8 8 = 1 𝑠𝑖𝑛 𝛼 = 8 8 2 = 1 2 2
- 13. B P J Trigonometric ration in Right Triangle
- 14. the sine of an angle is the length of the opposite side divided by the length of the hypotenuse. DEFINITION B P J sin 𝐽 = 𝑃𝐵 𝐵𝐽 the cosine of an angle is the length of the adjacent side divided by the length of the hypotenuse. 𝑐𝑜𝑠 𝐽 = 𝑃𝐽 𝐵𝐽 the tangent of an angle is the length of the opposite side divided by the length of the adjacent side. 𝑡𝑎𝑛 𝐽 = 𝑃𝐵 𝑃𝐽
- 15. the cosecant of an angle is the length of the hypotenuse divided by the length of the opposite side. Written : DEFINITION B P J cos𝑒𝑐 𝐽 = 𝐵𝐽 𝑃𝐵 the secant of an angle is the length of the hypotenuse divided by the length of the adjacent side.Written: 𝑠𝑒𝑐 𝐽 = 𝐵𝐽 𝑃𝐽 the tangent of an angle is the length of the adjacent side divided by the length of the opposite side. written : 𝑐𝑜𝑡 𝐽 = 𝑃𝐽 𝑃𝐵 cos𝑒𝑐 𝐽 = 1 sin 𝐽 𝑠𝑒𝑐 𝐽 = 1 cos 𝐽 𝑐𝑜𝑡 𝐽 = 1 tan 𝐽
- 16. S O H C A H T O A REMEMBER i n p p o s i t e y p o t e n u s e o s d j a c e n t y p o t e n u s e a n p p s o s i t e d j a c e n t
- 17. EXAMPLE Given right triangle ABC, right-angled at ∠ ABC. If the length of the side AB = 3 units, BC = 4 units. Determine sin A, cos A, and tan A. C BA 3 units 4 units
- 18. C BA 3 units 4 un From the figure below, 𝐴𝐶 = 𝐵𝐶2 + 𝐴𝐵2 = 32 + 42 = 5 𝑆𝑖𝑛 𝐴 = Cos 𝐴 = Tan 𝐴 = 𝑡ℎ𝑒 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒 𝑡ℎ𝑒 𝑎𝑛𝑔𝑙𝑒 𝐴 𝑡ℎ𝑒 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 = 𝑡ℎ𝑒 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑜𝑓 𝑎𝑛𝑔𝑙𝑒 𝐴 𝑡ℎ𝑒 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 𝑡ℎ𝑒 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒 𝑡ℎ𝑒 𝑎𝑛𝑔𝑙𝑒 𝐴 𝑡ℎ𝑒 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑜𝑓 𝑎𝑛𝑔𝑙𝑒 𝐴 4 5 = 3 5 = 4 3
- 19. Ratio for Specific Angles A(x,y) x y r Y O X Suppose point A (x, y), the length OA = r and the angle AOX = α. 𝑆𝑖𝑛 α = Cos 𝛼 = Tan 𝛼 = 𝑦 𝑟 𝑥 𝑟 𝑦 𝑥 𝛼 A(-x,y) -x y r Y O X 𝑆𝑖𝑛 α = Cos 𝛼 = Tan 𝛼 = 𝑦 𝑟 − 𝑥 𝑟 − 𝑦 𝑥 Quadrant II (90o-180o)Quadrant III (180o-270o) Y O X A(-x,-y) -x -y r 𝑆𝑖𝑛 α = Cos 𝛼 = Tan 𝛼 = − 𝑦 𝑟 − 𝑥 𝑟 𝑦 𝑥 O A(x,-y) x -y r Y X Quadrant IV (270o-360o) 𝑆𝑖𝑛 𝛼 = Cos 𝛼 = Tan 𝛼 = − 𝑦 𝑟 𝑥 𝑟 − 𝑦 𝑥
- 21. EXAMPLE Suppose given points A(-12,5) and ∠XOA = α. Determine the value of sin α, cos α and tan α SOLUTION x = -12 and y = 5. Quadrant II A(-12,5) 5 O Y X α Cos 𝐴 = − 12 13 Tan 𝐴 = − 5 12 𝑆𝑖𝑛 𝐴 = 5 13 12 𝑋𝑂 = 12 2 + 52 = 144 + 25 = 169 = 13 13
- 22. Trigonometric Ration For Special Angles 0o, 30°, 45°,60° and 90o 45o 45o 30o 60o 60o M K LP A B C 22 1 1
- 23. 45o 45o A B C 𝐴𝐶 = 𝐴𝐵2 + 𝐵𝐶2 = 1 + 1 = 2 1 1 2 sin 45 𝑜 = 1 2 = 1 2 2 𝑐𝑜𝑠 45 𝑜 = 1 2 = 1 2 2 𝑡𝑎𝑛 45 𝑜 = 1 1 = 1
- 24. 30o 60o M P L 2 1 𝑀𝑃 = 𝑀𝐿2 − 𝑃𝐿2 = 4 − 2 = 3 sin 30 𝑜 = 1 2 𝑐𝑜𝑠 30 𝑜 = 3 2 = 1 2 3 tan 30 𝑜 = 1 3 = 3 3 sin 60 𝑜 = 3 2 = 1 2 3 𝑐𝑜𝑠 60 𝑜 = 1 2 𝑡𝑎𝑛 60 𝑜 = 3 1 = 3 3
- 25. P(x,y) 1 1NO x y X Y ᶿ sin 𝜃 = 𝑦 1 = 𝑦 cos 𝜃 = 𝑥 1 = 𝑥 tan 𝜃 = 𝑦 𝑥 If 𝜃 = 0 𝑜, then P(1,0) • sin 0° = y = 0 • cos 0° = x = 1 • tan 0° = y/x = 0/1=0 • sin 90° = y = 1 • cos 90° = x = 0 • tan 90° =y/x =1/0, undefine If 𝜃 = 90 𝑜 , then P(0,1)
- 26. Trigonometric ratios of Special Angles 𝛼 0 𝑜 30 𝑜 45 𝑜 60 𝑜 90 𝑜 Sin 𝛼 0 1 2 1 2 2 1 2 3 1 Cos 𝛼 1 1 2 3 1 2 2 1 2 0 Tan 𝛼 0 1 3 3 1 3 ∞
- 27. Anzar want to determine Angle size from a trigonometric ratio. Given to her ratio as follows. sin 𝛼 = 1 2 , He must to determine the value of α (Angle size) PROBLEM
- 28. 1 30o sin 𝛼 = 1 2 , 𝑡ℎ𝑒𝑛 𝛼 = 𝑎𝑟𝑐 sin 1 2 = 30 𝑜 1 2