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- 1. passport to the world of Trigonometry CLICK TO CONTINUE
- 2. A triangle is a polygon made up of three connected line segments in such a way that each side is connected to the other two. CLICK TO CONTINUE
- 3. All these polygons are tri-gons and commonly called triangles CLICK TO CONTINUE
- 4. None of these is a triangle... Can you tell why not? CLICK TO CONTINUE
- 5. CLICK on each one that IS a triangle? CLICK TO CONTINUE
- 6. A right triangle is a special triangle that has one of its angles a right angle. You can tell it is a right triangle when when one angle measures 900 or the right angle is marked by a little square on the angle whose measure is 900. CLICK TO CONTINUE
- 7. Can you tell why? CLICK TO CONTINUE
- 8. These triangles are NOT right triangles. Explain why not? CLICK TO CONTINUE
- 9. Click on the triangle that is NOT a right triangle? CLICK TO CONTINUE
- 10. The longest side of a right triangle is the hypotenuse. The hypotenuse lies directly opposite the right angle. The legs may be equal in length or one may be longer than the other. CLICK TO CONTINUE
- 11. A right triangle has two legs and a hypotenuse... hyp ot leg leg enu s e CLICK TO CONTINUE
- 12. Click on the side that is the hypotenuse of the right triangle. side 1 side 3 side 2 CLICK TO CONTINUE
- 13. Click on the side that is the shor ter leg of the right triangle. side 1 side 3 side 2 CLICK TO CONTINUE
- 14. Click on the side that is the longer leg of the right triangle. side 1 side 3 side 2 CLICK TO CONTINUE
- 15. hyp o te nus e c The right triangle has a special property, called leg b 1 the Pythagorean Theorem, that can help a us find one side if we leg2 know the other two If the lengths of hypotenuse and legs are c, sides. a and b respectively, then c2 = a2 + b2 CLICK TO CONTINUE
- 16. Use the Pythagorean Theorem to find the length of the missing side. 10 x 14 c2 x2 x = a2 + b2 = 102 + 142 = 100 + 196 = 296 = sqrt(296) = 17.2 CLICK TO CONTINUE
- 17. Use the Pythagorean Theorem to find the length of the missing side. 15 10 x c2 152 225 x2 x = a2 + b2 = 102 + x2 = 100 + x2 = 125 = sqrt(125) = 11.18 CLICK TO CONTINUE
- 18. Find the hypotenuse of the given right triangle with the lengths of the legs known: Click on the selection that matches your answer: 6 8 A. 36 B. 10 C. 100 x D. 64 CLICK TO CONTINUE
- 19. Find the leg of the given right triangle with the lengths of the leg and hypotenuse known: Click on the selection that matches your answer: x A. 24 B. 6 C. 144 15 D. 12 9 CLICK TO CONTINUE
- 20. A In a right triangle, a given leg is called the adjacent side or the leg1 b opposite side, depending on the reference acute angle. hyp o te nus e c a leg2 CLICK TO CONTINUE
- 21. In a right triangle, a given leg is called the adjacent side or the opposite side, depending on the reference acute angle. A leg1 b hyp ote nus c e a leg2 leg2 is opposite to acute angle A leg1 is NOT opposite to acute angle A CLICK TO CONTINUE
- 22. A In a right triangle, a hyp given leg is called o te nus the adjacent side e c leg1 b or the opposite a side, depending on the reference leg2 acute angle. leg1 is adjacent to acute angle A leg2 is NOT adjacent to acute angle A CLICK TO CONTINUE
- 23. Click on the side that is opposite to angle B. hyp o leg1 ten use c b a leg2 B CLICK TO CONTINUE
- 24. Click on the side that is adjacent to angle B. hyp o leg1 ten use c b a leg2 B CLICK TO CONTINUE
- 25. The ratios of the sides of a right triangle have special names. There are three basic ones we will consider: • Sine cosine tangent CLICK TO CONTINUE
- 26. Let the lengths of legs be a and b, and the length of the hypotenuse be c. A is an acute angle. A leg1 hyp o ten use c b a leg2 With reference to angle A, the ratio of the length of the side opposite angle A to length of the hypotenuse is defined as: length of side opposite sine A = length of the hypotenuse ∠A = a/c Sine A is abbreviated Sin A. Thus, sin A = a/c. CLICK TO CONTINUE
- 27. Let the lengths of legs be a and b, and the length of the hypotenuse be c. A is an acute angle. A leg1 hyp o ten use c b a leg2 With reference to angle A, the ratio of the length of the side adjacent to angle A to length of the hypotenuse is defined as: length of side adjacent to angle A cosine A = length of the hypotenuse = b/c Cosine A is abbreviated to Cos A. Thus, cos A = b/c. CLICK TO CONTINUE
- 28. Let the lengths of legs be a and b, and the length of the hypotenuse be c. A is an acute angle. A leg1 hyp ot b c enu s a leg2 e With reference to angle A, the ratio of the length of the side opposite to angle A to length of the side adjacent to angle A is defined as: length of side opposite A tangent A = length of the adjacent = a/b tangent A is abbreviated Tan A. Thus, tan A = a/b CLICK TO CONTINUE
- 29. S = Sine This is a clever technique most O = Opposite H = Hypotenuse people use to remember these three basic trig ratios. C = Cosine SOH-CAH-TOA sounds strange? A = Adjacent H = Hypotenuse What if I told you it was the ancient oriental queen who T = Tangent O = Opposite loved Geometry? (not true!) A = Adjacent CLICK TO CONTINUE
- 30. Find the sine of the given angle. [SOHCAHTOA ] Sin B = Opposite/Hypotenuse sin 53.10 = 16/20 = 4/5 = 0.80 20 16 53.10 12 Cos B = Adjacent/Hypotenuse Cos 53.10 = 12/20 = 3/5 = 0.60 B Tan B = Opposite/Adjacent Tan 53.10 = 16/12 = 5/3 =1.67 CLICK TO CONTINUE
- 31. Find the value of sine, cosine, and tangent of the given acute angle. [SOHCAHTOA!] B Click to choose your answer from the choices sin 53.10 =? A. 5/3 53.10 15 B. 3/5 C. 4/3 D. 3/4 E. 4/5 F. 5/4 9 A. 5/3 B. 3/5 C. 4/3 D. 3/4 E. 4/5 F. 5/4 tan 53.10 =? A. 5/3 B. 3/5 C. 4/3 D. 3/4 E. 4/5 F. 5/4 cos 53.10 =? 12 CLICK TO CONTINUE
- 32. Does the trig ratio depend on the size of the angle or size of the side length? Let us consider similar triangles in our investigation. 9 6 3 4 36.870 A 5 12 8 36.870 A 36.870 10 15 A CLICK TO CONTINUE
- 33. 9 [Remember SOHCAHTOA!] • Compute the ratios and make a conjecture 6 12 3 36.870 8 4 36.870 36.870 5 A 10 A 15 A sin 36.870 ⅗ = 0.6 6/10 = 0.6 9/15 = 0.6 cos 36.870 ⅘ = 0.8 8/10 = 0.8 12/15 = 0.8 tan 36.870 ¾ =0.75 6/8 = 0.75 9/12 = 0.75 Conjecture: Trigonometric ratios are a property of similarity (angles) and not of the length of the sides of a right triangle. CLICK TO CONTINUE
- 34. • The trig ratios are used so often that • • technology makes these values readily available in the form of tables and on scientific calculators . We will now show you how to use your calculator to find some trig ratios. Grab a scientific calculator and try it out. CLICK TO CONTINUE
- 35. Each calculator brand may work a little dif ferently, but the results will be the same. Look for the trig functions on your calculator: sin, cos and tan select the trig ratio of your choice followed by the angle in degrees and execute (enter). • o • example: sin 30 will display 0.5 on some calculators you may have to type in the angle first then the ratio o example: 30 sin will display 0.5 CLICK TO CONTINUE
- 36. Use your calculator to verify that the sine, cosine and tangent of the following angles are correct (to 4 decimals): Angle A sin A cos A tan A 45o Sin 45 o =0.7071 Cos 45 o =0.7071 Tan 45 o =1.0000 60o Sin 60 o =0.8660 Cos 60 o =0.5000 Tan 60 o =1.7321 30o Sin 30 o =0.5000 Cos 30 o =0.8660 Tan 30 o =0.5774 82.5o Sin 82.5 o =0.9914 Cos 82.5 o =0.1305 Tan 82.5 o =7.5958 CLICK TO CONTINUE
- 37. Find the values of the following trig ratios to four decimal places: sin 34o = ? A. 0.8290 B. 0.6745 C. 0.5592 cos 56o = ? A. 0.5592 B. 0.8290 C. 1.4826 tan 72o = ? A. 0.3090 B. 0.9511 C. 3.0777 CLICK TO CONTINUE
- 38. • • We can use the inverse operation of a trig ratio to find the angle with the known trig ratio (n/m) The inverse trig ratios are as follows: • Inverse of sin (n/m) is sin -1 (n/m) • Inverse of cos (n/m) is cos -1 (n/m) • Inverse of tan (n/m) is tan -1 (n/m) CLICK TO CONTINUE
- 39. Suppose we know the trig ratio and we want to find the associated angle A. 4 5 A • From SOHCAHTOA, we know that from the angle A, we have the opposite side and the hypotenuse. • Therefore the SOH part helps us to know that we use sin A = O/H = 4/5 • The inverse is thus sin-1(4/5) = A • A = Sin-1 (4/5) = 53.13o CLICK TO CONTINUE
- 40. Suppose we know the trig ratio and we want to find the associated angle B. B 4 5 A • From SOHCAHTOA, we know that from the angle B, we have the adjacent side and the hypotenuse. • Therefore the CAH part helps us to know that we use cos A = A/H = 4/5 • The inverse is thus cos-1(4/5) = B • B = cos-1 (4/5) = 36.87o CLICK TO CONTINUE
- 41. Suppose we know the trig ratio and we want to find the associated angle A. 4 3 A • From SOHCAHTOA, we know that from the angle A, we have the opposite side and the adjacent side. • Therefore the TOA part helps us to know that we use tan A = O/A = 4/3 • The inverse is thus tan-1(4/3) = A • A = tan-1 (4/3) = 53.13o CLICK TO CONTINUE
- 42. Use a calculator to find the measure of the angles A and B. B 19.21 A 15 Use SOHCAHTOA as a guide to what ratio to use. m∠A =? A. 38.7 B. 51.3 C. 53.1 C m∠B =? A. 38.7 B. 51.3 C. 53.1 12 CLICK TO CONTINUE
- 43. Use trig ratios to find sides of a triangle. Remember SOHCAHTOA! a 12 300 b With reference to angle A, ●b is the length of side adjacent and ●a is the length of the side opposite the angle. ●the hypotenuse is given A Strategy: make an equation that uses only one leg and the hypotenuse at a time. CLICK TO CONTINUE
- 44. The tangent ratio may not easily help you figure out the legs a and b in this case. ( SOHCAHTOA!) a 12 300 b Using tangent: tan A = O/A Substituting values from the tgriangle: tan 30o = a/b From the calculator: tan 30o = 0.5774 Thus tan 30o = a/b 0.5774 = a/b A And, a = 0.5774(b) GETS YOU STUCK! CLICK TO CONTINUE
- 45. Using sine ratio to find the leg of a triangle. Remember SOHCAHTOA! a 12 300 b Using sine: sin A = O/H Substituting values from the tgriangle: Sin 30o = a/12 From the calculator: sin 30o = 0.5 Thus sin 30o = a/12 0.5 = a/12 A And a = 0.5(12) = 6 CLICK TO CONTINUE
- 46. Using the cosine ratio to find legs of a triangle. Remember SOHCAHTOA! a 12 300 b Using cosine: cos A = A/H Substituting values from the tgriangle: cos 30o = b/12 From the calculator: cos 30o = 0.8660 Thus cos 30o = b/12 0.866 = b/12 A And b = 0.866(12) = 10.39 CLICK TO CONTINUE
- 47. Find the lengths of the legs of the triangle and the third angle. Choose the correct answer. C a B b 25 o A m∠B = ? A. 90 B. 65 C. 25 a = ? A. 10.57 B. 22.66 C. 21 21 b = ? A. 21 B. 10.57 C. 22.66 CLICK TO CONTINUE
- 48. Use trig ratios to find the hypotenuse of a triangle. Remember SOHCAHTOA! c 12 300 b With reference to angle A, ●b is the length of side adjacent and ●12 is the length of the side opposite the angle. A ●c is the hypotenuse Strategy: make an equation that uses only one unkown at a time. CLICK TO CONTINUE
- 49. Use trig ratios to find the hypotenuse of a triangle. Remember SOHCAHTOA! c 12 300 b Since 12 is opposite to the angle, we use the sine ratio: Sine A = O/H Substituting values from the tgriangle: sin 30o = 12/c From the calculator: sin 30o = 0.5 A Thus, sin 30o = 12/c or 0.5 = 12/c c = 12/0.5 = 24 CLICK TO CONTINUE
- 50. Find the lengths of the hypotenuse, leg b and the third angle. Choose the best answer. b C b=? 13 55o B A m∠A = ? A. 35 B. 45 C. 55 A. 22.66 B. 10.57 C. 18.57 c c = ? A. 18.57 B. 10.57 C. 22.66 CLICK TO CONTINUE
- 51. • • We now have the tools we need to solve any right triangle (to determine the lengths of each and all sides and the angles, given minimal information) Remember SOHCAHTOA! Typically you get two pieces of information: • • One side length and one angle or CLICK TO CONTINUE Two sides’ lengths
- 52. Given one side length and one angle, determine the rest. Remember SOHCAHTOA ! B Find measure of angle B and side lengths AC and AB. c 12 C Since we know two angles (90 and 42) we can determine the 3rd from the Triangle Angle Sum Theorem: m∠B = 1800 –(900+420) = 480. 420 b A CLICK TO CONTINUE
- 53. Given one side length and one angle, determine the rest. Remember SOHCAHTOA ! B 12 C Strategy: side with length 12 is opposite to angle A. To find b, use tan A and to find c, use sin A sin A = O/H tan A = O/A sin 42 = 12/c tan 42 = 12/b c 0.6691 = 12/c 0.9004 = 12/b c = 12/0.6691 b = 12/0.9004 0 42 c = 17.93 A b = 13.33 b CLICK TO CONTINUE
- 54. Solve the triangle. Choose and check answer. m∠B = ? A. 46 C c b 440 B. 23.82 C. 23 b = ? A. 23 23 C. 23 c = ? A. 33.11 B B. 44 B. 23.82 C. 33.11 A CLICK TO CONTINUE
- 55. Given two side lengths, solve the triangle. Remember SOHCAHTOA ! A 10 C Strategy: •use Pythagorean Theorem to find the 3rd side length, a. •Use cosine ratio to find measure of angle A •Use the Triangle Angle Sum Theorem to find the measure of angle B. 17 a B CLICK TO CONTINUE
- 56. A Given two side lengths, solve the triangle. Remember SOHCAHTOA ! 10 C 17 a Using Pythagorean Theorem to find the 3rd side length, a. •c2 = a2 + b2 Pythagorean Theorem •172 = a2 + 102 Substituting values •289 = a2 + 100 Evaluating the squares •a2 = 289-100 Addition property of =. B •a2 = 189 Simplifying •a = sqrt(189) = 13.75 Taking square root. CLICK TO CONTINUE
- 57. Given two side lengths, solve the triangle. Remember SOHCAHTOA ! A 17 10 Using cosine ratio to find measure of angle A •cos A = A/H (the CAH part) •cos A = 10/17 (substituting values) •m∠A = cos-1(10/17) (inverse of cosine) •m∠A = 53.97o (Calculator) B a C CLICK TO CONTINUE
- 58. Solve the triangle. Click to check your answer… m∠A = ? A. 21.8 18 C c 45 C. 48.5 c = ? A. 21.8 B B. 68.2 B. 68.2 C. 48.5 m∠B = ? A. 21.8 B. 68.2 C. 48.5 A CLICK TO CONTINUE
- 59. Trigonometry is used to solve real life problems. The following slides show a few examples where trigonometry is used. Search the Internet for more examples if you like. CLICK TO CONTINUE
- 60. Measuring the height of trees What would you need to know in order to calculate the height of this tree? What trig ratio would you use? Click here to see if we agree. CLICK TO CONTINUE
- 61. Tall buildings (skyscrapers), towers and mountains… CLICK TO CONTINUE
- 62. Assume the line in the middle of the drawn triangle is perpendicular to the beach line. How far is the island from the beach? Click here to check my solution and compare with yours CLICK TO CONTINUE
- 63. Be proud of yourself! You have successfully completed a crash course in basic trigonometry and I expect you to be able to do well on this strand in the Common Core States Standards test. Print the certificate to show your achievement. CLICK TO CONTINUE
- 64. I hereby certify that _____________________________________ has satisfactorily completed a basic course in Introduction to Trigonometry on this day the ____________________ of the year 20___ The bearer is qualified to solve some real world problems using trigonometry. Signed: Nevermind E. Chigoba

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