1. By: Lee Lukasik and Bailey PotrzebowskiTRIGONOMETRY TEST REVIEW

2. Ohio Academic Content StandardsGrade 9Define the basic trigonometric ratios in righttriangles: sine, cosine, and tangent.Apply proportions and right triangle trigonometricratios to solve problems involving missing lengths andangle measures in similar figures. 3. Parts of a Triangle Consider angle A Side a- side opposite of A Side c- side adjacent to A Side b- the hypotenuse (always opposite of the 90 degree angle) 4. Trigonometric Functions Sine Cosine Tangent More Special Triangles SOH CAH TOA 5. SineRatio of the length of the side opposite the given angle to the lengthof the hypotenuse of a right-angled triangle Sine= Opp/Hyp Practice Problems 6. Sine Practice ProblemsPractice Problem #1 PracticeProblem #2Practice Problem #3 7. Practice Problem #1 Find side b and c for triangle ABC. A 40 cAnswer50 BC a= 3 8. Practice Problem #1 Answer We are given that angle A=40 degrees, angle B=90degrees, angle C=50 degrees and side a=6. We know that sine is opposite over hypotenuse sowe can take the sine of 40 degrees and make itequal to 3 divided by side b (sin40=3/b). Then wewould multiply both sides by b to get rid of thedenominator leaving us with b(sin40)=3. Finally wewould divide both sides by sin40 to get b by itself,b=3/sin40, giving us the answer that side b=4.67. Next we can find side c by taking the sine of 50degrees and set it equal to c divided by 4.67(sin50=c/4.67). We multiply both sides by 4.67 toget c by itself, giving us 4.67(sin50)= c. c= 3.58. 9. Practice Problem #2 A ship travels 10 km on a course heading 50 east of north.How far north, and how far east has the ship travelled at this point?Answer 10. Practice Problem #2 Answer To find the distance traveled north wewould take the sine of 40 degrees and set itequal to y divided by 10 km (sin40=y/10km).Next we would multiply both sides by 10 kmto get y by itself giving us 10km(sin40)=y. y=6.43km To find the distance traveled east we wouldtake the sine of 50 degrees and set it equalto x divided by 10km (sin50=x/10km). Nextwe would multiply both sides by 10km to getx by itself giving us 10km(sin50)=x.x=7.66km 11. Practice Problem # 3 Given that the sine of angle A= 0.6,calculate the length of side x.Answer 12. Practice Problem #3 Answer Since we know the sine of angle A is 0.6we can set that equal to 12cm divided byside AB (0.6=12cm/AB). We can concludethat AB= 12cm/0.6, giving us the side ofAB=20cm. Now using Pythagoras theorem we seethat x= the square root of 20 squaredminus 12 squared. Which equals 16cm. 13. Cosine Ratio of the adjacent side to thehypotenuse of a right-angled triangle. Cos=Adj/HypPracticeProblems 14. Cosine Practice ProblemsPractice Problem #1 PracticeProblem #2Practice Problem #3 15. Practice Problem #1In triangle ABC, angle C= 42 degrees, a=19 and b=26. Find the length of sideb.Ac=17Answer42B C a= 19 16. Practice Problem #1 Answer We are given that angle C=42 degrees and side a=19, so we can conclude that the cosine of 42 degrees is equal to 19 divided by side b(cos42=19/b). This gives us that b=19/cos42. b=25.6 17. Practice Problem #2 Find the cosine of angle A, and angle C. Ac=4 Answer B Ca= 3 18. Practice Problem #2 Answer We know that cosine is adjacent overhypotenuse. For angle A, side c=4 isadjacent, and the hypotenuse= 5; CosA=4/5 For angle C, side a=3 is adjacent, and thehypotenuse=5; Cos B=3/5. 19. Practice Problem #3 Find the value of x. B Answer36 C A x 20. Practice Problem #3 Answer We know that cosine is adjacent overhypotenuse. So we can agree that the cos36 degrees= x/17. To get x by itself we must multiply bothsides by 17 (36x17=x). Resulting withx=13.75 21. T a n g e n tTangent=Opposite Look at This video of Adjacenthow we use tangent Tan B = o aClick Here For aPracticeGraph of TangentProblems 22. How we can use Tangent 23. Notice:Vertical Asymptotes- the graph approaches a value butnever reaches itThe graph of tangent has vertical asymptotes atmultiples of 90 Tangent has NO value at these asymptotes! 24. Practice Problems Get Out Your Calculator! Going for Back toA HikeBasics At theAirport 25. Going on a Hike While on a hike, you come up to hill. The map you have says its 300 ft. until you reach the top of the hill. If you know that the incline of the hill is (10 degrees) determine the height of the hill (x).Tan 10 = xClick Here300 Check YourFor a HintAnswerTan (10)52.898 ft= .17632x10 26. Basic Tangent Problem HINT Find The Tangent of the angle labeled F 3050 is the side opposite angle F and 2000 is adjacent Check Your Answer 1.525 27. At The AirportAn airplane takes off at an angle of 25. Beforethe plane changes its angle, it is at a height of5000ft. How far has the airplane traveled alongthe ground?Click HereFor Answer! Click the airplane for a hint!10722.535 ft Solve for x Tan(25) = 5000x 5000 ftx25 28. Triangles that have angle measurements of 30-60-90 or 45-45-90 have side length ratios that are easy to remember. Ifyou know one side length the other two on the triangle willfollow30 60 90 Triangles45 45 90 Triangles 29. Solving With Special Triangles Length ofthe longer The length leg is 3 of the legstimes the are thesmaller leg same Thehypotenuse is The2 times thehypotenuselength of a is 2 timesleg the length ofthe smallestleg 30. SOH CAH TOA Q: What is SOH CAH TOA? A: an easy way to remember what makes up the trigonometry functions Said as "so - cuh - toe uh 31. References1st slide Picture from Powerpoint clip art2nd slide Picture from powerpoint clip artOhio standards fromhttp://www.ode.state.oh.us/GD/Templates/Pages/ODE/ODEDetail.aspx?Page=3&TopicRelationID=1704&Content=86689Tangent http://www.onlinemathlearning.com/tangent-problems.htmlinfo/vidTangent All created information by self as well as pictures created by selfproblem in program paintslidesSpecial http://www.onlinemathlearning.com/image-files/special-rt-triangles triang-454590-1.gifVideo 2 http://www.yourteacher.com/geometry/306090triangle.phpAll other Created using powerpoint toolspictures