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By: Lee Lukasik and Bailey PotrzebowskiTRIGONOMETRY TEST REVIEW
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Ohio Academic Content StandardsGrade 9“Define 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.”
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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)
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Trigonometric Functions Sine Cosine Tangent More Special Triangles SOH CAH TOA
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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
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Sine Practice Problems Practice Problem #1 Practice Problem #2 Practice Problem #3
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Practice Problem #1 Find side b and c for triangle ABC. A 40 c Answer 50 B C a= 3
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Practice Problem #1 Answer We are given that angle A=40 degrees, angle B=90 degrees, angle C=50 degrees and side a=6. We know that sine is opposite over hypotenuse so we can take the sine of 40 degrees and make it equal to 3 divided by side b (sin40=3/b). Then we would multiply both sides by b to get rid of the denominator leaving us with b(sin40)=3. Finally we would 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 50 degrees and set it equal to c divided by 4.67 (sin50=c/4.67). We multiply both sides by 4.67 to get c by itself, giving us 4.67(sin50)= c. c= 3.58.
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Practice Problem #2 A ship travels 10 km on a course heading 50º east of north. How far north, and how far east has the ship travelled at this point? Answer
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Practice Problem #2 Answer To find the distance traveled north we would take the sine of 40 degrees and set it equal to y divided by 10 km (sin40=y/10km). Next we would multiply both sides by 10 km to get y by itself giving us 10km(sin40)=y. y= 6.43km To find the distance traveled east we would take the sine of 50 degrees and set it equal to x divided by 10km (sin50=x/10km). Next we would multiply both sides by 10km to get x by itself giving us 10km(sin50)=x. x=7.66km
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Practice Problem # 3 Given that the sine of angle A= 0.6, calculate the length of side x. Answer
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Practice Problem #3 Answer Since we know the sine of angle A is 0.6 we can set that equal to 12cm divided by side AB (0.6=12cm/AB). We can conclude that AB= 12cm/0.6, giving us the side of AB=20cm. Now using Pythagoras’ theorem we see that x= the square root of 20 squared minus 12 squared. Which equals 16cm.
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Cosine Ratio of the adjacent side to the hypotenuse of a right-angled triangle. Cos=Adj/Hyp Practice Problems
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Cosine Practice Problems Practice Problem #1 Practice Problem #2 Practice Problem #3
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Practice Problem #1In triangle ABC, angle C= 42 degrees, a=19 and b=26. Find the length of sideb. A c=17 Answer 42 B C a= 19
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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
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Practice Problem #2 Find the cosine of angle A, and angle C. A c=4 Answer B C a= 3
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Practice Problem #2 Answer We know that cosine is adjacent over hypotenuse. For angle A, side c=4 is adjacent, and the hypotenuse= 5; Cos A=4/5 For angle C, side a=3 is adjacent, and the hypotenuse=5; Cos B=3/5.
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Practice Problem #3 Find the value of x. B Answer 36 C A x
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Practice Problem #3 Answer We know that cosine is adjacent over hypotenuse. So we can agree that the cos 36 degrees= x/17. To get x by itself we must multiply both sides by 17 (36x17=x). Resulting with x=13.75
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T a n g e n tTangent=Opposite Look at This video of Adjacent how we use tangent Tan B = o aClick Here For a PracticeGraph of Tangent Problems
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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!
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Practice Problems Get Out Your Calculator!• Going for • Back to A Hike Basics • At the Airport
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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 Here 300 Check YourFor a Hint Answer Tan (10) 52.898 ft = .17632 x 10°
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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
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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 Here For Answer! Click the airplane for a hint! 10722.535 ft Solve for x Tan(25) = 5000 x 5000 ft x 25°
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Triangles that have angle measurements of 30-60-90 or 45- 45-90 have side length ratios that are easy to remember. If you know one side length the other two on the triangle will follow30 60 90 Triangles 45 45 90 Triangles
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Solving With Special Triangles • Length of the longer • The length leg is √3 of the legs times the are the smaller leg same• The hypotenuse is • The √2 times the hypotenuse length of a is 2 times leg the length of the smallest leg
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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”
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References1st slide Picture from Powerpoint clip art2nd slide Picture from powerpoint clip art Ohio standards from http://www.ode.state.oh.us/GD/Templates/Pages/ODE/ODEDetai l.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 “paint”slidesSpecial 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
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