Unit 2 Trigonometry LessonsDone - OAME

Possible solutionopposite side to Badjacent side to Blength of side btanB length of side a2000tan35 Å s represents the distance between the sailboat and the cliff.s20000.700207538 stanB 0.700207538 s 2000s 2856.2960143 the sailboat is approximately 2856 m away from the cliff.(or almost 3 km)Problem 3:A sailboat that is 2 km due west of a lighthouse sends a signal to thelighthouse that it is in distress. The lighthouse quickly signals a rescueplane that is 7 km due south of the lighthouse. What heading from due northshould the plane take in order to intercept the troubled sailboat?2 kmBC7 kmPossible Solution:opposite side to Aadjacent side to Alength of side atan A length of side b2tan A 7tan A tan AA 0.2857142857 A 15.9453959 the plane should take a heading of about 16 west of north to intercept and rescuethe sailboat.

ConsolidateDebriefSmall Groups Æ PresentationPlace the small groups together that have the same problem to check eachothers work. Each of the groups prepares for the presentation.Randomly select one of the groups for each problem to present to the class.Home Activity or Further Classroom ConsolidationApplicationStudents complete BLM2.3.2

MBF3CBLM2.3.1Trigonometry Application ProblemsProblem 1:While walking to school you pass a barn with a silo. Looking up to thetop of the silo you estimate the angle of elevation to the top of the silo tobe about 14 . You continue walking and find that you were around40 m from the silo. Using this information and your knowledge oftrigonometric ratios calculate the height of the silo.A14 B40 mCProblem 2:A sailboat is approaching a cliff. The angle of elevation from the sailboatto the top of the cliff is 35 0 . The height of the cliff is known to be about2000 m. How far is the sailboat away from the base of the cliff?A2000 m35 CB

MBF3CBLM2.3.1Trigonometry Application Problems (Continued)Problem 3:A sailboat that is 2 km due west of a lighthouse sends a signal to thelighthouse that it is in distress. The lighthouse quickly signals a rescueplane that is 7 km due south of the lighthouse. What heading from duenorth should the plane take in order to intercept the troubled sailboat?2 kmBC7 kmA

MBF3CBLM2.3.2Name:Date:Round ’s to whole degrees. Length answers should be rounded to 1 decimal place andinclude units.1.The top of a lighthouse is 100 m above sea level. The angle of elevation from the deck ofthe sailboat to the top of the lighthouse is 28 . Calculate the distance between thesailboat and the lighthouse.100 m28 CB2.An archer shoots and gets a bulls-eye on the target. From the archer’s eye level the angleof depression to the bulls-eye is 5 . The arrow is in the target 50 cm below the archer’seye level. Calculate the distance the arrow flew to hit the target (the dotted line).C5 B50 cmAFor the following questions you will need to create your own diagrams. Draw themcarefully and refer to the written description to ensure you find the correct solution.3.Two islands A and B are 3 km apart. A third island C is located due south of A and duewest of B. From island B the angle between islands A and C is 33 . Calculate how farisland C is from island A and from island B.4.The foot (bottom) of a ladder is placed 1.5 m from a wall. The ladder makes a 70 anglewith the level ground. Find the length of the ladder. (Round to one decimal place.)5.A tow truck raises the front end of a car 0.75 m above the ground. If the car is 2.8 m longwhat angle does the car make with the ground?6.A construction engineer determines that a straight road must rise vertically 45 m over a250 m distance measured along the surface of the road (this represents the hypotenuse ofthe right triangle). Calculate the angle of elevation of the road.Solutions:1. 188.1 m4. 4.4 m2. 573.7 cm5. 16 3. Distance A to C: 1.6 km Distance B to C: 2.5 km6. 10

Unit 2 Day 4: Trigonometry – Sine LawMBF 3CMaterialsDescriptionThis lesson introduces the Sine Law to the students.Minds On Computer Labwith Geometer’sSketchPad.BLM2.4.1, 2.4.2AssessmentOpportunitiesWhole Class Æ DiscussionDraw the following diagram on the board:Bx1060A0500CPose the following question to the students:With what we know from the diagram, can we find the value of x?Ask the students, if more information was provided like the followingdiagram if that would help them solve for x? Lead students throughthe following solution:h sin 60 010Bx10A600h500CIntendedResponse:No, too hard forus!The studentsare discoveringthe sine lawusing numericalvalues. h 8.66cmh sin 50 0Now,xx 8.66sin 50 0x 11.3 cm0Observation: h 10sin60and h 11.3sin50 0in this caseAction!ac sin A sin CPose the question to them: Is this true all the time?Pairs Æ InvestigationStudents use Geometer’s Sketchpad to explore the sine law followingBLM2.4.1.If the schooluses LanSchoolor somebroadcastcapability for theteacher then theteacher couldbroadcast thesteps first andlet the studentstry it afterward.

Whole Class Æ Guided InstructionDemonstrate to students how to use the Sine Law to solve triangles.Example 1: In ΔABC, given that B 48 , C 25 , and side a (namedas BC) BC 36 cm. Find the length of AB and AC correct to 1 decimal place.(Help by drawing the diagram included below.)A48 25 B36 cmCAccording to the Sine Law we need a ratio of the sine of an angle and itscorresponding side, currently we don’t have this, however, we do have Band C so we are able to solve for A. A 180 - (48 25 ) A 43 We have BC 36 cm and we need to find AB and AC.So using the Sine Lawabc sin A sin B sin Cwe have:36ABAC sin 43 sin 48 sin 25 Solving separately:36AB sin 43 sin 48 andHere, multiply both sides by sin 48 andGives:sin 48 36 ABsin 43 and36AC sin 43 sin 25 here, multiply both sides by sin 25 sin 25 36 ACsin 43 Finishing off using a calculator: AB 39.2 cmandAC 22.3 cmHave students working with a partner solve example 2.Example 2: Solve for the value of h inthe following diagram:h43 56 68m

Possible Solution:Let’s label the diagram A at the peak and BC on the base. Thus, inabc sin A sin B sin CWe get:bc68 sin 81 sin 43 sin 56 We can solve for either of b or c and then use the primary trigonometric ratiosto complete the solution for h.68b sin 81 sin 43 or68c sin 81 sin 56 Solving for the chosen side following the same steps as above we get:b 47.0 morc 57.1 m Finally to solve for h use the sine trigonometric ratio.Sin 56 ConsolidateDebrief47hh 38.9 m57horsin 43 orh 38.9 mWhole Class Æ DiscussionSummarize when to use the Sine Law.The Sine Law:Using the same triangle above you could construct another perpendicular line from Bor C to the opposite side and create a similar expression for sin A and itscorresponding side a. In general the Sine Law takes the form:sin A sin B sin C abcorabc sin A sin B sin CHome Activity or Further Classroom ConsolidationConcept PracticeStudents complete BLM2.4.2.Emphasize thatthe two equalsigns constitutesthree equations.

MBF 3CBLM 2.4.1Name:Date:Investigation: Sine Law – Geometer’s Sketchpad1. Load Geometer’s Sketchpad.2. Start with a new document (default).3. Select the Straightedge Tool(4th button down the toolbar)4. Draw three lines – making sure thateach new line starts from a previousline and that the last point returns tothe first completing the triangle.(shown right)5. Switch to the selection tool(1st button on the toolbar)6. Select and right-click on each vertexand from the short-cut menu select“Show Label” (also shown right)7. Next select any line and from theMeasure menu (or from the right-clickshort-cut) select “Length”.This should display m AB (shown)8. Repeat Step 6 for the other lines, makingsure to unselect before selecting a newline. (If anything else is selected lengthmay not appear on the menu.)9. Next select in the following order thevertices: A, B then C – then click theMeasure menu and choose “Angle”.This should display m ABC and themeasure of that angle.10. Now repeat Step 8 but for angles BACand ACB. (shown)11. If you select any point you can drag the point to a new location and all of themeasurements update automatically. (You can also select and move an entireline.)12. Try this and adjust the position of the triangle toleave more room below our measurements.13. We will now add some calculations namely thevalues for the Sine Law:abc sin A sin B sin C14. To do this select the Measure menu and select“Calculate ”. A new dialogue box appears(shown right) where we will enter our calculation.15. First click on the measurement for side a (in thiscase it is m BC ), then click on the division sign

and type “si” for the sine function, next click on the measurement for A (in thiscase it is m BAC (depending on the size of your triangle you will see differentresults.) Click OK16. This will add a new measurement to your document, repeat step 15 for side band side c. For side b use m CA and sin(m ABC) for side c use m AB andsin(m ACB). Calculations are shown in the bottom diagrams.17. Now change the position of your vertices; this will change the lengths and anglesin your triangle – make note of what happens to all three of the calculation boxesabc sin A sin B sin C . (two variations shown below)for the Sine Law:18. Next create three more calculations for the other version of the Sine Law:sin A sin B sin C abc(shownright)19. Experiment with more positions of thetriangle vertices.20. Notice that the set of three values ineither version of the Sine Law remainthe same. This shows that the ratioof any side to the sine of thecorresponding angle in a triangleremains equal to the ratio of any otherside to the sine of the correspondingangle. Eithersin A sin B sin C abcorabc sin A sin B sin C

MBF3CBLM2.4.2Name:Date:1.Solve for the given variable (correct to 1 decimal place) in each of the following:(a)a10 sin 35 sin 40 2.b65 sin 75 sin 48 (b)(c)c75 sin 55 sin 80 For each of the following diagrams write the equation you would use to solve for theindicated variable:CB(b)(c)b75 c23.6 cm15 35 aA(a)53 AB46 36 cm14.2 mC73 C3.Solve for each of the required variables from Question #2.4.For each of the following triangle descriptions you should make a sketch and then findthe indicated side rounded correctly to one decimal place.(a)In ΔABC, given that A 57 , B 73 , and AB 24 cm. Find the length of AC(b)In ΔABC, given that B 38 , C 56 , and BC 63 cm. Find the length of AB(c)In ΔABC, given that A 50 , B 50 , and AC 27 m. Find the length of AB(d)In ΔABC, given that A 23 , C 78 , and AB 15 cm. Find the length of BC(e)In ΔABC, given that A 55 , B 32 , and BC 77 cm. Find the length of AC(f)In ΔABC, given that B 14 , C 78 , and AC 36 m. Find the length of BCSolutions:1. (a) 8.9 units (b) 50.0 units (c) 90.2 units2. (a)abc3623.614.2 (b)(c)sin 53 sin 81 sin 35 sin 70 sin 15 sin 73 3. (a) 29.1 cm (b) 38.7 cm (c) 52.5 m4. (a) 30.0 cm (b) 52.4 cm (c) 34.7 m (d) 6.0 cm (e) 49.8 cm (f) 148.7 mB

Unit 2 Day 5 - Trigonometry - Cosine LawMBF AssessmentOpportunitiesThis lesson introduces the Cosine Law to the students.Minds On Whole Class Æ DiscussionADraw the following diagram on the board:12 cmc60 BC26 cmAsk students to solve for c in the triangle above using the Sine Law:Results:c2612 sin A sin B sin 60 The solution is stalled at this point since each part of the ratio has some missinginformation. We cannot solve the triangle. We need to develop a new formula – thisformula is called the Cosine LawAction!Whole Class Æ Guided InstructionPose to the students: Haven’t you always dreamed about using the5Pythagorean Theorem for all triangles?Recall the most famous Pythagorean triangle35 2 32 4 24ForΔ ABC,A4BxDh2 x 2 42andh2 42 x 2From1and26h1aC5– xh 2 (5 x )2 62h 2 6 2 (5 x )4 2 x 2 6 2 (5 x) 24 2 x 2 6 2 (5 2 10 x x 2 )4 2 x 2 5 2 x 2 10 x 6 24 2 5 2 10(4 cos A) 6 2c 2 b 2 2bc cos A a 2Is this true for all triangles?22x4 4 cos A xcosA

Guide students through example 1 and 2 to solve for a missing side and a missingangle using the Cosine Law.Example 1: Now we can solve for c from our first problem (redrawn below)Ac2 a2 b2 – 2ab · cos C222c (26) (12) – 2 (26) (12) · cos 60 c2 676 144 – 624 · (0.5)12 cmcc2 676 144 – 312c2 50860 c 2 508 c 22.54 cmB26 cmCExample 2: In ΔABC, given BC 7 cm, AC 8 cm and AB 10 cm. Find themeasure of A to the nearest degree.(Help by drawing the diagram included below.)A10 cm8 cmB7 cmCWe are asked to find the measure of A to the nearest degree so we should use theCosine Law formula that is most appropriate:a2 b2 c2 – 2bc · cos AFilling in what we know gives:(7)2 (8)2 (10)2 – 2(8)(10) · cos A49 64 100 – 160 · cos ARearranging:160 · cos160 · cos160 · cos A A A 64 100 64 100 115– 49– 49Finally:115160cos A cos A 0.71875Use your calculator to solve for A A 44.048625674 ConsolidateDebriefConcept Practice A to the nearest degree is 44 Whole Class Æ DiscussionAsk students to set up the equations using the Cosine Law for questions #1a,and #1b on BLM2.5.1. Verify that they have the correct set up.Home Activity or Further Classroom ConsolidationStudents complete BLM2.5.1.

MBF3CBLM2.5.1Name:Date:1.For each of the following diagrams write the equation you would use to solve for theindicated variable:B(a)(b)(c)C28.4 cmc26 cma23.6 cmA B14.2 m33.2 cm53 22.4 mB75 A36 cmCCB2.Solve for each of the required variables from Question #1.3.For each of the following triangle descriptions you should make a sketch and then findthe indicated value.(a)In ΔABC, given that AB 24 cm, AC 34 cm, and A 67 . Find the length of BC(b)In ΔABC, given that AB 15 m, BC 8 m, and B 24 . Find the length of AC(c)In ΔABC, given that AC 10 cm, BC 9 cm, and C 48 . Find the length of AB(d)In ΔABC, given that A 24 , AB 18.6 m, and AC 13.2 m. Find the length of BC(e)In ΔABC, given that AB 9 m, AC 12 m, and BC 15 m. Find the measure of B.(f)In ΔABC, given that AB 18.4 m, BC 9.6 m, and AC 10.8 m. Find the measure of A.Solutions:(a) a2 (36)2 (26)2 – 2(36)(26) · cos 53 (b) (28.4)2 (23.6)2 (33.2)2 – 2(23.6)(33.2) · cos B(c) c2 (22.4)2 (14.2)2 – 2(22.4)(14.2) · cos 75 2. (a) 29.1 cm (b) 57 (c) 23.2 m3. (a) 33.1 cm (b) 8.4 m (c) 7.8 cm (d) 8.5 m (e) 53 (f) 24 1.

Unit 2 Day 6 Trigonometry – Applying the Sine and Cosine LawMBF 3CDescriptionMaterialsThis lesson has students solving triangles by choosing between the SineLaw and the Cosine Law.BLM2.6.1,2.6.2AssessmentOpportunitiesMinds On Whole Class Æ Four CornersPost four signs, one in each corner labelled – Sine Law, Cosine Law,Pythagorean Theorem, Trigonometric Ratio (SOHCAHTOA).Provide each of the students with one of the triangles on BLM2.6.1.Instruct the students to make a decision as to which method they would use tosolve the missing angle or side and to stand in the corner where it is labelled.Once students are all placed have the students discuss amongst themselves toconfer that they have selected an appropriate method. Allow students tomove to a different location after discussion. Ask one representative fromeach corner to explain why their triangle(s) would be best solved using thatparticular method.Ask students to complete a Frayer model (BLM2.6.2) for their method. Addto word wall or class math dictionary.Action!Pairs Æ PracticeAsk students to individually solve the triangle they were given for the previousactivity. Then with an elbow partner check each others work.Possible solutions for the triangles on teachers copy of the Four Corners.ConsolidateDebriefWhole Class Æ DiscussionClarify any problems from the pairs solving for their missing side or angle.Highlight what the phrase “solving a triangle” means (solve for all sides and anglesin a triangle).Pairs Æ PracticeAsk pairs to trade questions and solve for the remaining parts of their triangle.BLM2.6.1Answers:1.Sine Law2.Cosine Law3.Cosine Law4.Sine Law5.PythagoreanTheorem6.Trig RatioSupport studentswith thecharacteristics. Maywant to includeitems like:3 sides Æ you needto use Cosine Lawto solve for angles.2 sides and 1 angleÆ it depends onwhere the angle islocatedif the angle isbetween the twosides (contained)then you need touse the Cosine Law– usually to find lastside.ii) if the angle is notcontained then theSine Law can beused usually to findanother angle.1 side and 2 anglesÆ then you need touse the Sine Law,usually to find aside.ApplicationHome Activity or Further Classroom ConsolidationStudents complete BLM2.6.3

MBF3CBLM2.6.1Four Corners – TrianglesB1)2)Ca56 ABa14 m80 A43 38 cm22 mC3)4)C23 cm AAB28 cm26 cmB33 cm37 cm53 AC5)A6)Ac125 cm123 cm BC65 cmBC45 cmBA

MBF3CBLM2.6.1Four Corners – Triangles (Teacher)B1)2)Ca56 A14 min aNow we can solve for a using the Sine Lawab sin A sin Ba38 sin 56 sin 81 a2 b2 c2 – 2bc · cos Aa2 (14)2 (22)2 – 2(14)(22) · cos 80 2a 196 484 – 616 · 0.1736481782a 196 484 – 106.96727744a2 573.03272256 38 sin 56 sin 81 38a 0.829037570.98768834a 31.9 cmc c 38 sin 43 sin 81 38 0.681998360.98768834c 26.2 cm 573.03272256a 23.9 msin B Solving for the other sides:bc sin B sin C38c sin 81 sin 43 a2Solving for the other angles.sin A sin B absin 80 sin B 23.914sin 80 14 sin B23.9a vs.22 mC B we can do this using the sum of the anglestriangle. B 180 - 56 - 43 Therefore, B 81 Sine Law80 A43 38 cmFirst we need to find BaCosine Lawsin B 0.57687483c2 a2 b2 – 2ab · cos C c2 (31.9)2 (38)2 – 2(31.9)(38) cos 43 c2 1017.61 1444 – 2424.4(0.7313537)c2 688.5160858c 26.2 cm0.98480775 1423.9 B 35.231034 B 35 Æ this gives us C 65 (i.e. 180 - 80 - 35 )

MBF3CBLM2.6.1Four Corners – Triangles (Teacher)3)4)CA26 cm28 cm23 cm AB8341518cos A 0.54940711 A 56.67365194 A 57 Solving for the other angles:sin A sin B absin 57 sin B 2823sin 57 23 sin B280.83867057sin B 2328sin B 0.688907967 B 43.543727 B 44 Æ this gives us C 79 (i.e. 180 - 57 - 44 )37 cm53 33 cma2 b2 c2 – 2bc · cos A(28)2 (23)2 (33)

1 Review Trigonometry – Solving for Sides Review Gr. 10 2 Review Trigonometry – Solving for Angles Review Gr. 10 3 Trigonometry in the Real World C2.1 4 Sine Law C2.2 5 Cosine Law C2.3 6 Choosing between Sine and Cosine Law C2.3 7 Real World Problems C2.4 8 More Real World Problems C2.4 9