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SLO The basics of Trigonometry http://www.youtube.com/watch?v=lJqhTr0SI4k (The basics and how to label the sides of a triangle) 1 of 43 Where did it come from? •Trigonometry was probably invented for use in astronomy. •The origins of trigonometry can be traced to the civilizations of ancient Egypt, Mesopotamia and the Indus valley, more than 4000 years ago. 2 of 43 Copy into your notes Words to Learn Angle of Elevation: angle from horizontal upwards Angle of Depression: Angle from horizontal downwards dp: decimal place Trigonometry: The word comes from Greek: trigōnon "triangle" + metron "measure“ (sine, cosine and tangent) Vertical: up/down Horizontal: left/right Plane: flat surface Ascend: To go up Descend: To go down 3 of 43 Copy into your notes Naming parts of a triangle The side opposite the marked angle is called the opposite side. Take care, the opposite and adjacent can ‘swap’ places The longest side is called the hypotenuse. O P P O S I T E H Y P O T E N U S E θ This letter is called Theta (a Greek letter) ADJACENT The side next to the marked angle is called the adjacent side. 4 of 43 Label the sides 5 of 43 Copy into your notes Using a calculator Finding the value of sin 65°? First make sure that your calculator is set to degrees. Key in: sin 6 5 = This is important Your calculator should display 0.906307787 Only round answers at the final step 6 of 43 Copy into your notes The inverse of sin Finding the value of sin–1 0.5 To work this out use the sin–1 key on the calculator. shift sin 0 . 5 = The answer to the above is 30. sin–1 is the inverse or opposite of sin. It is sometimes called arcsin. sin 30° 0.5 sin–1 7 of 43 Your Turn: Find the following. 1) sin 79° = 0.982 2) cos-1 0.4 = 3) tan 65° = 4) cos 11° = 0.982 2.14 66.42 5) sin-10.6 = 36.87 6) tan 84° = 7) tan 49° = 1.15 8) sin 62° = 0.883 9) tan-11.3= 52.43 10) cos 56° = 0.559 8 of 43 9.51 SLO Introduction to SOHCAHTOA http://www.youtube.com/watch?v=54hXSHe1SQE (Remembering sohcahtoa) 9 of 43 Copy into your notes SOHCAHTOA You MUST Some Old Hippy Caught Another Hippy Tripping On Acid remember this 10 of 43 Copy into your notes What do the letters of SOHCAHTOA mean? Guess what the letters stand for Sin SOHCAHTOA Adjacent Opposite Opposite Hypotenuse Tangent Cos 11 of 43 Adjacent Hypotenuse Copy into your notes From SOHCAHTOA to triangles (click to show how to form triangles) SOH θ 12 of 43 CA H θ TOA θ Copy into your notes How to use the SOHCAHTOA triangles (same method as Distance/time triangles in science) If one letter is above another letter, divide O SxH θ If one letter is next to another letter, multiply 13 of 43 How to use a SOHCAHTOA triangle Basic Rule: Put finger over what you want to find E.g. 1) What does the opposite equal? O SxH θ Opposite = Sin (angle) x Height 14 of 43 Copy into your notes How to use the SOHCAHTOA triangles Basic Rule: Put finger over what you want to find E.g. 2) What does the Hypotenuse equal to? O Sθ H Hypotenuse = Opposite ‚ Sin(angle) 15 of 43 Your turn: Use the triangle given below to find how you would calculate the given lengths 1) Adjacent A Cx H θ Cos(Angle) x Hypotenuse 2) Hypotenuse Adjacent ‚ Cos(Angle) 16 of 43 Your turn: Use the triangle given below to find how you would calculate the given lengths 1) Adjacent O T xA θ 17 of 43 Opposite ‚ Tan(Angle) 2) Opposite Tan(Angle) x Adjacent SLO How to decide which part of SOHCAHTOA to use 18 of 43 Copy into your notes SOH or CAH or TOA O A O S H C H T A When we use Trigonometry we need to use the correct part of SOHCAHTOA Step 1: Identify what sides you have and what sides you have been asked to find. Step 2: Find the SOHCAHTOA triangle with these 2 sides. 19 of 43 SOH or CAH or TOA E.g. for the following triangle do you use sin, cos or tan to find the length x? Step 1: Label what is given and what is to find Find: Hypotenuse x 390 Step 2: Which part of SOH CAH TOA uses H and A 27cm A Given: Adjacent 20 of 43 C H Copy into your notes SOH or CAH or TOA E.g. for the following triangle do you use sin, cos or tan to find the length x? Step 1: Label what is given and what is to find Find: Opposite Step 2: Which part of SOH CAH TOA uses O and A x 390 O 27cm Given: Adjacent 21 of 43 T A Your Turn: For the following triangle do you use sin, cos or tan to find the length x? 19cm O x 250 S H 67cm A 470 x 22 of 43 C H Your Turn: For the following triangle do you use sin, cos or tan to find the length x? x O 8cm 250 470 S H O x T A 83cm 23 of 43 Questions to do from the books SOH or CAH or TOA Achieve Gamma IAS 1.7 24 of 43 P402 Ex28.01 Q1 Merit Excellence Copy into your notes SLO Finding lengths of sides Note Trigonometry is for right angled triangles only Trigonometry must involve angles http://www.youtube.com/watch?v=D38S1sgErjQ (Finding sides using SOHCAHTOA) 25 of 43 Copy into your notes E.g. 1: Find the length of the side marked X [1] Label given side and side to find. 380 [2] Use information from the diagram to choose which part of SOHCAHTOA to use 48m [4] check answer reasonable 26 of 43 O X O S [3] Use SOHCAHTOA triangle to find length (Cover up what you want to find) H H X Sin38 48 X = 29.6m Opposite is smaller than Hypotenuse therefore good answer E.g. 2: Find the length of the side marked X in the diagram below. [1] Label given side and side to find. [2] Use information from the diagram to choose which part of SOHCAHTOA to use 48m A 570 H X A C H [3] Use SOHCAHTOA triangle to find length (Cover up what you want to find [4] Check answer reasonable 27 of 43 X = 48 ‚ cos57 X = 88.1 m Hypotenuse bigger than adjacent, therefore good answer. Your Turn: Find the length of the side marked X [1] Label given side and side to find. 17m 290 [2] Use information from the diagram to choose which part of SOHCAHTOA to use [3] Use SOHCAHTOA triangle to find length (Cover up what you want to find [4] Check answer reasonable Hard to check this one. 28 of 43 A O X O T A X = Tan29 x 17 X = 9.4 m Your Turn: Find the length of the side marked X 15m X [1] Label given side and side to find. H O 340 [2] Use information from the diagram to choose which part of SOHCAHTOA to use S [3] Use SOHCAHTOA triangle to find length (Cover up what you want to find X = 15 ‚ sin34 X = 26.8 m [4] Check answer reasonable 29 of 43 O H Hypotenuse is longer than opposite therefore good answer. Your Turn: Find x to 2 decimal places. O 12 cm x S H x = 12 × sin 56° x = 9.95 cm 56° x 50° 6.5 cm 30 of 43 O T A x = tan50 x 6.5 x = 7.75 cm Your Turn: Find x to 2 decimal places. x 32° 455 m O S H x O T A 31 of 43 x = 455 ‚ sin32 x = 858.62 m x = 7 ‚ tan43 x = 7.51m Questions to do from the books Achieve Gamma P402 Ex28.01 Q3 – 5 IAS 1.7 P8 Q19 – 40 Merit Excellence P12 Q41 – 49 Some harder Achieve questions to follow but first practice the basics 32 of 43 Applications of Trigonometry A 5 m long ladder is resting against a wall. It makes an angle of 70° with the ground. What is the distance between the base of the ladder and the wall? 5m We are given the hypotenuse and we want to find the length of the side adjacent to the angle, so we use: A 70° x C H x = 5 × cos 70° = 1.71 m 33 of 43 Your Turn: A ladder 4.7 m long is leaning against a wall. The angle between the wall and ladder is 27°. Draw a diagram and find the height the ladder extends up the wall. (Click for diagram) 27° Ladder (4.7 m) Wall (x) A A C H H x = cos27 x 4.7 x = 4.19 m 34 of 43 Your Turn: A building that is 5.6 m high is to have a rope hooked from it’s top to make an angle of 70° with the ground. How long will the rope need to be? (Click for diagram) O H rope 5.6m building O S H 70 o Rope = 5.6 sin(70) = 5.96 m 35 of 43 Your Turn: To find the height of the school flagpole, the angle from a person’s eye-level to the top of the flagpole is measured. This angle is 23°. If the person is standing level with the base of the flagpole and 3.5 m from it, and their eye is 1.63 m above the ground, how tall is the flagpole? (Click for diagram) 23° A = 3.5 m x=0 O T A 1.63 m 3.5 m 36 of 43 flag = 1.63 + 3.5 x tan(23) = 3.1 m Your Turn: When crossing a river which is 4.2 metres wide with parallel river banks, a boat forms an angle of 55° with the bank where it starts. How far down the river will the boat reach the opposite bank? (Click for diagram) 4.2 m 55 o Starting point 37 of 43 A O O T A A = 4.2 ÷ tan(55) = 2.9 m Your Turn: 7. A metal brace is to be used to support a frame with an angle of 70° on a base of 1.2 m as shown. (a) How long should the metal brace be? (b) How high up the frame will the brace meet the vertical frame? 70° 1.2 m (b) (a) A O C H T A d = 1.2 ÷ cos(70) Metal brace length = 3.5m 38 of 43 ht = 1.2 x tan(70) ht = 3.3m Finding side lengths 39 of 43 Your Turn: Find the height of the tree x 23 1.72m 12m O tan A x tan 23 12 x 12 tan 23 x 5.09m(3sf ) height of the tree 1.72 5.09 6.8m(2sf ) The tree is approximately 7m tall 40 40 of 43 Your Turn: When throwing the discus, Anna’s throw is in line with the centre of the throwing circle and a distance of 25.6 m straight out in the middle of the allowable sector. Bernice’s throw is on an angle of 12o from the middle and is a perpendicular distance of 24.8 m.` Who threw the greatest throw and by how much? (click for clue) Step 1: label sides: Step 2: C H 24.8 m d A 12° H Step 3: d = 24.8 ‚ cos(12) = 25.35m Anna wins by (25.6 – 25.35) = 0.25 m 41 of 43 distance perpendicular middle line A A B Questions to do from the books Achieve Gamma P402 Ex28.01 Q3 – 5 IAS 1.7 P8 Q19 – 34 42 of 43 Merit P11 Q35 – 49 Excellence SLO Finding angles http://www.youtube.com/watch?v=6wG3_x40YBA (Finding angles using SOHCAHTOA) 43 of 43 Copy into your notes SOHCAHTOA to find angles O A O S H C H T A Step 1: Label the 2 sides you are given using the angle you have been asked to find. Step 2: Find the SOHCAHTOA triangle with the 2 given sides. Step 3: Finger over S or C or T (this is the angle). Step 4: Use inverse button and SOHCAHTOA triangle to find angle (you will need to use brackets). 44 of 43 Copy into your notes E.g. 1: Find the size of angle x METHOD [1] Label 2 given sides. 27m O [2] Choose from sin/cos/tan. x A 42m O 45 of 43 [3] Finger over S, C or T T A [4] Calculate angle using inverse button ( -1 ): remember brackets. X = tan-1(27 ‚ 42) X = 32.70 E.g. 2: Find the size of angle x METHOD [1] Label 2 given sides. x H 61m [2] Choose from sin/cos/tan. O 52m O [3] Finger over S, C or T [4] Calculate angle using inverse button ( -1 ): remember brackets. 46 of 43 S H X = sin-1(52 ‚ 61) X = 58.50 E.g. 3: Find the size of angle x METHOD [1] Label 2 given sides. 34m H [2] Choose from sin/cos/tan. x A 17m A [3] Finger over S, C or T [4] Calculate angle using inverse button ( -1 ): remember brackets. 47 of 43 C H X = cos-1(17 ‚ 34) X = 600 Your Turn: 8 cm Find θ to 2 decimal places. 5 cm θ We are given the lengths of the sides opposite and the adjacent to the angle, so we use: O T A 48 of 43 θ = tan–1 (8 ‚ 5) = 57.99° (to 2 d.p.) Your Turn 1) Find angle A 16.1mm O 2) Find angle B 23.4 mm H A O S H A = sin-1(16.1 ‚ 23.4) A = 43.5° (1 d.p.) 49 of 43 B 2.15m A 4.07 m H A C H B = cos-1(2.15 ‚ 4.07) B = 58.1° (1 d.p.) Your Turn: A vertical mast is held by a 48 m long wire. The wire is attached to a point 32 m up the mast. Draw a diagram and find the angle the wire makes with the mast. (click for diagram) A C H H A 32 m A 48 m A = cos-1(32 ‚ 48) A = 48.2° (1 d.p.) 50 of 43 Your Turn: A swimming pool is to be dug out of the ground. The side view cross-section is a trapezium shape, with a length of 25 m. The pool is to be 1.2 m deep at one end and 2.8 m deep at the other end. If digging starts at the shallow (1.2 m) end, what angle should be used to ensure that the pool will be 2.8 m deep when the 25 m length is dug? Step 1: Label sides Step 2: 25 m 1.2 m O T A Step 3: x = tan–1(0.064) = 3.7° (to 2 s.f.) 51 of 43 xo A 2.8 m O Finding angles 52 of 43 Angles of elevation 53 of 43 Angles of depression 54 of 43 Questions to do from the books Angles Achieve Gamma P404 Ex 28.02 Q3 IAS 1.7 P15 Q50 – 59 55 of 43 Merit P16 Q60 – 67 Excellence 3D Shapes Most 3D questions involve Pythagoras, Trigonometry and 2 steps 56 of 43 Pythagoras Revision Find the length of the side marked P 1230 m P² = 1230² + 960² = 2 434 500 p P = 1560 (to 3 s.f.) 57 of 43 960 m Pythagoras Revision Find the length of the side marked y 13.2 m y² = 32.7² – 13.2² = 895.05 y = 895.05 = 29.9 (to 3 s.f.) 58 of 43 32.7 m y Questions to do from the books (Pythagoras) Achieve Gamma P386 Ex27.01 Q4–15 Merit Excellence P387 Ex27.01 Q16 – 17 P394 Ex27.03 P389 Ex27.02 IAS 1.7 59 of 43 P4 Q1 - 6 P4 7 – 18 P412 Ex 29.01 Q8 – 11 The following questions are in 3D and involve a combination of Pythagoras and Trigonometry 60 of 43 Copy into your notes E.g. Calculate the size of: a) length x, G O 7m H b) length w, c) angle CHE and d) angle GCH length x 2 = 52 + 62 x A B x = √(52 + 62) x = √61 F x = 7.8 m (1 d.p.) w Make sure length w 2 = 72 + 7.82 you use w D C whole w = √(72 + 7.82) Ax 5m answer for w = √110 O x in w = 10.5 m (1 d.p.) E 6 mA calculation angle CHE O T A CHE = tan-1(5 ‚ 6) CHE = 39.8° (1 d.p.) 61 of 43 angle GCH O T A GCH = tan-1(7 ‚ 7.8) GCH = 41.9° (1 d.p.) Copy into your notes Two step problems E.g. Find angle α Clue: Draw separate triangles (Click for first half of answer) G 10m A 8m D F 6m a 2 b2 h2 (8) 2 (10) 2 h 2 164 h 2 h 12.8m(1dp ) AG = 12.8m (1dp) 62 of 43 A 12.8m G O tan A 6 tan 12.8 6 tan 1 12.8 25.1(1dp) Your Turn: B Find the angle θ 15cm A G F 10cm 15cm H E 10cm H O A C 12 tan 18.03 12cm 1 12 tan G 18.03 33.6(1dp) tan a 2 b2 h2 (10) 2 (15) 2 EG 2 100 225 EG 2 E 18.03cm(2dp) 63 of 43 G To find EG EG 325 12cm D (Click for half the answer) E C 18.0cm Your Turn: A ramp is to be used to drive a wheel of a vehicle up a height of 0.7 m. The angle the ramp forms with the ground is 15° and the ramp is to be 0.5 m wide. A diagonal brace is to be used as shown How long will this brace be? (click for first half of answer) Side face can be used to find slant length, d d 0.7 m H Step 1: Label sides O Step 2: 15 o 0.5 m O h, b t g n e le S H Step 3: d = 0.7 ÷ sin(15) Slant length = 2.7 m (to 2 s.f.) Now find brace length c bra slant length d Use the right-angle triangle on slant b² = 2.7² + 0.5² = 7.5648 b = 8.73 = 2.75 Brace length = 2.8 m (to 2 s.f.) 64 of 43 Your Turn: A chair has a square seat with 4 legs 65 cm long and 80 cm apart. A bar joining two of the adjacent legs is bolted 2 cm from the floor. If the bar is to form an angle of 36o with a line parallel to the floor, find: (a) how far up the leg the bolt hole should be drilled (b) the length of the bar between bolt holes. bar 65 cm b H O d 36 36o o 80 cm A 80 cm (a) Step 1: Label sides Step 2: TOA Step 3: d = 80 x tan (36) = 58.1 (to 3 s.f.) Distance up leg = 58.1 + 2 = 60.1 cm 65 of 43 36 o A 80 cm (b) Step 1: Label sides Step 2: CAH Step 3: b = 80 ÷ cos (36) Length of bar = 98.9 cm (to 3 s.f.) Your Turn: A chair has a square seat with 4 legs 65 cm long and 80 cm apart. A bar parallel to the floor joining two diagonally opposite legs is bolted 2 cm from the floor. Another bar is slanted to join diagonally opposite legs 2 cm from the floor and 2 cm from the seat on the other leg. What angle is there between the horizontal bar and the slanted bar? (Click for halfway workings) First step: we need Floor Parallel Bar length ar db nte sla 65 cm r floor parallel ba Floor bar’s length = d. d² = 80² + 80² = 12 800 d = 12 800 = 113 cm (3 s.f.) d cm 80 cm 80 cm Floor bar length =113 cm (3 s.f.) Step 1: Label sides Step 2: Step 3: x = tan–¹(61 ÷ 113) = 28.4° 66 of 43 O T A b cm 61 cm O 113 cmA 80 cm Angles in a cuboid 67 of 43 Lengths in a square-based pyramid 68 of 43 Questions to do from the books 3D shapes Achieve Merit Gamma P414 Ex29.02 Q6 – 11 P420 Ex29.05 IAS 1.7 P21 Q72 – 73 P26 Q78 – 79 69 of 43 Excellence P19 Q68 – 70 P24 Q74 – 77 Revision http://www.youtube.com/watch?v=E1MAUC5TLOI (Revision of Trigonometry + Pythagoras) http://www.youtube.com/watch?v=hG-1RO0TghA (Good revision of Trigonometry(sides and angles). Achieve: watch first 17 minutes, Ignore: 17.10 to 21.05 Merit: watch 21.05 onwards 70 of 43 Achieve Trigonometry Revision (Angles and sides) 71 of 43 Find the length of the side marked x: Step 1: Label sides Step 2: 16 cm A C H H 39 o xcm x = 16 x cos (39) = 12.4 (to 3 s.f.) 72 of 43 A Find the size of the angle, Theta: o Step 1: Label sides Step 2: O S H Step 3: Θ = sin–¹(32.8 ÷ 45.8) = 45.7 (to 3 s.f.) 73 of 43 H 45.8 m O 32.8 m Find the angle marked k: k Step 1: Label sides Step 2: O S H Step 3: k = sin–¹(0.4) = 23.6 (to 1 d.p.) 74 of 43 o H O 2.8 7 Find the length of the side marked c: Step 1: Label sides Step 2: O S H Step 3: c = sin(23) x 8 = 3.1 mm (to 2 s.f.) 75 of 43 23 O c H o 8 mm Find the length of the side marked d: Step 1: Label sides Step 2: 7 O A T A Step 3: d = 7 x tan(43) = 6.5 (to 2 s.f.) 76 of 43 dO 43 o Find the angle marked y: Step 1: Label sides Step 2: A C H 75 m H y Step 3: y = cos-1 (50 ÷ 75) = 48.2 (to 1 d.p.) 77 of 43 o A 50 m Find the length of the side marked y: Step 1: Label sides Step 2: O S H 50.0 m O H Step 3: y = sin(75) x 50.0 = 48.3 m (to 3 s.f.) 78 of 43 y 75 o Find the angle marked p: 9 cm o p Step 1: Label sides Step 2: O T A Step 3: p = tan–¹(9 ÷ 16) = 29.4 (to 1 d.p.) 79 of 43 A 16 cm O Find the angle marked t: Step 1: Label sides Step 2: O S H Step 3: t = sin–¹(8 ÷ 18) = 26.4 (to 1 d.p.) 80 of 43 8 mm t O H o 18 mm Find the angle marked k: k Step 1: Label sides Step 2: O S H Step 3: k = sin–¹(0.4) = 23.6 (to 1 d.p.) 81 of 43 o H O 2.8 7 108 m Find the size of x: xm Step 1: Label sides Step 2: O T A Step 3: x = 108 x tan(18) = 35.1 (to 3 s.f.) 82 of 43 O Ao 18 Find the size of Y: y Step 1: Label sides Step 2: A C H Step 3: y = cos(34) x 75 = 62.2 (to 3 s.f.) 83 of 43 34 o A 75 H Find the size of h: Step 1: Label sides Step 2: ho 75 m A O T A Step 3: h = tan–¹(59 ÷ 75) = 38.2 (to 3 s.f.) 84 of 43 O 59 m