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PYTHAGORAS & TRIGONOMETRY

PYTHAGORAS & TRIGONOMETRY. PYTHAGORAS. Pythagoras Theorem states:. Can only occur in a right angled triangle. h 2 = a 2 + b 2. hypotenuse. h. - is opposite the right angle and is ALWAYS the longest side. a. right angle. b. Calculating the Hypotenuse :. e.g. 7.65 mm. e.g. x. 3 m.

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PYTHAGORAS & TRIGONOMETRY

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  1. PYTHAGORAS & TRIGONOMETRY

  2. PYTHAGORAS Pythagoras Theorem states: Can only occur in a right angled triangle h2 = a2 + b2 hypotenuse h - is opposite the right angle and is ALWAYS the longest side. a right angle b Calculating the Hypotenuse: e.g. 7.65 mm e.g. x 3 m 11.3 mm x 6 m x2 = 7.652 + 11.32 square root undoes squaring x2 = 32 + 62 x2 = 186.2125 x2 = 45 x = √186.2125 x = √45 x = 13.65 mm (2 d.p.) x = 6.71 m (2 d.p.)

  3. Calculating the smaller sides: e.g. y2 + 42 = 92 92 = y2 + 42 9 cm - 42 y2 = 92 – 42 - 42 y y2 = 65 y = √65 4 cm y = 8.06 cm (2 d.p.) smaller sides should always be smaller than the hypotenuse e.g. 8.6 mm y2 + 8.62 = 9.42 9.42 = y2 + 8.62 - 8.62 y2 = 9.42 – 8.62 - 8.62 y y2 = 14.4 9.4 mm y = √14.4 y = 3.79 mm (2 d.p.)

  4. PYTHAGOREAN TRIPLES - Special sets of whole numbers that fit into the Pythagoras equation e.g. 3, 4 and 5 x 2 - Each of these sets can be multiplied by numbers to find further triples. 6, 8 and 10 5, 12 and 13 x 3 15, 36 and 39 7, 24 and 25 x 5 35, 120 and 125 PYTHAGOREAN APPLICATIONS e.g. A ladder 5 m long is placed against the wall. The base of the ladder is 2 m from the wall. Draw a diagram to show this information and calculate how high up the wall the ladder reaches. Wall (x) Ladder (5 m) x2 + 22 = 52 52 = x2 + 22 x2 = 52 - 22 -22 -22 x2 = 21 x = √21 x = 4.58 m (2 d.p.) Base (2 m)

  5. TRIGONOMETRY (SIN, COS & TAN) - Label the triangle as follows, according to the angle being used. to remember the trig ratios use SOH CAH TOA Hypotenuse (H) Opposite (O) and the triangles A Always make sure your calculator is set to degrees!! Adjacent (A) O A O S H C H T A means divide 1. Calculating Sides means multiply e.g. O 7.65 m x H h T A O O 29° 50° h = tan50 x 6.5 6.5 cm A O x = sin29 x 7.65 h = 7.75 cm (2 d.p.) x = 3.71 m (2 d.p.) S H

  6. e.g. d = 455 ÷ sin32 O d H 455 m d = 858.62 m (2 d.p.) O S H 32° Inverses - are used when calculating angles. - are found above the sin, cos and tan buttons so the ‘shift’ or 2nd function is needed e.g. Find angle A if sin A = 0.1073 sin-1 undoes sin sin-1 A = sin-1(0.1073) sin-1 A = 6.2° (1 d.p.) tan-1 undoes tan e.g. Find angle B if tan B = ¾ tan-1 B = tan-1(¾) tan-1 If using the fraction button, you may need to use brackets! B = 36.9° (1 d.p.)

  7. 2. Calculating Angles -Same method as when calculating sides, except we use inverse trig ratios. e.g. B 4.07 m 23.4 mm H sin-1 undoes sin 2.15 m 16.1 mm H A O A cosB = 2.15 ÷ 4.07 sinA = 16.1 ÷ 23.4 A O B = cos-1(2.15 ÷ 4.07) A = sin-1(16.1 ÷ 23.4) C H S H B = 58.1° (1 d.p.) A = 43.5° (1 d.p.) Don’t forget brackets, and fractions can also be used

  8. TRIGONOMETRY APPLICATIONS e.g. 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. A Wall (x) C H 27° A Ladder (4.7 m) x = cos27 x 4.7 H x = 4.19 m (2 d.p.) A e.g. 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. C H H A 48 m cosA = 32 ÷ 48 32 m A = cos-1(32 ÷ 48) A A = 48.2° (1 d.p.)

  9. SCALE DIAGRAMS - are accurate plans of objects that can be smaller or larger in real life. - the scale gives information on how sizes on the plan and object relate. e.g. If a map has a scale of 1 cm = 200 km, how much would 4 cm on the map equate to in real life? 4 × 200 = 800 km e.g. Draw a right angled triangle with shorter sides of 3 cm and 6 cm and then accurately measure the hypotenuse. Is your hypotenuse approx 6.7 cm long? e.g. Wheelchair ramps must be no steeper than 10°. If a ramp starts 3 m away from a door that is 0.25 m above the ground, will it be safe? Use a scale of 2 cm = 1 m. You will need a ruler and a protractor. door ramp 0.5 cm 6 cm Now measure the angle between the ramp and ground. Is it safe? Why?

  10. VECTORS - Vectors describe a movement (translation). To describe vectors as a column vector: - top number describes sideways movement (negative = left and positive = right) - bottom number describe up/down movement (negative = down and positive = up) e.g. Draw the vector q = b vectors can start anywhere q e.g. Draw the vector b = e.g. Write vector CD as a column vector e.g. Write vector AB as a column vector

  11. BEARINGS - Bearings are used to indicate directions - Are measured clockwise from North - Must be expressed using 3 digits (i.e. 000° to 360°) - Compass directions such as NW give directions but are not bearings 000° e.g. The compass points and their bearings: 315° 045° e.g. Draw a bearing of 051°: NW NE N W E 270° 090° 51° SW SE 225° 135° e.g. What is the bearing of R from N? S N 180° Bearing = 180 + 37 = 217° 37°

  12. MAGNITUDE AND BEARINGS OF VECTORS - The magnitude of a vector is its length and is calculated using Pythagoras - The direction of a vector (bearing) is calculated using Trigonometry e.g. Calculate the magnitude and bearing of the vector : N Magnitude: x2 = 42 + 62 x = √42 + 62 x = √52 x = 7.2 units (1 d.p.) x O 6 Bearing: tanA = 6 ÷ 4 A = tan-1(6 ÷ 4) O A A = 56° Bearing = 56 + 270 T A A 4 = 326°

  13. e.g. A plane flying at 500km/hr heads North. A wind blows from the west at 50 km/hr. Find the actual direction the plane ends up heading on. Also calculate its final speed. 50 km/hr O N Final Speed: x2 = 502 + 5002 Actual flight path (x) x = √502 + 5002 500 km/hr x = √252500 A A x = 502.5 km/hr (1 d.p.) Bearing: tanA = 50 ÷ 500 A = tan-1(50 ÷ 500) O A = 5.7° (1 d.p.) Bearing = 006° T A

  14. GRID REFERENCE - Grid referencing is a six figure system of map co-ordinates used to give locations - The first three figures refer to the horizontal scale. - The last three figures refer to the vertical scale. e.g. A location has a grid reference of 295868. How is it found? On the horizontal scale we look for the 29 and then move 5 tenths further right On the vertical scale we look for the 86 and then move 8 tenths further up RAPID NUMBERING - RAPID stands for Rural Address Property IDentification - It accurately gives the location of rural properties - It is based on the distance a property is from the beginning of the road - The distance is measured in metres with the final measurement being divided by 10. If the property is on the right side, the number is rounded to the nearest even whole number and if it is on the left side, to the nearest odd number. e.g. What is the RAPID number for a property located 527 metres up the left side of a road? 527 ÷ 10 = 52.7 Left side means we round to nearest odd number = 53

  15. 3D FIGURES - Pythagoras and Trigonometry can be used in 3D shapes e.g. Calculate the length of sides x and w and the angles CHE and GCH A B x2 = 52 + 62 x = √52 + 62 G F x = √61 x = 7.8 m (1 d.p.) w O Make sure you use whole answer for x in calculation w2 = 72 + 7.82 D C 7 m w = √72 + 7.82 A x 5 m w = √110 O w = 10.5 m (1 d.p.) H E 6 m A tanCHE = 5 ÷ 6 tanGCH = 7 ÷ 7.8 O O CHE = tan-1(5 ÷ 6) GCH = tan-1(7 ÷ 7.8) T A T A CHE = 39.8° (1 d.p.) GCH = 41.9° (1 d.p.)

  16. 1. The Angle Between Two Planes - is the smallest possible angle between the planes. - is defined by the rays on each plane perpendicular to the line of intersection. e.g. Find the angle between the planes CHEB and ABCD B C 1. First define the two planes 2. Define the line of intersection 3. Define the rays perpendicular to the line of intersection F G A 4. The angle is located between the two rays 7 m A D 5 m O tanHCD = 5 ÷ 7 O E H 6 m HCD = tan-1(5 ÷ 7) T A HCD = 35.5° (1 d.p.)

  17. 2. Angle Between a Line and a Plane - is the smallest angle between the line and the projection of that line onto the plane. e.g. Find the angle between the line BH and plane ABFE B C 1. Define the line and plane F 2. Look towards plane and line G x 3. Project line onto the plane A 4. The angle is located between the line and its projection 7 m A D 5 m E H 6 m O Make sure you use whole answer for x in calculation First need to find length of projection (x) x2 = 52 + 72 tanEBH = 6 ÷ 8.6 O x = √52 + 72 EBH = tan-1(6 ÷ 8.6) x = √74 T A EBH = 34.9° (1 d.p.) x = 8.6 m (1 d.p.)

  18. NON-RIGHT ANGLED TRIANGLES 1. Naming Non-right Angled Triangles - Capital letters are used to represent angles - Lower case letters are used to represent sides e.g. Label the following triangle c A The side opposite the angle is given the same letter as the angle but in lower case. B b C a

  19. 2. Sine Rule a = b = c . SinA SinB SinC a) Calculating Sides Only 2 parts of the rule are needed to calculate the answer e.g. Calculate the length of side p p = 6 . Sin52 Sin46 52° A × Sin52 p = 6 × Sin52 Sin46 × Sin52 B 46° 6 m b p = 6.57 m (2 d.p.) p a To calculate you must have the angle opposite the unknown side. Re-label the triangle to help substitute info into the formula

  20. b) Calculating Angles For the statement: 1 = 3 is the reciprocal true? 2 6 Yes as 2 = 6 1 3 Therefore to calculate angles, the Sine Rule is reciprocated so the unknown angle is on top and therefore easier to calculate. You must calculate Sin51 before dividing by 6 (cannot use fractions) SinA = SinB = SinC a b c a = b = c . SinA SinB SinC e.g. Calculate angle θ Sinθ = Sin51 7 6 θ A Sinθ = Sin51 × 7 6 × 7 × 7 B 51° 6 m b θ = sin-1( Sin51 × 7) 6 7 m a θ = 65.0° (1 d.p.) To calculate you must have the side opposite the unknown angle Re-label the triangle to help substitute info into the formula

  21. 3. Cosine Rule -Used to calculate the third side when two sides and the angle between them (included angle) are known. a2 = b2 + c2 – 2bcCosA a) Calculating Sides e.g. Calculate the length of side x x2 = 132 + 112 – 2×13×11×Cos37 13 m b x2 = 61.59 A 37° x = √61.59 x x = 7.85 m (2 d.p.) a 11 m c Remember to take square root of whole, not rounded answer Re-label the triangle to help substitute info into the formula

  22. b) Calculating Angles - Need to rearrange the formula for calculating sides CosA = b2 + c2 – a2 2bc Watch you follow the BEDMAS laws! e.g. Calculate the size of the largest angle P CosR = 132 + 172 – 242 2×13×17 24 m a Q CosR = -0.267 13 m b A R = cos-1(-0.267) 17 m c R = 105.5° (1 d.p.) R Re-label the triangle to help substitute info into the formula Remember to use whole number when taking inverse

  23. 4. Area of a triangle - can be found using trig when two sides and the angle between the sides (included angle) are known Area = ½abSinC e.g. Calculate the following area Area = ½×8×9×Sin39 9 m b C Area = 22.7 m2 (1 d.p.) 39° 52° 8 m a 89° Re-label the triangle to help substitute info into the formula Calculate size of missing angle using geometry (angles in triangle add to 180°)

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