1. PYTHAGORAS & TRIGONOMETRY

2. PYTHAGORAS Can only occur in a right angled triangle Pythagoras Theorem states: hypotenuse right angle e.g. square root undoes squaring smaller sides should always be smaller than the hypotenuse h 2 = a 2 + b 2 h a b x 7.65 m 11.3 m 9.4 cm y 8.6 cm x 2 = 7.65 2 + 11.3 2 x 2 = 186.2125 x = √186.2125 x = 13.65 m (2 d.p.) 9.4 2 = y 2 + 8.6 2 y 2 + 8.6 2 = 9.4 2 - 8.6 2 y 2 = 9.4 2 – 8.6 2 y 2 = 14.4 y = √14.4 y = 3.79 cm (2 d.p.)

3. PYTHAGOREAN TRIPLES - Special sets of whole numbers that fit into the Pythagoras equation e.g. 3, 4 and 5 5, 12 and 13 7, 24 and 25 - Each of these sets can be multiplied by numbers to find further triples. x 2 6, 8 and 10 15, 36 and 39 35, 120 and 125 x 3 x 5 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) Base (2 m) 5 2 = x 2 + 2 2 x 2 + 2 2 = 5 2 -2 2 x 2 = 5 2 - 2 2 x 2 = 21 x = √21 x = 4.58 m (2 d.p.)

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

5. e.g. d 455 m 32° H OS O H d = 455 ÷ sin32 d = 858.62 m (2 d.p.) 2. Calculating Angles -Same method as when calculating sides, except we use inverse trig ratios. A 16.1 mm 23.4 mm e.g. O H S O H sinA = 16.1 ÷ 23.4 sin -1 undoes sin A = sin -1 (16.1 ÷ 23.4) A = 43.5° (1 d.p.) Don’t forget brackets, and fractions can also be used B 2.15 m 4.07 m H A C A H cosB = 2.15 ÷ 4.07 B = cos -1 (2.15 ÷ 4.07) B = 58.1° (1 d.p.)

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

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

8. 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 e.g. The compass points and their bearings: E S W NE SWSE NW 000° 090° 180° 270° 045° 135°225° 315° e.g. Draw a bearing of 051°: N 51° e.g. What is the bearing of R from N? N 37° Bearing= 180 + 37 = 217°

9. 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 : Magnitude:x 2 = 4 2 + 6 2 x = √4 2 + 6 2 x = √52 x = 7.2 units (1 d.p.) x Bearing: N 4 6 T O A A O A tanA = 6 ÷ 4 A = tan -1 (6 ÷ 4) A = 56° Bearing= 56 + 270 = 326°

10. 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. N 500 km/hr 50 km/hr Actual flight path (x) Final Speed:x 2 = 50 2 + 500 2 x = √50 2 + 500 2 x = √252500 x = 502.5 km/hr (1 d.p.) A Bearing: T O A tanA = 50 ÷ 500 A = tan -1 (50 ÷ 500) A = 5.7° (1 d.p.) Bearing= 006° O A

11. 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

12. 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 x w 6 m 7 m H GF E DC BA 5 m x 2 = 5 2 + 6 2 x = √5 2 + 6 2 x = √61 x = 7.8 m (1 d.p.) w 2 = 7 2 + 7.8 2 w = √7 2 + 7.8 2 w = √110 w = 10.5 m (1 d.p.) Make sure you use whole answer for x in calculation O A tanCHE = 5 ÷ 6 CHE = tan -1 (5 ÷ 6) CHE = 39.8° (1 d.p.) T O A O A T O A tanGCH = 7 ÷ 7.8 GCH = tan -1 (7 ÷ 7.8) GCH = 41.9° (1 d.p.)

13. 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 6 m 7 m H G F E D CB A 5 m 1. First define the two planes 2. Define the line of intersection 3. Define the rays perpendicular to the line of intersection 4. The angle is located between the two rays T O A tanHCD = 5 ÷ 7 HCD = tan -1 (5 ÷ 7) HCD = 35.5° (1 d.p.) O A

14. 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 H G F E D CB A 6 m 7 m 5 m 1. Define the line and plane 2. Look towards plane and line 3. Project line onto the plane 4. The angle is located between the line and its projection x x 2 = 5 2 + 7 2 x = √5 2 + 7 2 x = √74 x = 8.6 m (1 d.p.) First need to find length of projection (x) O A T O A tanEBH = 6 ÷ 8.6 EBH = tan -1 (6 ÷ 8.6) EBH = 34.9° (1 d.p.) Make sure you use whole answer for x in calculation

15. 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 a B C The side opposite the angle is given the same letter as the angle but in lower case. b c A

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

17. 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. a = b = c. SinA SinB SinC SinA = SinB = SinC a b c e.g. Calculate angle θ 7 m 6 m θ 51° Sinθ = Sin51 7 6 To calculate you must have the side opposite the unknown angle × 7 Sinθ = Sin51 × 7 6 θ = sin -1 ( Sin51 × 7) 6 θ = 65.0° (1 d.p.) You must calculate Sin51 before dividing by 6 (cannot use fractions) Re-label the triangle to help substitute info into the formula A B a b

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

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

20. 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 52° 89° 8 m 9 m Re-label the triangle to help substitute info into the formula C a b 39° Calculate size of missing angle using geometry (angles in triangle add to 180°) Area = ½×8×9×Sin39 Area = 22.7 m 2 (1 d.p.)