TRIGONOMETRY 2.4

PYTHAGORAS h2 = a2 + b2 Can only occur in a right angled triangle hypotenuse Pythagoras Theorem states: h a h2 = a2 + b2 b right angle e.g. x 9.4 cm 7.65 m y 11.3 m 8.6 cm x2 = 7.652 + 11.32 y2 + 8.62 = 9.42 9.42 = y2 + 8.62 square root undoes squaring smaller sides should always be smaller than the hypotenuse x2 = 186.2125 - 8.62 y2 = 9.42 – 8.62 - 8.62 x = √186.2125 y2 = 14.4 x = 13.65 m (2 d.p.) y = √14.4 y = 3.79 cm (2 d.p.)

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

e.g. d = 455 ÷ sin32 O d H 455 m d = 858.62 m (2 d.p.) O S H 32° 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

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 C H H A 48 m cosA = 32 ÷ 48 32 m A = cos-1(32 ÷ 48) A A = 48.2° (1 d.p.)

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

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

For the statement: 1 = 3 is the reciprocal true? 2 6 Yes as 2 = 6 1 3 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

Sine Rule Applications e.g. A conveyor belt 22 m in length drops sand onto a cone-shaped heap. The sides of the cone measure 7 m and the cone’s sides make an angle of 32° with the ground. Calculate the angle that the belt makes with the ground (θ), and the diameter of the cone’s base (x). SinA = SinB = SinC a b c Conveyor belt : 22 m 116° A b Sinθ = Sin148 7 22 b 7 m 7 m a 32° B 32° 148° B A θ Sinθ = Sin148 × 7 22 × 7 × 7 x a a = b = c . SinA SinB SinC θ = sin-1( Sin148 × 7) 22 x = 7 . Sin116 Sin32 θ = 9.7° (1 d.p.) × Sin116 x = 7 × Sin116 Sin32 × Sin116 x = 11.87 m (2 d.p.)

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

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

Cosine Rule Applications e.g. A ball is hit a distance of 245 m on a golf hole. The distance from the ball to the hole is 130 m. The angle between the hole and tee (from the ball) is 60 °. Calculate the distance from the tee to the hole (x) and the angle (θ) at which the golfer hit the ball away from the correct direction. b x a Hole Tee a2 = b2 + c2 – 2bcCosA A θ x2 = 1302 + 2452 – 2×130×245×Cos60 x2 = 45075 130 m A 245 m 60° x = √45075 b a c c Ball x = 212.31 m (2 d.p.) CosA = b2 + c2 – a2 2bc Cosθ = 212.312 + 2452 – 1302 2×212.31×245 Remember to use whole number from previous question! Cosθ = 0.848 θ = cos-1(0.848) θ = 32.0° (1 d.p.)

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 tanCHE = 7 ÷ 7.8 O O CHE = tan-1(5 ÷ 6) CHE = tan-1(7 ÷ 7.8) T A T A CHE = 39.8° (1 d.p.) CHE = 41.9° (1 d.p.)

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°)

Area Applications e.g. A ball is hit a distance of 245 m on a golf hole. The distance from the ball to the hole is 130 m. The angle between the hole and tee (from the ball) is 60 °. Calculate the area contained in between the tee, hole and ball. Hole Tee Area = ½abSinC C Area = ½×130×245×Sin60 130 m 245 m 60° a b Area = 13791.5 m2 (1 d.p.) Ball