Note 1: Pythagoras Theorem. The longest side is always opposite the right angle and is called the hypotenuse (H).

Transcription

1 Trigonometry

2 Note 1: Pythagoras Theorem The longest side is always opposite the right angle and is called the hypotenuse (H). O H x

3 Note 1: Pythagoras Theorem In a right-angled triangle the square of the hypotenuse is equal to the sum of the squares on the other two sides. a H hypotenuse a 2 + b 2 = H 2 b

4 e.g. Pythagoras Theorem Find the side marked d a 2 + b 2 = H 2 7 cm 4 cm d d d 2 = d 2 = 49 d 2 = d 2 = 33 d = 33 d = 5.74 cm (3sf)

5 e.g. Pythagoras Theorem cone has a base diameter of 10 cm and a slant height of 15 cm. What is its vertical height? a 2 + b 2 = H 2 x x 2 = x 2 = 225 x 2 = 200 x = 200 5cm x = 14.1 cm (3 sf) 10 cm

6 Visual Proof of Pythagorean Theorem

7 e.g. Pythagoras Theorem Solve for x. x (x-2) (x-2) = x 2 x 2-4x = x 2-4x = 0-4x + 68 = 0-4x = -68 x = x = 17 8 GMM Pg Ex.27.01

8 1. Find the length of a diagonal of a rectangular box of length 12cm, width 5cm and height 4 cm. 2. ship sails 20 km due North and then 35 due East. How far is it from its starting point? 40.3 km 13.6 cm 3. The diagonal of a rectangle exceeds the length by 2 cm. If the width of the rectangle is 10 cm, find the length. 24 cm 4. n aircraft flies equal distances SE and then SW to finish 120 km due South of its starting point. How long is each part of its journey? 84.9 km

9 Note 1: Trig Ratios (Sine, Cosine, Tangent) Recall: The longest side is always opposite the right angle and is called the hypotenuse (H). The side opposite the marked angle of 35 is called the opposite (O) The other side is called the adjacent () O H 35

10 Note 1: Sine, Cosine, Tangent How we label our triangle depends on which angle we are concerned with. Now..the side opposite the marked angle of 55 is called the opposite (O) The other side is called the adjacent () 55 H O

11 Note 1: Sine, cosine, tangent In similar triangles, it is clear that the ratio will be the same in both triangles O H 6 = 2 = Opp Hyp

12 Note 1: Sine, Cosine, Tangent Three important functions are: sin θ = O H H cos θ = tan θ = O S O H C H T O θ O H For any angle x the values for sin x, cos x and tan x can be found using either a calculator or tables

13 e.g. Label the sides of these triangle as opposite to θ (O), adjacent () or hypotenuse. O B θ H θ H C O θ O H H θ D O

14 Try these! Write trigonometric ratios (in fraction form) for each of the following triangles 3 α 9 5 B β C 5 5 θ x 5 D

15 Starter (Extention) - Pythagoras Theorem Solve for x. 4 (x + 2) [4(x+2)] 2 + (x+3) 2 = 25 2 (4x + 8) 2 + (x+3) 2 = 25 2 (16x 2 +64x + 64) + (x 2 +6x+9) = x 2 +70x + 73 = x 2 +70x -552 = 0 (x+3) 25 Using quadratic formula or GDC, X = 4 G-Solv x = , x = 4 GMM Pg Ex Ex 27.03

16 Using Technology scientific or graphics calculator can be used to obtain accurate values of trig ratios. Use a calculator to find the value of each of the following correct to 4 decimal places. a.) sin 30 b.) cos54 c.) tan89 = = =

17 Using Technology To find an angle, when you know the ratio of two sides we use the inverse trig functions. a.) sin θ = b.) cos θ = c.) tan θ = ¾ θ =sin θ =cos θ =tan θ = 6.2 θ = 56.9 θ = 36.9

18 Note 2: Find side length of a right angled triangle If the size of one angle and the length of one side of a right angled triangle are given, the length of any other side can be found using: S O H C H T O Find x in the equation cos 20 = 3 x 3 x 20 x = 3cos 20 x = 2.82 multiply both sides of the equation by 3 evaluate using the calculator

19 e.g. Calculate the length of the labelled sides 7 cm 29 x y x cos 29 = x both sin 29 = 7 sides by 7 7cos 29 = x x = cm 7sin 29 = y y 7 y = cm 4 sf 58 s cos 58 = 50 s sin 58 = t m t 50cos 58 = s s = m 50sin 58 = t t = m

20 e.g. Calculate the length of the labelled side x tan 25.4 = x H O 3 sf x tan 25.4 = cm x = 4.75 cm z cm sin 31.3 = 7.4 Z z sin 31.3 = 7.4 H O z = 7.4 sin 31.3 z = 14.2 cm

21 e.g. Calculate the length of the labelled sides 30 cm 52 x y x y cos 52 = sin 52 = cos 52 = x x = cm 30sin 52 = y y = cm 4 sf.5 m 78 s t cos 78 = s 0.5 sin 78 = t cos 78 = s.5 sin 78 = t s = m t = m GMM Pg 402 Ex 28.01

22 e.g. Find the length marked x a.) Find BD from triangle BDC BD tan 32 = 10 b.) Now find x from from triangle BD 10 tan 32 = BD x B sin 38 = BD x 38 D cm C x = BD sin 38 x = 10 tan 32 sin 38 x = 3.85 cm (3 sf)

23 Note 3: Finding an unknown angle S O H C H T O If we know the length of any two sides in a right angled triangle, it is possible to calculate the size of the other angles: 3 5 x 1.) Choose the correct trig formula to use based on what sides are given 2.) Substitute side lengths into formula 3.) Change fraction to a decimal 4.) Work out angle using one of the inverse trig keys Find x in the equation sin x = 5 3 sin x = 0.6 x = sin -1.6 x = 36.9 (1 dp)

24 e.g. Finding an unknown angle 7 sin x = sin x = 0.7 x Find x (2 dp) 22 tan x = 16 tan x = S O H C H T O x = tan -1 ( ) x x = sin -1 ( ) x = (2 dp) x = (2 dp)

25 e.g. Finding an unknown angle S O H C H T O Find x 5 x cos x = cos x = tan x = 12 tan x = x = tan -1 ( ) x x = cos -1 ( ) x = 60.0 (1 dp) x = 71.1 (1 dp) GMM Pg 404 Ex 28.02

26 Starter a.) Use pythagoras theorem to find the unknown side. b.) Solve for θ 6 8 θ θ 24

27 Starter - Finding ngles ramp is 10 m long. It has been constructed so that it rises to a point 1.2 m above the ground. a.) Draw a diagram and place the measurements 10 m and 1.2 m on the correct sides. b.) Use trig to calculate the angle between the ramp and the ground. 1. sinθ = 10 2 θ = sin θ = m θ 1.2 m

28 e.g. Calculate the height of this regular pyramid = D = D 2 D = 20 m x = 26 2 x 2 = x 2 = x 2 = 576 x = 24 x Gamma Ex28.03 pg odd Ex29.01 pg odd

29 Note 4: Bearings bearing is an angle measured clockwise from North. It is given using 3 digits. e.g. The bearing of B from is 052 The bearing of from B is 232 N N 52 B 232

30 4 km 5 km θ tan θ = 4 5 = 0.8 θ = tan = 38.7 = 039 ( or ) Remember that bearings always have 3 digits (between 000 and 360 )

31 e.g. ship sails 22 km from on a bearing of 042, and a further 30 km on a bearing of 090 to arrive at B. What is the distance and bearing of B from? a.) Draw a clear diagram and label all points D N E 30 km B 42 b.) Find DE and D DE D sin 42 = cos 42 = sin 42 = DE 22 cos 42 = D DE = km F D = km

32 e.g. ship sails 22 km from on a bearing of 042, and a further 30 km on a bearing of 090 to arrive at B. What is the distance and bearing of B from? a.) Draw a clear diagram and label all points D N E 42 c.) Using ΔBF, 30 km B 2 = F 2 + BF 2 (Pythagoras Theorem) B 2 = B 2 = B = 47.6 km (3 sf) B F F = DE + EB = = km BF = D = km

33 D e.g. ship sails 22 km from on a bearing of 042, and a further 30 km on a bearing of 090 to arrive at B. What is the distance and bearing of B from? N 42 a.) Draw a clear diagram and label all points E 30 km B km km d.) The bearing of B from is given by the angle DB. <DB = <BF F tan BF = = BF = BF = tan B is 47.6 km from on BF = 69.9 a bearing of F NuLake Read pg Pg

34 Starter β First find the angle, β 130 tan β = 58 β = tan T β = 66 + β = 90 = = 24