2. 1 Contents 1.General InformationGeneral Information 2.Finding SidesFinding Sides 3.Finding AnglesFinding Angles 4.BearingsBearings 5.Sine RuleSine Rule 6.Area FormulaArea Formula 7.Cosine Rule - SideCosine Rule - Side 8.Cosine Rule – AngleCosine Rule – Angle 9.Radial SurveysRadial Surveys Press ‘Ctrl-A’ © G Dear 2008 (Free to use. May not be sold)

3. Press Ctrl-A ©G Dear2008 – Not to be sold/Free to use Area of Circles and Elipses Stage 6 - Year 12 General Mathematic (HSC)

4. 3 3.1 Calculating Trig Ratios (1/11) You can use your calculator to find trigonometric ratios. When finding the angle we need to show working Example: Sin θ = 0.649 θ = Sin -1 (0.649) = 40.4662 o Use DMS or o ’ ’’ = 40 o 27’ 58’’ Degrees Minutes Seconds

5. 4 3.1 The Trigonometric Ratios (2/11) α OppositeAdjacentHypotenuse

6. 5 3.1 The Trigonometric Ratios (3/11) OppositeAdjacentHypotenuse β

7. 6 3.1 The Trigonometric Ratios (4/11) θ o h a S Sin θ = C Cos θ = T Tan θ = ohoh ahah oaoa S 0 H C A H T 0 A ome ld ags an lways ide heir ld ge

8. 7 3.1 The Trigonometric Ratios (5/11) θ o h a θ o h a

9. 8 3.1 The Trigonometric Ratios (6/11) 45 o 13cm x h Cos 45 o = x 13 a ah Because we have a and h Cos we must use Cos. x13 13x x = 13 x Cos 45 o ≈ 9.192 388 ≈ 9.2cm

10. 9 3.1 The Trigonometric Ratios (7/11) 35 o 10m x h Sin 35 o = x 10 o oh Because we have o and h Sin we must use Sin. x10 10x x = 10 x Sin 35 o ≈ 5.735 764 ≈ 5.7m

11. 10 3.1 Finding an Unknown Side (8/11) 40 o 12m p a Tan 40 o = p 12 o oa Because we have o and a Tan we must use Tan. x12 12x p = 12 x Tan 40 o ≈ 10.069 195 ≈ 10.1m

12. 11 3.1 Finding an Unknown Side (9/11) 50 o 3m d h Sin 50 o = 3d3d o oh Because we have o and h Sin we must use Sin. x d d x d = 3 ÷ Sin 50 o ≈ 3.916 221 ≈ 3.9 m d x Sin 50 o = 3 ÷ Sin 50 o

13. 12 3.1 Finding an Unknown Side (10/11) 60 o 7m m a Cos 60 o = 7m7m h ah Because we have a and h Cos we must use Cos. x m m x m = 7 ÷ Cos 60 o ≈ 14.0 m m x Cos 60 o = 7 ÷ Cos 60 o

14. 13 3.1 Finding an Unknown Side (11/11) 60 o 7m w a Tan 60 o = 10 m o oa Because we have o and a Tan we must use Tan. x w w x w = 10 ÷ Tan 60 o ≈ 5.773 502 ≈ 5.8 m w x Tan 60 o = 10 ÷ Tan 60 o

15. 14 3.1 Finding Angles (1/3) θoθoθoθo 7m 3m h o oh Because we have o and h Sin we must use Sin. Sin θ o = 3737 θ o = Sin -1 ( ) 3737 = 25.376 933 525Use DMS or o ’ ’’ Degrees Minutes Seconds = 25 o 22’ 37’’

16. 15 3.1 Finding Angles – (2/3) θoθoθoθo 5m 2m h a ah Because we have a and h Cos we must use Cos. Cos θ o = 2525 θ o = Cos -1 ( ) 2525 = 66.421 821 522Use DMS or o ’ ’’ Degrees Minutes Seconds = 66 o 25’ 31’’

17. 16 3.1 Finding Angles (3/3) θoθoθoθo 6m 3m a o oa Because we have o and a Tan we must use Tan. Tan θ o = 3636 θ o = Tan -1 ( ) 3636 = 26.565 051 177Use DMS or o ’ ’’ = 26 o 33’ 54’’

18. 17 3.2 Bearing (1/4) Bearings are used to describe direction. Compass bearings have four main directions and four middle directions NE SESW NW E N S W True Bearings are more specific, using a 3-digit angle clockwise from North. WebWeb|FlashFlash WebWeb|FlashFlash Practice 1Practice 2 WebWeb|FlashFlash Practice 3

19. 18 3.2 Bearing Ex 1 (2/4) A Ship A is 17km west of a lighthouse L. A.A.A.A..B.B.B.B.L.L.L.L B Ship B is due south of the lighthouse L. BSE Ship B is SE of ship A. 45 o N 17 km A Calculate distance d from Ship A to ship B. d Cos 45 o = 17 d d = 17 ÷ Cos 45 o ≈ 24 km

20. 19 3.2 Bearing Ex 2 (3/4) sails for 20 km on a bearing of 130 o. A Ship sails for 20 km on a bearing of 130 o. 130 o ? How far south is the ship from its starting point? 20 km d 50 o Cos 50 o = d 20 d = 20 x Cos 50 o ≈ 12.9 km

21. 20 3.2 Bearing Ex 3 (4/4) Mary walks 3.4 km east, then 1.3 km south. 3.4 km 1.3 km θoθoθoθo Find Mary’s bearing from her starting position. Tan θ o = 1.3 3.4 θo =θo = Tan -1 ( ) 1.3 3.4 90 o ≈ 21 o Bearing = 90 o + 21 o = 111 o

22. 21 3.3 The Sine Rule – Finding a Side (1/1) a Sin A A a B b C c = b Sin B = c Sin C WebWeb|FlashFlash To use the Sine Rule to find a side … we need two angles and … the sides opposite them.

23. 22 3.4 The Sine Rule – Finding an Angle (1/2) Sin Aa A a B b C c = Sin Bb = Sin Cc To use the Sine Rule to find an angle … we need two angles and … the sides opposite them.

24. 23 3.4 The Sine Rule – Finding an Angle (2/2) Sin Aa A 5 cm 40 o 4 cm = Sin Bb Sin A5 = Sin 404 Sin A = 5 x Sin 404 A = 4 Sin -1 ( ) X 5 ≈ 54 o

25. 24 3.5 Find the Area using Sine (1/) a b C WebWeb|FlashFlash Area = ab x Sin C 1212

26. 25 3.5 Find the Area using Sine (2/) 1212 = x 4 x 4 1212 x Sin 60 o ≈ 6.9 cm 2 60 o

27. 26 3.5 Find the Area using Sine (3/) 1212 = x 4 1212 = 7.2 cm 2 x 3.6 Area = bh

28. 27 3.5 Find the Area using Sine (3/) Why the Difference? Area = ab x Sin C 1212 0.5 x 3.95 x 3.95 x Sin 59.5 o a = 4,b = 4,c = 60 o Lower Limit = ≈ 6.7 0.5 x 4.05 x 4.05 x Sin 60.5 o Upper Limit = ≈ 7.1 Area = bh 1212 0.5 x 3.95 x 3.55 b = 4,h = 3.6 Lower Limit = = 7.0 0.5 x 4.05 x 3.65 Upper Limit = = 7.4 The error in both calculations overlap so both answers could be correct within the error limits. Error!!!

29. 28 3.6 The Cosine Rule – Find a Side (1/2) A a B b C c WebWeb|FlashFlash c 2 = a 2 + b 2 – 2ab Cos C b 2 = a 2 + c 2 – 2ac Cos B a 2 = b 2 + c 2 – 2bc Cos A

30. 29 3.6 The Cosine Rule – Find a Side (2/2) 7 6 30 o c WebWeb|FlashFlash c 2 = a 2 + b 2 – 2ab Cos C c 2 = 6 2 + 7 2 – 2 x 6 x 7 x Cos 30 c = 6 2 + 7 2 – 2 x 6 x 7 x Cos 30 = 3.500552254 ≈ 3.5

31. 30 3.7 The Cosine Rule – Find an Angle (1/2) A a B b C c Cos C = a 2 + b 2 – c 2 2ab Cos B = a 2 + c 2 – b 2 2ac Cos A = b 2 + c 2 – a 2 2bc

32. Cos -1 ( ) 2x7x5 31 3.7 The Cosine Rule – Find an Angle (2/2) A 7 B 5 CoCoCoCo 6 Cos C = a 2 + b 2 – c 2 2ab Cos C = + 5 2 7272 – 6 2 Cos C = + 5 2 7272 – 6 2 Cos -1 C = 57.12165044 o C ≈ 57 o

33. 32 3.8 The Radial Survey (1/3) A survey where angles and distances are measured from a point 030 o 140 o 245 o 310 o N 45m 65m 35m 50m 110 o 105 o 65 o 360 o -310 o +30 o 65 o

34. 33 3.8 The Radial Survey – Perimeter (2/3) 030 o 140 o 245 o 310 o 45m 65m 35m 50m 110 o 105 o 65 o d1d1 d2d2 d3d3 d4d4 d 2 = a 2 + b 2 – 2ab Cos C d 1 2 = 50 2 + 45 2 – 2x50x45 Cos 65 d 1 = 50 2 + 45 2 – 2x50x45 Cos 65 d 1 ≈ 51.217358 d 1 ≈ 51 m d 2 d 3 d 4 Repeat for d 2, d 3 and d 4. d 2 d 2 ≈ 91 m d 3 d 3 ≈ 81 m d 4 d 4 ≈ 47 m d 1 +d 2 + d 3 + d 4 Perimeter ≈ d 1 + d 2 + d 3 + d 4 ≈ 51 + 91 +81 +47 ≈ 270 m

35. 34 3.8 The Radial Survey – Area (3/3) 030 o 140 o 245 o 310 o 45m 65m 35m 50m 110 o 105 o 65 o A1A1 A2A2 A3A3 A4A4 Area = ab x Sin C 1212 A 1 = x 50 x 45 x Sin 65 1212 ≈ 1019.596 260 ≈ 1020 A 2 A 3 A 4 Repeat for A 2, A 3 and A 4. A 2 A 2 ≈ 1374 m 2 A 3 A 3 ≈ 1099 m 2 A 4 A 4 ≈ 793 m 2 A 1 +A 2 + A 3 + A 4 Area ≈ A 1 + A 2 + A 3 + A 4 ≈ 1020 + 1374 +1099 +793 ≈ 4286 m 2