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60º

60º. 2. 2. 2. 60º. 60º. 2. 60º. Exact Values. Some special values of Sin, Cos and Tan are useful left as fractions, We call these exact values. 30º. 3. 1. This triangle will provide exact values for sin, cos and tan 30º and 60º. 30º. 2. 3. 60º. 1. 1. 0. 1. 0. ∞. 0.

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60º

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  1. 60º 2 2 2 60º 60º 2 60º Exact Values Some special values of Sin, Cos and Tan are useful left as fractions, We call these exact values 30º 3 1 This triangle will provide exact values for sin, cos and tan 30º and 60º

  2. 30º 2 3 60º 1 1 0 1 0 ∞ 0 Exact Values ½ ½ 3

  3. Exact Values Consider the square with sides 1 unit 45º 2 1 1 45º 1 1 We are now in a position to calculate exact values for sin, cos and tan of 45o

  4. 45º 2 1 45º 1 x 0º 30º 45º 60º 90º Sin xº 1 0 Cos xº 1 0 Tan xº 0 Exact Values 1

  5. Naming the sides of a Triangle B a c C b A 27-Oct-14

  6. Area of ANY Triangle Key feature To find the area you need to know 2 sides and the angle in between (SAS) Remember: Sides use lower case letters of angles The area of ANY triangle can be found by the following formula. B a Another version C c Another version b A If you know A, b and c If you know B, a and c If you know C, a and b 27-Oct-14

  7. Area of ANY Triangle Example : Find the area of the triangle. The version we use is B B a 20cm c C C 30o 25cm b A A

  8. Area of ANY Triangle Example : Find the area of the triangle. The version we use is E d 10cm 60o f 8cm F e D

  9. What Goes In The Box ? (1) 12.6cm 23o 15cm (2) 5.7m 71o 6.2m Key feature Remember (SAS) Calculate the areas of the triangles below: A =36.9cm2 A =16.7m2

  10. Finding the Angle given the area Q 95 cm2 12 cm R P 17 cm Find the size of angle P p r Area = ½qrsinP 95 = ½ 17 × 12sinP 95 = 102sinP q sinP = 95 ÷ 102 = 0∙931 P = sin-1 0∙931 = 68∙6° 27-Oct-14

  11. The side opposite angle A is labelled a The side opposite angle C is labelled c The side opposite angle B is labelled b B C A The Sine Rule c a b The Rule

  12. R x 10m 34o 41o P Q 10 x sin 34° sin 41° Calculating Sides Using The Sine Rule Example 1 Find the length of x in this triangle. Now cross multiply.

  13. F 10m 133o D 37o E x 10 x sin 37° sin 133° Example 2 Find the length of x in this triangle. Cross multiply = 12.14m

  14. Problem The balloon is anchored to the ground as shown in the diagram. x 25 sin 35° sin 70° sin 75° Calculate the distance between the anchor points. A 35° 25 m b c 70° 75° Cross multiply xm B C a

  15. 12cm y (1) (2) 47o 32o x 72o 93o 16mm 17m (4) (3) 143o a 89m 12o g 87o 35o Find the unknown side in each of the triangles below: x = 6.7cm y = 21.8mm a = 51.12m g = 49.21m

  16. Y 45m 38m 23o ao Z X 45 38 sin aº sin 23º Calculating Angles Using The Sine Rule Example 1. Find the angle ao Cross multiply = 0.463 Use sin-1

  17. 75m bo 143o 38m Example 2. Find the size of the angle bo = 0.305

  18. (1) 8.9m 100o (2) ao 12.9cm bo 14.5m 14o (3) 93o 49mm 14.7cm co 64mm Calculate the unknown angle in the following: bo = 16o ao = 37.2o c = 49.9o

  19. Sine Rule Basic Examples

  20. In the examples which follow use the sine rule to find the required side or angle. B D 3.2 E 80° 5.8 C 5.7 A 74° 34° 4.6 A 76° C Find AB Find angle F F 9.6 Find angle B B

  21. In the examples which follow use the sine rule to find the required side or angle. C M 7 86° N 8 12 72° 72° Find angle N T P A B 13.8 Find AB 36° R Find PQ 127° Q

  22. B 3.8 S R 60° 9 75° 78° H C Find RT 130° 32° T Find BC 4.1 A 27° I G X Find GI 13 B 10 C 8 9 39° Y 70° Find angle A Z Find angle Z A

  23. Sine Rule (Bearings) Trigonometry

  24. In the examples which follow use the sine rule to find the required side or angle. 1. Consider two radar stations Alpha and Beta. Alpha is 140 miles west of Beta. The bearing of an aero plane from Alpha is 032° and from Beta it is 316°. How far is the aeroplane from each of the radar stations?

  25. Two ports Dundee and Stonehaven are 54 miles apart with Dundee approximately due south of Stonehaven. • The bearing of a ship at sea from Stonehaven is 098° • and from Dundee it is 048°. • How far is the ship from Dundee? • A ship is sailing north. At noon its bearing from a lighthouse is 240°. Five hours later the ship is 84km further north and its new bearing from the lighthouse is 290°. • How far is the ship from the lighthouse now?

  26. Two ports P and Q are 35 miles apart with P due east of Q. The bearings of a ship from P and Q are 190° and 126° respectively. • a) What is the bearing of port Q from the ship? • b) How far is the ship from port P • The bearing of an aeroplane from Aberdeen is 068° and from Dundee it is 030°. Dundee is 69 miles south of Aberdeen. • How far is the aero plane from Aberdeen?

  27. Two oil rigs are 80 miles apart with rig • B being 80 miles east of rig A. • The bearing of a ship from rig B is 194° and from A the bearing is 140°. • Find the distance of the ship from rig A. • At noon a ship lies on a bearing 070° from Dundee. • The ship sails a distance of 45 miles south and at 3pm its • new bearing from Dundee is 130°. • How far is the ship from Dundee now? • If the ship maintains the same speed, • how long will it take to return to Dundee?

  28. An aeroplane leaves Leuchers airport and flies 80 miles north. • It then changes direction and flies 120 miles east. • The plane now turns onto a bearing of 126° • and flies a further 164 miles. • Calculate the bearing and distance of • the plane from Leuchers.

  29. Note: the last question requires the cosine rule as well as the sine rule.

  30. Cosine Rule The Cosine Rule can be used with ANY triangle as long as we have been given enough information. B a c C b A Given angle C Given angle A Given angle B 27-Oct-14

  31. Using The Cosine Rule C 5m x 43o A B 12m Example 1 : Find the unknown side in the triangle below: Identify sides a, b, c and angle Ao x 43o a = b = 5 c = 12 Ao = Write down the Cosine Rule for a a2 = b2 + c2 - 2bc cos A Substitute values x2 = 52 + 122 - 2 x 5 x 12 cos 43o x2 = 81.28 Square root to find “x”. x = 9.02m

  32. P 17.5 m 12.2 m 137o R y Q Using The Cosine Rule Example 2 : Find the length of side QR Identify the sides and angle. p = y r = 12.2 q = 17.5 P= 137o Write down Cosine Rulefor p p2 = q2 + r2 – 2pq cos P Substitute y2 = 12.22 + 17.52 – 2 x 12.2 x 17.5 x cos 137o y2 = 767.227 y = 27.7m

  33. 43cm (1) 78o 31cm p Find the length of the unknown side in the triangles: b A C a c B a2 = b2 + c2 – 2bc cosA p2 = 432 + 312 – 2 × 43× 31× cos78° p2 = 2255∙7 p= 47∙5 cm

  34. 112º (2) 28 mm 17 mm m 5∙2m 38o k 8m Find the length of the unknown side in the triangles: m = 5∙05m k = 37∙8 mm (3)

  35. 4.7cm B A A 39° 6cm Find a C Find c 5.6cm C 32° C 6cm B Cosine Rule Basic Examples 1. Finding a side. 1. 2. A Find b C Find a B 2.7cm 3. C 4. 2.4cm 7.4cm 3.8cm 36° 104° B A

  36. Find c A A 4.5cm B 22° 1.7cm 8.7cm 44° 6.6cm Find a C C Q 4 4 134° R P Find q V 11.4cm 40° T 8.8cm Find v U Find a. B 6cm A 5. B 67° 7. 6. 8cm C M 8. 3cm 71° 5cm 10. N Find m P 9.

  37. Finding Angles Using The Cosine Rule The Cosine Rule formula can be rearranged to allow us to find the size of an angle This formula is cyclic, depending on the angle to be found

  38. Finding Angles Using The Cosine Rule Example 1 : Calculate the unknown angle, xo . F d e Label and identify angles and sides E D f d = 11 e = 9 f = 16 D = xo e 2 + f 2 - d 2 cos D = Write the formula for cos D 2ef 92 + 162 - 112 cos x = Substitute values into the formula. 2 x 9 x 16 cos x= 0.75 Use cos-1 0∙75 to find x x = 41∙4o

  39. Example 2: Find the unknown angle in the triangle: B Label and identify the sides and angle. B = yo b = 26 a = 13 c = 15 C A a 2 + c 2 - b 2 cos B = Write down the formula for cos B 2ac - 132 + 152 262 Substitute values cos y = 2 x 13 x 15 cos y= - 0∙723 The negative tells you the angle is obtuse. Use cos-1 -0∙723 to find y y = 136.3o

  40. (2) bo 7m 5m 10m (1) 12.7cm 8.3cm 7.9cm Calculate the unknown angles in the triangles below: C A ao ao = 37.3o B B bo =111.8o C D

  41. Basic Examples 2. Finding an angle. B 4.7cm C A A 3.6cm B 6cm 4.1 cm 1. 2. 5.6 cm C 3. 5cm 5.2cm 3.2cm Find C Find A 6cm Find B B A C A 10.5cm B C 1.9cm C 4cm 2.7 cm 4. C 3.6cm 5. 6. 5.8 cm 7.4 cm B 2.4cm A 4.7cm A Find B Find A B Find C

  42. 6.9cm Y 6cm L X B A 4.5cm 8. 9. 8.7cm 7. 2.8 cm 6.6cm M 7cm 8cm Find X Find the largest angle. 4.5cm Z K C Find L M 3cm 11.9cm V T 5cm N 11. 12. Q 8.8 cm 4.5cm 9.6 cm 4 4 Find the smallest angle Q 10. U R P Find the largest angle 7.6cm Find all the angles

  43. Cosine Rule or Sine Rule 1. Do you know the length of ALL the sides? 2. Do you know 2 sides and the angle in between? How to determine which rule to use Two questions OR SAS If YES to either of the questions then Cosine Rule Otherwise use the Sine Rule

  44. Calculate the size of x in each of these diagrams

  45. The Sine Rule The angle of elevation of the top of a building measured from point A is 25o. At point D which is 15m closer to the building, the angle of elevation is 35o Calculate the height of the building. Application Problems T 15 m 145o 35o 25o A B D 10o 36.5 Angle TDA = 180 – 35 = 145o Angle DTA = 180 – 170 = 10o

  46. The Sine Rule T C 5o 25o 20o A B 50 m The angle of elevation of the top of a column measured from point A, is 20o. The angle of elevation of the top of the statue is 25o. Find the height of the statue when the measurements are taken 50 m from its base 180 – 115 = 65o Angle ATC = 180 – 70 = 110o 180 – 110 = 70o Angle ACT = Angle BCA = 65o 110o 70o =5.1 m 53.21 m

  47. The Cosine Rule Application Problems L N 57 miles 24 miles H 40 miles 20° A B • A fishing boat leaves a harbour (H) and travels due East for 40 miles to a marker buoy (B). At B the boat turns left and sails for 24 miles to a lighthouse (L). It then returns to harbour, a distance of 57 miles. • Make a sketch of the journey. • Find the bearing of the lighthouse from the harbour. (nearest degree) Bearing = 90 – 20 = 070°

  48. a2 = b2 + c2 – 2bcCosA The Cosine Rule P An AWACS aircraft takes off from RAF Waddington (W) on a navigation exercise. It flies 530 miles North to a point (P) as shown, It then turns left and flies to a point (Q), 670 miles away. Finally it flies back to base, a distance of 520 miles. Find the bearing of Q from point P. Not to Scale 670 miles 530 miles Q 520 miles W P = 48.7° (49°) Bearing = 180 + 49 = 229°

  49. COSINE RULE 1. Find the length of the third side of triangle ABC when i) b = 2, c = 5,  A = 60° ii) a = 2, b = 5,  A = 65° iii) a = 2, c = 5,  B = 115° iv) b = 6, c = 8,  A = 50° • Find QR in ∆PQR in which PR = 4, PQ = 3,  P = 18° • Find XY in ∆XYZ in which YZ = 25, XZ = 30,  Z = 162° • In ∆ABC, a = 7, b = 4,  C = 53°. Calculate c.

  50. 5. In ∆ABC, b = 4·2, c = 6·5,  A = 24°. Calculate a. • In ∆ABC, a = 1·64, c = 1·64,  B = 110°. Calculate b. • In ∆ABC, a = 18·5, b = 22·6,  C = 72·3°. Calculate c. • 8. In ∆ABC, a = 100, b = 120,  C = 15°.Calculate c. • 9. In ∆ABC, b = 80, c = 100, A = 123°. Calculate a.

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