5-May-15 Exact Values Angles greater than 90 o Trigonometry Useful Notation & Area of a triangle Using Area of Triangle Formula Cosine Rule Problems Sine Rule Problems Mixed Problems

5-May-15 Starter Questions

5-May-15 Exact Values Learning Intention Success Criteria 1.Recognise basic triangles and exact values for sin, cos and tan 30 o, 45 o, 60 o. 1.To build on basic trigonometry values. 2.Calculate exact values for problems.

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

x0º30º45º60º90º Sin xº Cos xº Tan xº  ½ ½ 33  3 2  Exact Values

11 45º 2 2 Consider the square with sides 1 unit 1 1 We are now in a position to calculate exact values for sin, cos and tan of 45 o

x0º30º45º60º90º Sin xº Cos xº Tan xº  ½ ½ 33  3 2  Exact Values 1  2 1

5-May-15 Now try Exercise 1 Ch8 (page 94) Exact Values

5-May-15 Starter Questions

5-May-15 Learning Intention Success Criteria Angles Greater than 90 o 1.Introduce definition of sine, cosine and tangent over 360 o using triangles with the unity circle. 1.Find values of sine, cosine and tangent over the range 0 o to 360 o. 2.Recognise the symmetry and equal values for sine, cosine and tangent.

5-May x y r Angles Greater than 90 o We will now use a new definition to cater for ALL angles. O x-axis r y-axis y x A o New Definitions P(x,y)

5-May-15 Trigonometry Angles over 90 0 (1.2, 1.6) 53 o The radius line is 2cm. The point (1.2, 1.6). Find sin cos and tan for the angle. Check answer with calculator Example 1

5-May-15 Trigonometry Angles over 90 0 (-1.8, 0.8) 127 o The radius line is 2cm. The point (-1.8, 0.8). Find sin cos and tan for the angle. Check answer with calculator Example 1

1)Sin 135 o 2)Cos 150 o 3)Tan 135 o 4)Sin 225 o 5)Cos 270 o What Goes In The Box ? Write down the equivalent values of the following in term of the first quadrant (between 0 o and 90 o ): sin 45 o 1)Sin 300 o 2)Cos 360 o 3)Tan 330 o 4)Sin 380 o 5)Cos 460 o -cos 45 o -tan 45 o -sin 45 o -cos 90 o - sin 60 o cos 0 o - tan 30 o sin 20 o - cos 80 o

5-May-15 Now try Exercise 2 Ch8 (page 97) Trigonometry Angles over 90 0

5-May-15 Trigonometry Angles over 90 0 Extension for unit 2 Trigonometry GSM Software

All +ve Sin +ve Tan +ve Cos +ve 180 o - x o 180 o + x o 360 o - x o Angles Greater than 90 o (0,1) (-1,0) (0,-1) (1,0) Two diagrams display same data in a different format

5-May-15 Starter Questions Starter Questions 8cm 3cm

5-May-15 Learning Intention Success Criteria Area of a Triangle 1.To show the standard way of labelling a triangle. 2. Find the area of a triangle using basic trigonometry knowledge. 1.Be able to label a triangle properly. 2.Find the area of a triangle using basic trigonometry knowledge.

5-May-15 Labelling Triangles A B C A a B b C c Small letters a, b, c refer to distances Capital letters A, B, C refer to angles In Mathematics we have a convention for labelling triangles.

F E D F E D 5-May-15 Labelling Triangles d e f Have a go at labelling the following triangle.

5-May-15 Area of a Triangle A B 12cm C 10cm Example 1 : Find the area of the triangle ABC. 50 o (i)Draw in a line from B to AC (ii)Calculate height BD D (iii)Area 7.66cm

5-May-15 Area of a Triangle Q P 20cm R 12cm Example 2 : Find the area of the triangle PQR. 40 o (i)Draw in a line from P to QR (ii)Calculate height PS S (iii)Area 7.71cm

5-May-15 Now try Exercise 3 Ch8 (page 99) Constructing Pie Charts

5-May-15 Starter Questions Starter Questions

5-May-15 Learning Intention Success Criteria 1.Know the formula for the area of any triangle. 1. To explain how to use the Area formula for ANY triangle. Area of ANY Triangle 2.Use formula to find area of any triangle given two length and angle in between.

General Formula for Area of ANY Triangle Consider the triangle below: AoAo BoBo CoCo a b c h Area = ½ x base x height What does the sine of A o equal Change the subject to h. h = b sinA o Substitute into the area formula

5-May-15 Area of ANY Triangle A B C A a B b C c The area of ANY triangle can be found by the following formula. Another version Another version Key feature To find the area you need to knowing 2 sides and the angle in between (SAS)

5-May-15 Area of ANY Triangle A B C A 20cm B 25cm C c Example : Find the area of the triangle. The version we use is 30 o

5-May-15 Area of ANY Triangle D E F 10cm 8cm Example : Find the area of the triangle. The version we use is 60 o

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

5-May-15 Now try Exercise 4 Ch8 (page 100) Area of ANY Triangle

5-May-15 Starter Questions Starter Questions

5-May-15 Learning Intention Success Criteria 1.Know how to use the sine rule to solve REAL LIFE problems involving lengths. 1. To show how to use the sine rule to solve REAL LIFE problems involving finding the length of a side of a triangle. Sine Rule

C B A 5-May-15 Sine Rule a b c The Sine Rule can be used with ANY triangle as long as we have been given enough information. Works for any Triangle

Deriving the rule B C A b c a Consider a general triangle ABC. The Sine Rule Draw CP perpendicular to BA P This can be extended to or equivalently

Calculating Sides Using The Sine Rule 10m 34 o 41 o a Match up corresponding sides and angles: Now cross multiply. Solve for a. Example 1 : Find the length of a in this triangle. A B C

Calculating Sides Using The Sine Rule 10m 133 o 37 o d = 12.14m Match up corresponding sides and angles: Now cross multiply. Solve for d. Example 2 : Find the length of d in this triangle. C D E

What goes in the Box ? Find the unknown side in each of the triangles below: (1) 12cm 72 o 32 o a (2) 93 o b 47 o 16mm a = 6.7cm b = 21.8mm 5-May-15

Now try Ex 6&7 Ch8 (page 103) Sine Rule

5-May-15 Starter Questions Starter Questions

5-May-15 Learning Intention Success Criteria 1.Know how to use the sine rule to solve problems involving angles. 1. To show how to use the sine rule to solve problems involving finding an angle of a triangle. Sine Rule

Calculating Angles Using The Sine Rule Example 1 : Find the angle A o AoAo 45m 23 o 38m Match up corresponding sides and angles: Now cross multiply: Solve for sin A o = Use sin to find A o

Calculating Angles Using The Sine Rule 143 o 75m 38m BoBo = Example 2 : Find the angle B o Match up corresponding sides and angles: Now cross multiply: Solve for sin B o Use sin to find B o

What Goes In The Box ? Calculate the unknown angle in the following: (1) 14.5m 8.9m AoAo 100 o (2) 14.7cm BoBo 14 o 12.9cm A o = 37.2 o B o = 16 o

5-May-15 Now try Ex 8 & 9 Ch8 (page 106) Sine Rule

5-May-15 Starter Questions Starter Questions

5-May-15 Learning Intention Success Criteria 1.Know when to use the cosine rule to solve problems. 1. To show when to use the cosine rule to solve problems involving finding the length of a side of a triangle. Cosine Rule 2. Solve problems that involve finding the length of a side.

C B A 5-May-15 Cosine Rule a b c The Cosine Rule can be used with ANY triangle as long as we have been given enough information. Works for any Triangle

Deriving the rule A B C a b c Consider a general triangle ABC. We require a in terms of b, c and A. Draw BP perpendicular to AC b P x b - x BP 2 = a 2 – (b – x) 2 Also: BP 2 = c 2 – x 2  a 2 – (b – x) 2 = c 2 – x 2  a 2 – (b 2 – 2bx + x 2 ) = c 2 – x 2  a 2 – b 2 + 2bx – x 2 = c 2 – x 2  a 2 = b 2 + c 2 – 2bx*  a 2 = b 2 + c 2 – 2bcCosA *Since Cos A = x/c  x = cCosA When A = 90 o, CosA = 0 and reduces to a 2 = b 2 + c 2 1 When A > 90 o, CosA is positive,  a 2 > b 2 + c 2 2 When A b 2 + c 2 3 The Cosine Rule The Cosine Rule generalises Pythagoras’ Theorem and takes care of the 3 possible cases for Angle A. a 2 > b 2 + c 2 a 2 < b 2 + c 2 a 2 = b 2 + c 2 A A A Pythagoras + a bit Pythagoras - a bit Pythagoras

a 2 = b 2 + c 2 – 2bcCosA Applying the same method as earlier to the other sides produce similar formulae for b and c. namely: b 2 = a 2 + c 2 – 2acCosB c 2 = a 2 + b 2 – 2abCosC A B C a b c The Cosine Rule The Cosine rule can be used to find: 1. An unknown side when two sides of the triangle and the included angle are given. 2. An unknown angle when 3 sides are given. Finding an unknown side.

5-May-15 Cosine Rule How to determine when to use the Cosine Rule. Works for any Triangle 1. Do you know ALL the lengths. 2. Do you know 2 sides and the angle in between. SAS OR If YES to any of the questions then Cosine Rule Otherwise use the Sine Rule Two questions

Using The Cosine Rule Example 1 : Find the unknown side in the triangle below: L 5m 12m 43 o Identify sides a,b,c and angle A o a =Lb =5c =12A o =43 o Write down the Cosine Rule. a 2 =b2b2 +c2c2 -2bccosA o Substitute values to find a 2. a 2 = x 5 x 12 cos 43 o a 2 = (120 x0.731 ) a 2 =81.28 Square root to find “a”. a = L = 9.02m Works for any Triangle

Example 2 : Find the length of side M. 137 o 17.5 m 12.2 m M Identify the sides and angle. a = Mb = 12.2C = 17.5A o = 137 o Write down Cosine Rule a 2 =b2b2 +c2c2 -2bccosA o a 2 = – ( 2 x 12.2 x 17.5 x cos 137 o ) a 2 = – ( 427 x – ) Notice the two negative signs. a 2 = a 2 = a = M = 27.7m Using The Cosine Rule Works for any Triangle

What Goes In The Box ? Find the length of the unknown side in the triangles: (1) 78 o 43cm 31cm L (2) 8m 5.2m 38 o M L = 47.5cm M = 5.05m

5-May-15 Now try Ex 11.1 Ch11 (page 142) Cosine Rule

5-May-15 Starter Questions Starter Questions 54 o

5-May-15 Learning Intention Success Criteria 1.Know when to use the cosine rule to solve problems. 1.Know when to use the cosine rule to solve REAL LIFE problems. 1. To show when to use the cosine rule to solve REAL LIFE problems involving finding an angle of a triangle. Cosine Rule 2. Solve problems that involve finding an angle of a triangle. 2. Solve REAL LIFE problems that involve finding an angle of a triangle.

C B A 5-May-15 Cosine Rule a b c The Cosine Rule can be used with ANY triangle as long as we have been given enough information. Works for any Triangle

Finding Angles Using The Cosine Rule Consider the Cosine Rule again: a 2 =b2b2 +c2c2 -2bc cosA o We are going to change the subject of the formula to cos A o Turn the formula around: b 2 + c 2 – 2bc cos A o = a 2 Take b 2 and c 2 across. -2bc cos A o = a 2 – b 2 – c 2 Divide by – 2 bc. Divide top and bottom by -1 You now have a formula for finding an angle if you know all three sides of the triangle. Works for any Triangle

AoAo 16cm 9cm11cm Write down the formula for cos A o Label and identify A o and a, b and c. A o = ? a = 11b = 9c = 16 Substitute values into the formula. Calculate cos A o. Cos A o =0.75 Use cos to find A o A o = 41.4 o Example 1 : Calculate the unknown angle x o. Finding Angles Using The Cosine Rule Works for any Triangle

Example 2: Find the unknown Angle in the triangle: 26cm 15cm 13cm yoyo Write down the formula. Identify the sides and angle. A o = y o a = 26b = 15c = 13 Find the value of cosA o cosA o = The negative tells you the angle is obtuse. A o = y o =136.3 o Finding Angles Using The Cosine Rule Works for any Triangle

What Goes In The Box ? Calculate the unknown angles in the triangles below: (1) 10m 7m 5m AoAo BoBo (2) 12.7cm 7.9cm 8.3cm A o =111.8 o B o = 37.3 o

5-May-15 Now try Ex 11.2 Ch11 (page 143) Cosine Rule

5-May-15 Starter Questions Starter Questions 61 o

5-May-15 Learning Intention Success Criteria 1.Be able to recognise the correct trigonometric formula to use to solve a problem involving triangles. 1. To use our knowledge gained so far to solve various trigonometry problems. Mixed problems

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

A The angle of elevation of the top of a column measured from point A, is 20 o. The angle of elevation of the top of the statue is 25 o. Find the height of the statue when the measurements are taken 50 m from its base 50 m Angle BCA = 70 o Angle ACT = Angle ATC = 110 o 65 o m B T C 180 – 110 = 70 o 180 – 70 = 110 o 180 – 115 = 65 o 20 o 25 o 5o5o

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. (a)Make a sketch of the journey. (b)Find the bearing of the lighthouse from the harbour. (nearest degree) H 40 miles 24 miles B L 57 miles A

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. P 670 miles W 530 miles Not to Scale Q 520 miles

5-May-15 Now try Ex 14 Ch8 (page 117) Mixed Problems