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TRIGONOMETRY GENERAL MATHS UNIT TWO RIGHT ANGLED TRIANGLES – BACKGROUND INFO The internal angles within the triangle add up to 180° The square in the corner of the triangle means that the angle here is 90° and means that this is a right angled triangle This means that the other 2 angles add up to 90° RIGHT ANGLED TRIANGLES So we know this triangle is right angled (square in the corner). If we are given any two of: An angle (other than the 90° angle) and/or any 2 sides lengths, we can then find every other length and angle of the triangle. θ Opposite Adjacent We label the sides of the triangle based around the position of the angle in given in a problem. Opposite θ Adjacent I have labelled the triangle based on the location of the angle given by θ In problems we are usually told 2 measurements and are asked to an unknown. We are asked to find θ. The ‘opposite’ length and ‘hypotenuse’ are known we use? We are asked to find θ. The ‘adjacent’ length and ‘hypotenuse’ are known we use? We are asked to find ‘opposite’ length. The θ and ‘hypotenuse’ are known we use? RIGHT ANGLED TRIANGLES Applying the rule: eg. Find the length of 𝑥 correct to 2 decimal places. Step 1: Identify and label the parts of the triangle that are given in the question. Step 2: Choose which formula to use to solve the unknown 𝑥. 𝑂 𝐻 Step 3: Apply the formula 𝑥 sin 50 = 4 𝑥 = 4 sin 50 𝜃 𝑥 = 3.06m RIGHT ANGLED TRIANGLES Applying the rule: eg2. Find the length of 𝑥 correct to 2 decimal places. Step 1: Identify and label the parts of the triangle that are given in the question. Step 2: Choose which formula to use to solve the unknown 𝑥. 𝜃 𝐴 𝐻 Step 3: Apply the formula 𝑥 c𝑜𝑠 60 = 10 𝑥 = 10cos 60 𝑥 = 5cm RIGHT ANGLED TRIANGLES Applying the rule: eg3. Find the length of 𝑥 correct to 2 decimal places. Step 1: Identify and label the parts of the triangle that are given in the question. Step 2: Choose which formula to use to solve the unknown 𝑥. 𝜃 𝐴 𝑂 Step 3: Apply the formula 𝑥 tan 55 = 8 𝑥 = 8tan 55 𝑥 = 11.43cm RIGHT ANGLED TRIANGLES Applying the rule: eg4. Find the angle given by 𝑥 (round to 2 decimal places) Step 1: Identify and label the parts of the triangle that are given in the question. Step 2: Choose which formula to use to solve the unknown 𝑥. 𝜃 Step 3: Apply the formula 𝐴 tan 𝑥 𝑂 6 = 8 6 −1 tan 8 𝑥= 𝑥 = 36.87° NOW DO EXERCISE 7A Questions 1, 2AFG, 3ABJ, 7, 10, 11, 13, 15, 16, 19 ANGLES OF ELEVATION & DEPRESSION Angles of Elevation and Depression are used when dealing with directions which require us to look up or down at something. The angle of elevation is the angle between the horizontal and an an object which is higher than the observer. (eg. The top of a mountain or flagpole) The angle of depression is the angle between the horizontal and an object which is lower than the observer. (eg. A boat at sea when the observer is on a cliff) ANGLES OF ELEVATION & DEPRESSION ANGLES OF ELEVATION & DEPRESSION ANGLES OF ELEVATION & DEPRESSION eg. Standing on a cliff 50 metres high, the angle of depression of a boat at sea is 12°. How far is the boat from the base of the cliff? Sketch the scenario. Then fill in anything else we can: • Angle of Depression = Angle of Elevation • We want to know how far the boat is from the base So our triangle to solve is: Opposite 𝜃 Adjacent SOH – CAH – TOA 𝑂𝑝𝑝𝑜𝑠𝑖𝑡𝑒 tan 𝜃 = 𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡 50 tan 12 = 𝑥 50 𝑥= tan(12) 𝑥 = 235.23 𝑚𝑒𝑡𝑟𝑒𝑠 𝟏𝟐° 50m 𝟏𝟐° 𝒙 eg. Standing on a cliff 80 metres high, the angle of depression of a boat at sea is 23°. Over time it moves to a new position with an angle of depression of 28°. How far has the boat moved in this time? 23° 28° Sketch the scenario. Then fill in anything else we can: • Angle of Depression = Angle of Elevation • We want to know how far the boat has moved So we have 2 triangles to solve: SOH – CAH – TOA 𝑂𝑝𝑝𝑜𝑠𝑖𝑡𝑒 tan 𝜃 = 𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡 80 tan 23 = 𝑎 80 𝑎= tan(23) 𝑎 = 188.47 𝑚𝑒𝑡𝑟𝑒𝑠 80m 𝟐𝟑° 𝒙 𝟐𝟖° 𝑂𝑝𝑝𝑜𝑠𝑖𝑡𝑒 tan 𝜃 = 𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡 80 tan 28 = 𝑏 80 𝑏= tan(28) 𝑏 = 150.46 𝑚𝑒𝑡𝑟𝑒𝑠 ∴ 𝑥 = 𝑎 − 𝑏 = 188.47 − 150.46 = 38.01 𝑚𝑒𝑡𝑟𝑒𝑠 NOW DO EXERCISE 7B Questions 1-8 BEARINGS 360° 0° Bearings measure the direction of one object to another. 40° There are two systems used for describing bearings: True Bearings – measured in a clockwise direction starting 270° 90° at North (0° T) all the way around to 360°T 040°T Compass (or Conventional) Bearings – are measured: First relative to North and South, then relative to East or West N40°E 180° BEARINGS 360° 0° Bearings measure the direction of one object to another. There are two systems used for describing bearings: True Bearings – measured in a clockwise direction starting at North (0° T) all the way around to 360°T (180 + 30) = 210°T Compass (or Conventional) Bearings – are measured: First relative to North and South, then relative to East or West S30°W 270° 90° 30° 180° BEARINGS 360° 0° Eg 3. Find the True and Compass Bearings of the following: True Bearings – measured in a clockwise direction starting 60° 270° 90° at North (0° T) all the way around to 360°T (360 – 60) = 300°T Compass (or Conventional) Bearings – are measured: First relative to North and South, then relative to East or West N60°W E 180° BEARINGS 360° 0° Eg 4. Find the True and Compass Bearings of the following: True Bearings – measured in a clockwise direction starting at North (0° T) all the way around to 360°T 270° 90° 70° (180 – 70) = 110°T Compass (or Conventional) Bearings – are measured: First relative to North and South, then relative to East or West S70°E 180° BEARINGS 360° 0° Eg 5. Find the True and Compass Bearings of the following: True Bearings – measured in a clockwise direction starting 270° 90° 75° at North (0° T) all the way around to 360°T (180 + 75) = 255°T Compass (or Conventional) Bearings – are measured: First relative to North and South, then relative to East or West S75°W 180° BEARINGS 360° 0° Eg 6. Find the True and Compass Bearings of the following: True Bearings – measured in a clockwise direction starting 25° 270° at North (0° T) all the way around to 360°T (90 - 25) = 065°T Compass (or Conventional) Bearings – are measured: First relative to North and South, then relative to East or West N65°E 180° 90° BEARINGS - APPLICATIONS Eg 1. A ship sails 20km in a direction of N60°W. How far West from the starting point has the ship sailed? Step 1: Draw a diagram of the situation, labelling compass points and the given information Step 2: Find the triangle that we are interested in and label H, O, A, θ O Step 3: Decide which rule to apply & solve the problem Θ = 60° Hypotenuse = 20km Opposite = ? sin 𝜃 = 𝑜𝑝𝑝 ℎ𝑦𝑝 sin 60 = 𝑥 20 𝑥 = 20 sin 60 𝑥 = 17.32𝑘𝑚 H θ Eg 2. A boat race circuit consists of four legs: Leg 1: 4km West. Leg 2: 6km South. Leg 3: 2km East. Leg 4: Back to the starting point. a) How long is the final leg (Leg 4)? b) On what bearing must the final leg be sailed? a) Step 1: Draw a diagram using the given information Step 2: Find the triangle that we are interested in and label relevant points Step 3: Decide which rule to apply & solve the problem 𝑐 2 = 𝑎2 + 𝑏 2 𝑥= 22 + 62 𝑥 = 40 ∴ 𝑥 ≅ 6.32𝑘𝑚 Eg 2. A boat race circuit consists of four legs: N Leg 1: 4km West. Leg 2: 6km South. Leg 3: 2km East. Leg 4: Back to the starting point. 𝜽 b) On what bearing must the final leg be sailed? Step 1: Use the diagram from part a to choose how to solve This is our starting point which we are considering W E Step 2: Find the triangle that we are interested in S Opp and label relevant points 𝑜𝑝𝑝 tan 𝜃 = 𝑎𝑑𝑗 Step 3: Decide which rule to apply & solve the problem Adj 𝜃= 𝑡𝑎𝑛−1 ( 2 ) 6 𝜃 = 18.43° So the bearing to sail at is 18.43°𝑇 𝑜𝑟 𝑁18.43°𝐸 NOW DO EXERCISE 7B Questions 10 – 18; 20 NON RIGHT ANGLED TRIANGLES - THE SINE RULE So far we have been solving right angled triangles (where one corner is 90°). Now we look at solving every other triangle. The internal angles still add up to 180°. One method is the sine rule THE SINE RULE Two angles and a side length are known. Find the unknown length 𝑥 correct to 2 decimal points. Step 1: Label the triangle. A Step 2: Identify variables & choose which rule to use. 𝑩 = 𝟏𝟎𝟐° 𝑨 = 𝟓𝟒° 𝒃=𝟖 𝒂 =? 𝑎 sin 𝐴 = B c b 𝑏 sin 𝐵 Step 3: Substitute in values and solve the unknown variable. 𝑎 8 = sin(54) sin(102) a = sin 54 × 8 sin 102 𝒙a = 6.62 C THE SINE RULE Two angles and a side length are known. a) Find the unknown length 𝑥 correct to 2 decimal points. Step 1: Label the triangle. Step 2: Identify variables and choose equation to use. 𝒂 =? 𝑨 = 𝟒𝟔° 𝒃 = 𝟏𝟐 𝑩 = 𝟖𝟕° 𝑎 sin 𝐴 = 𝑏 sin 𝐵 B c b 12 sin 87 = 8.64 b) Then solve for the remaining unknowns on the triangle. Internal angles = 180. So, A + B + C =180. C = 180 – 46 – 87 = 47° 𝑏 𝑐 12 𝑐 = → = → 𝑐 = sin 47 × sin 𝐵 sin 𝐶 sin 87 sin 47 47° A Step 3: Substitute in values and solve. 𝑎 12 = → a = sin 46 sin 46 sin 87 a 𝒙 8.64 8.79 12 sin 87 = 8.79 C THE SINE RULE Two angles and a side length are known. a) Find the unknown length 𝑥 correct to 2 decimal points. Step 1: Label the triangle. Step 2: Identify variables and choose which equation to use. 𝒃 = 𝟑𝟎 𝑩 = 𝟕𝟓° 𝑪 = 𝟏𝟖 𝒄 = ? 𝑏 sin 𝐵 = 𝑐 sin 𝐶 9.60 sin 𝐵 sin 87 sin 75 a 87° A C b 30 sin 75 = 9.60 b) Then solve for the remaining unknowns on the triangle. Internal angles = 180. So, A + B + C =180°. A = 180 – 75 – 18 = 87° 𝑎 𝑏 𝑎 30 = → = → 𝑐 = sin 87 × sin 𝐴 31.06 c𝒙 Step 3: Substitute values into equation and solve. 30 𝑐 = → c = sin 18 sin 75 sin 18 B 30 sin 75 = 31.06 THE SINE RULE One angle and 2 side lengths are known. a) Find the unknown angle 𝑥 correct to 2 decimal points. Step 1: Label the triangle. Step 2: Identify variables & choose equation to use. 𝑨 =? 𝒃 = 𝟐𝟒 𝒂 = 𝟏𝟕 𝑩 = 𝟗𝟒° 𝑏 sin 𝐵 = 41.04° A 𝒙 44.96° = 17 sin 𝐴 → sin 94 24 = sin 𝐴 17 a 15.80c 𝑎 sin 𝐴 B Step 3: Substitute values into equation and solve. 24 sin 94 C b → sin 𝐴 = 17 sin 94 24 → 𝐴= −1 17 sin 94 sin ( ) 24 ∴ 𝐴 = 44.96° b) Now solve for the remaining unknowns on the triangle. Internal angles = 180. So, A + B + C =180°. ∴ C = 180 – 94 – 44.96 = 41.04° 𝑐 sin 𝐶 = 𝑏 sin 𝐵 → 𝑐 sin 41.04 = 24 sin 94 → 𝑐 = sin 41.04 × 24 sin 94 = 15.80 THE SINE RULE Two side lengths and an angle are known. a) Find the unknown angle 𝑥 correct to 2 decimal points. c Step 1: Label the triangle. Step 2: Identify variables and choose the appropriate equation.A 67.2° 𝒃 = 𝟐𝟎 𝑩 = 𝟖𝟎° 𝒄 = 𝟏𝟏 𝑪 =? B 18.72 a 32.8° 𝒙 b 𝑏 𝑐 = sin 𝐵 sin 𝐶 Step 3: Substitute given values into the equation. Solve the unknown variable θ. 20 sin 80 = 11 sin 𝐶 → sin 80 20 = sin 𝐶 11 → 11sin 80 20 = sin 𝐶 → 𝐶 = −1 11sin 80 sin ( 20 b) Then solve for the remaining unknowns on the triangle. Internal angles = 180. So, A + B + C =180°. A = 180 – 32.8 – 80 = 67.2° 𝑎 𝑏 𝑎 20 = → = → 𝑎 = sin 67.2 × sin 𝐴 sin 𝐵 sin 67.2 sin 80 20 sin 80 ) = 32.8° = 18.72 C NOW DO EXERCISE 7C Questions 1 - 3; 6 - 8; 12, 14, 18, 21 NON RIGHT ANGLED TRIANGLES - THE COSINE RULE THE COSINE RULE Eg1. Use the cosine rule to find the angle 𝑥 (to 2 decimal places) b C Step 1. Label the triangle A, B, C, a, b, c. A Step 2. Identify variables and choose the equation to use. c a 𝒙 B 𝒂 = 𝟏𝟎 𝒄 = 𝟏𝟐 𝒃=𝟖 𝑩 =? Step 3. Substitute in given values and solve. 102 + 122 − 82 cos 𝐵 = ( ) 2 × 10 × 12 𝐵= 102 +122 −82 −1 cos ( ) 2 ×10×12 = 100+144−64 −1 cos ( ) 240 = 41.41° THE COSINE RULE Eg2. Use the cosine rule to find the angle 𝑥 (to 2 decimal places) B Step 1. Label the triangle A, B, C, a, b, c. Step 2. Identify variables and choose the equation to use. c a 𝒂 = 𝟐𝟑 𝒄 = 𝟐𝟏 Step 3. Solve A 𝒙 b C 𝒃 = 𝟏𝟔 𝑨 =? cos 𝐴 = 162 +212 −232 ( ) 2 ×16×21 2 + 212 − 232 16 256 + 441 − 529 −1 −1 A = cos ( ) = cos ( ) 2 × 16 × 21 672 ∴𝐴 = 168 −1 cos ( ) 672 = 75.52° THE COSINE RULE B Eg3. Use the cosine rule to find the length 𝑥 (to 2 decimal places) Step 1. Label the triangle A, B, C, a, b, c. 𝒙a c Step 2. Identify variables and choose the equation to use. C A b 𝒂 =? 𝒄 = 𝟏𝟒 𝒃 = 𝟏𝟏 𝑨 = 𝟔𝟒° Step 3. Substitute in values and solve. 𝑎2 = 112 + 142 − 2 11 14 cos(64) 𝑎2 = 317 − 135.01 = 181.98 ∴ 𝑎 = 181.98 = 13.49 Eg4. a) Use the cosine rule to find the length 𝑥 correct to 2 decimal places. b) Use your answer to then solve for all other unknown values on the triangle. a) Step 1. Label the triangle A, B, C, a, b, c. B Step 2. Identify variables and choose which equation to use. a c C 5.57 𝒙 𝒂=𝟔 𝒃=𝟕 𝒄 =? 𝑪 = 𝟓𝟎° Step 3. Substitute in given values and solve. 𝑐 2 = 62 + 72 − 2 6 7 cos 50 → 𝑐 2 = 36 + 49 − 83 cos 50 𝑐 = 36 + 49 − 53.99 → ∴ 𝑐 = 31.01 = 5.57 b A b) cos 𝐴 = 72 + 5.572 −(62 ) 2×7×5.57 →𝐴= 49+31.01−36 −1 cos ( ) 77.96 = cos −1 0.56 = 55.63° Sum of Internal Angles = 180° = A+B+C → B =180° - 50°- 55.63° = 74.37° NOW DO EXERCISE 7D Questions 1 – 9 & 13. FINDING THE AREA OF THE TRIANGLE B If two length’s and the angle between them are known, we find the area using: 𝟏 𝑨𝒓𝒆𝒂 = 𝒃𝒄 𝐬𝐢𝐧(𝑨) 𝟐 Which can also be represented as: 𝑨𝒓𝒆𝒂 = or 𝑨𝒓𝒆𝒂 = 𝟏 𝒂𝒄 𝒔𝒊𝒏(𝑩) 𝟐 𝟏 𝒂𝒃 𝒔𝒊𝒏(𝑪) 𝟐 c A a b C FINDING THE AREA OF THE TRIANGLE B Find the area for the triangle pictured. Step 1 : Label A, B, C, a, b, c. c6m a A 𝟏 𝑨𝒓𝒆𝒂 = 𝒃𝒄 𝐬𝐢 𝐧 𝑨 𝟐 𝟏 𝑨𝒓𝒆𝒂 = 𝒂𝒄 𝒔𝒊 𝒏 𝑩 𝟐 𝟏 𝑨𝒓𝒆𝒂 = 𝒂𝒃 𝒔𝒊𝒏(𝑪) 𝟐 72° 𝟏 𝑨𝒓𝒆𝒂 = 𝒃𝒄 𝐬𝐢𝐧(𝑨) 𝟐 𝐴 = 72° 𝑐 = 6𝑚 𝑏 = 9𝑚 9m b Step 2 : Identify what we know on the triangle and decide which rule to use. C Step 3 : Substitute values into equation and solve. 𝑨𝒓𝒆𝒂 = 𝟏 𝒃𝒄 𝟐 𝐬𝐢𝐧(𝑨) 𝟏 ( 𝟐 = × 𝟗 × 𝟔) sin 𝟕𝟐 = 27 sin 72 = 25.68𝑚2 FINDING THE AREA OF THE TRIANGLE – HERON’S FORMULA B If all 3 length’s are known, we can use 𝑨𝒓𝒆𝒂 = where c 𝒔(𝒔 − 𝒂)(𝒔 − 𝒃)(𝒔 − 𝒄) 𝒔= 𝟏 (𝒂 𝟐 a + 𝒃 + 𝒄) A b C FINDING THE AREA OF THE TRIANGLE HERON’S FORMULA 𝑨𝒓𝒆𝒂 = 𝒔(𝒔 − 𝒂)(𝒔 − 𝒃)(𝒔 − 𝒄) Where, 𝒔 = (𝒂 + 𝒃 + 𝒄) 𝟏 𝟐 Find the area for the triangle pictured. B a c Step 1 : Label A, B, C, a, b, c. Step 2 : Identify what we know on the triangle and decide 𝑎 = 6𝑚 which rule to use. 𝑏 = 8𝑚 𝑐 = 12𝑚 C Step 3 : Substitute values into equation and solve. A b 𝟏 𝟐 First find 𝐬; 𝒔= Then solve Area; 𝑨𝒓𝒆𝒂 = = 𝒂+𝒃+𝒄 = 𝟏 𝟐 𝟔 + 𝟖 + 𝟏𝟐 = 𝟏 𝟐 × 𝟐𝟔 = 𝟏𝟑 𝒔(𝒔 − 𝒂)(𝒔 − 𝒃)(𝒔 − 𝒄) 𝟏𝟑(𝟏𝟑 − 𝟔)(𝟏𝟑 − 𝟖)(𝟏𝟑 − 𝟏𝟐) = 𝟏𝟑 × 𝟕 × 𝟓 × 𝟏 = 21.33𝑐𝑚2 FINDING THE AREA OF THE TRIANGLE HERON’S FORMULA 𝑨𝒓𝒆𝒂 = 𝒔(𝒔 − 𝒂)(𝒔 − 𝒃)(𝒔 − 𝒄) Where, 𝒔 = (𝒂 + 𝒃 + 𝒄) 𝟏 𝟐 Find the area for the triangle pictured, correct to 2 decimals. B a c Step 1 : Label A, B, C, a, b, c. Step 2 : Identify what we know on the triangle and decide 𝑎 = 5𝑚 which rule to use. 𝑏 = 4𝑚 𝑐 = 7𝑚 C Step 3 : Substitute values into equation and solve. A b 𝟏 𝟐 First find 𝐬; 𝒔= Then solve Area; 𝑨𝒓𝒆𝒂 = = 𝒂+𝒃+𝒄 = 𝟏 𝟐 𝟓+𝟒+𝟕 = 𝟏 𝟐 × 𝟏𝟔 = 𝟖 𝒔(𝒔 − 𝒂)(𝒔 − 𝒃)(𝒔 − 𝒄) 𝟖(𝟖 − 𝟓)(𝟖 − 𝟒)(𝟖 − 𝟕) = 𝟖×𝟑×𝟒×𝟏 = 9.80𝑚2 = 𝟗𝟔 B 𝑨𝒓𝒆𝒂 = 𝒔(𝒔 − 𝒂)(𝒔 − 𝒃)(𝒔 − 𝒄) Where, 𝒔 = (𝒂 + 𝒃 + 𝒄) 𝟏 𝟐 Find the area for the triangle pictured. 𝟏 𝑨𝒓𝒆𝒂 = 𝒃𝒄 𝐬𝐢 𝐧 𝑨 𝟐 𝟏 𝑨𝒓𝒆𝒂 = 𝒂𝒄 𝒔𝒊 𝒏 𝑩 𝟐 𝟏 𝑨𝒓𝒆𝒂 = 𝒂𝒃 𝒔𝒊𝒏(𝑪) 𝟐 Step 1 : Label A, B, C, a, b, c. c a 𝟏 𝑨𝒓𝒆𝒂 = 𝒂𝒃 𝐬𝐢𝐧(𝑪) 𝟐 C = 62° 𝑎 = 6𝑚 𝑏 = 7𝑚 A b Step 2 : Identify what we know on the triangle and decide which rule to use. 62° C Step 3 : Substitute values into equation and solve. 𝑨𝒓𝒆𝒂 = 𝟏 𝒂𝒃 𝟐 𝐬𝐢𝐧(𝑨) 𝟏 ( 𝟐 = × 𝟔 × 𝟕) sin 𝟔𝟐 = 21 sin 62 = 18.54𝑚2 NOW DO EXERCISE 7E Questions 1, 3, 8, 10, 12, 15, 16, 18 – 20