10 Trigonometry II (Path) PDF

10 Trigonometry II (Path) LESSON SEQUENCE 10.1 Overview...............................................................................................................................................................504 10.2 The sine rule...............................................................

10 Trigonometry II (Path) LESSON SEQUENCE 10.1 Overview...............................................................................................................................................................504 10.2 The sine rule........................................................................................................................................................507 10.3 The cosine rule...................................................................................................................................................515 10.4 Area of triangles.................................................................................................................................................520 10.5 The unit circle..................................................................................................................................................... 527 10.6 Trigonometric functions.................................................................................................................................. 536 10.7 Solving trigonometric equations................................................................................................................. 543 10.8 Review................................................................................................................................................................... 548 LESSON 10.1 Overview Why learn this? Trigonometry is the branch of mathematics that describes the relationship between the angles and side lengths in triangles. The ability to calculate distances using angles has long been critical. As early as the third century BCE, trigonometry was being used in the study of astronomy. Early explorers, using rudimentary calculations and the stars, were able to navigate their way around the world. They were even able to map coastlines along the way. Cartographers use trigonometry when they are making maps. It is essential to be able to calculate distances that can’t be physically measured. Astronomers use trigonometry to calculate distances such as that from a particular planet to Earth. Our explorations have now turned towards the skies and outer space. Scientists design and launch space shuttles and rockets to explore our universe. By applying trigonometry, they can approximate the distances to other planets. As well as in astronomy and space exploration, trigonometry is widely used in many other areas. Surveyors use trigonometry in setting out a land subdivision. Builders, architects and engineers use angles, lengths and forces in the design and construction of all types of buildings, both domestic and industrial. In music, a single note is a sine wave. Sound engineers manipulate sine waves to create the desired effect. Trigonometry has many real-life applications. Hey students! Bring these pages to life online Watch Engage with Answer questions videos interactivities and check solutions Find all this and MORE in jacPLUS Reading content Extra learning and rich media, resources including interactivities and videos for Differentiated every concept question sets Questions with immediate feedback, and fully worked solutions to help students get unstuck 504 Jacaranda Maths Quest 10 Stage 5 NSW Syllabus Third Edition Exercise 10.1 Pre-test 1. MC From the following options select the exact value of sin(30°). √ 1 1 3 A. √ B. C. D. 1 2 2 2 2. Solve for x, correct to 2 decimal places. A 42° C 60° x 7m B 3. Solve for y, correct to 2 decimal places. y 75° 55° 10 cm x Choose the values of the angles B and B′ in the below triangle, to the nearest degree. (Assume BC = B′ C.) 4. MC C 19 11 B′ 25° B A B A. 47° and 133° B. 46° and 134° C. 47° and 153° D. 25° and 155° 5. Calculate the perimeter of the following triangle, correct to 2 decimal places. C 7 5 63° A B x 6. Solve for x, correct to 1 decimal place. x 12 m 10 m 37° TOPIC 10 Trigonometry II (Path) 505  7. MC Calculate the area of the triangle shown. A. 13.05 cm2 B. 18.64 cm2 7 cm C. 22.75 cm2 D. 26.1 cm2 55° 6.5 cm 8. State in which quadrant of the unit circle is the angle 203° located. 9. Determine the value of cos(60°) using part of the unit circle. y 1 0.8 0.6 0.4 0.2 0 x 0.5 1 If cos(x°) = p for 0 ≤ x ≤ 90°, then sin(180 − x°) in terms of p is: B. 180 − p C. 1 − p 1 − p2 10. MC √ A. p D. Select the amplitude and period, respectively, of y = −2 sin(2x) from the following options. A. −2, 360° B. −2, 180° 11. MC C. 2, 2 D. 2, 180° 12. MC Select the correct equation for the graph y A. y = 4 cos(2x) shown. 4 B. y = −4 cos(2x) 3 C. y = −4 sin(2x) 2 D. y = 4 sin(2x) 1 0 x –1 90° 180° 270° 360° –2 –3 –4 Select the correct solutions for the equation sin(x) = for x over the domain 0 ≤ x ≤ 360°. 1 13. MC A. x = 30° and x = 150° B. x = 30° and x = 210° 2 C. x = 60° and x = 120° D. x = 60° and x = 240° 14. MC Select the correct solutions for the equation cos (2x) = − for x over the domain 0 ≤ x ≤ 360°. √ 2 A. x = 22.5° and x = 337.5° B. x = 45° and x = 315° 2 C. x = 67.5° and x = 112.5° D. x = 157.5° and x = 202.5° 506 Jacaranda Maths Quest 10 Stage 5 NSW Syllabus Third Edition  15. Using the graph shown, solve the equation 7 sin(x) = −7 for 0 ≤ x ≤ 360°. y 1 0.5 0 x 90° 180° 270° 360° –0.5 –1 LESSON 10.2 The sine rule LEARNING INTENTION At the end of this lesson you should be able to: apply the sine rule to evaluate angles and sides of triangles recognise when the ambiguous case of the sine rule exists apply the sine rule to evaluate angles involving the ambiguous case. 10.2.1 The sine rule eles-6287 In any triangle, the angles are named by the vertices A, B and C and the B corresponding opposite sides as a, b and c as shown in the diagram at right. B right-angled triangles, ΔADB and ΔCDB. Let BD be the perpendicular line from B to AC, of length h, giving two c a Using ΔADB: Using ΔCDB: A C A C b sin(A) = sin(C) = h h B h = c sin(A) h = a sin(C) c a c h a Equating the values of h: c sin(A) = a sin(C) A C D b h– = sin(A) and h– = sinC) c a giving: = B c a B sin(C) sin(A) c h Similarly, if a perpendicular line is drawn from vertex A to BC, then: = C c b A b C sin(C) sin(B) h = c sin(B) and h = b sin(C) TOPIC 10 Trigonometry II (Path) 507  The sine rule can be used to solve non-right-angled triangles if we are given: 1. two angles and one side 2. two sides and an angle opposite one of these sides. Sine rule The sine rule for any triangle ABC is: B = = c B a a b c A C sin(A) sin(B) sin(C) A b C WORKED EXAMPLE 1 Determining unknown angles and sides of a given triangle In the triangle ABC, a = 4 m, b = 7 m and B = 80°. Calculate the values of A, C and c. Round angles to the nearest minute and lengths to 2 decimal places. THINK WRITE/DRAW 1. Draw a labelled diagram of the triangle ABC B and fill in the given information. c 80° a = 4 A C A b=7 C 2. Check that one of the criteria for the sine The sine rule can be used since two side lengths rule has been satisfied. and an angle opposite one of these side lengths have been given. 3. Write down the sine rule to calculate A. To calculate angle A: = a b sin(A) sin(B) = 4 7 4. Substitute the known values into the rule. sin(A) sin(80°) 5. Transpose the equation to make sin(A) 4 sin(80°) = 7 sin(A) sin(A) = the subject. 4 sin(80°) 7 A = sin −1 4 sin(80°) ( ) 6. Evaluate and write your answer. ≈ 34.246 004 71° 7 7. Round off the answer to degrees and minutes. A = 34°15′ C ≈ 180° − (80° + 34°15′ ) = 65°45′ 8. Determine the value of angle C using the fact that the angle sum of any triangle is 180°. 9. Write down the sine rule to calculate the To calculate side length c: = value of c. c b sin(C) sin(B) 508 Jacaranda Maths Quest 10 Stage 5 NSW Syllabus Third Edition  = c 7 ′ 10. Substitute the known values into the rule. sin(65°45 ) sin(80°) 7 sin(5°45′ ) 11. Transpose the equation to make c the subject. c= sin(80°) 12. Evaluate. Round off the answer to 2 decimal ≈ 6.48 m places and include the appropriate unit. 10.2.2 The ambiguous case eles-5005 If two side lengths and an angle opposite one of these side lengths are given, then B For example, if a = 10, c = 6 and C = 30°, two possible triangles could two different triangles may be drawn. B a = 10 c=6 A 30° be created. A C In the first case angle A is an acute angle, while in the second case angle A is an obtuse angle. When using the sine rule to determine an angle, the inverse sine function is used. B In lesson 10.7, you will see that the sine of an angle between 0° and 90° has the B a = 10 For example, sin(40°) ≈ 0.6427 and sin(140°) ≈ 0.6427. c=6 same value as the sine of its supplement. A 30° A C WORKED EXAMPLE 2 Solving triangles and checking for the ambiguous case In the triangle ABC, a = 10 m, c = 6 m and C = 30°. Determine two possible values of A, and hence two possible values of B and b. Round angles to the nearest minute and lengths to 2 decimal places. Case 1 THINK WRITE/DRAW 1. Draw a labelled diagram of the triangle ABC B and fill in the given information. B a = 10 c=6 A 30° A C 2. Check that one of the criteria for the sine rule The sine rule can be used since two side lengths has been satisfied. and an angle opposite one of these side lengths have been given. = a c 3. Write down the sine rule to determine A. To determine angle A: sin(A) sin(C) = 10 6 4. Substitute the known values into the rule. 10 sin(30°) = 6 sin(A) sin(A) sin(30°) sin(A) = 10 sin(30°) 5. Transpose the equation to make sin(A) the subject. 6 TOPIC 10 Trigonometry II (Path) 509  A = sin −1 10 sin(30°) ( ) 6. Evaluate angle A. ≈ 56.442 690 24° 6 7. Round off the answer to degrees and minutes. A = 56°27′ B ≈ 180° − (30° + 56°27′ ) = 93°33′ 8. Determine the value of angle B, using the fact that the angle sum of any triangle is 180°. 9. Write down the sine rule to calculate b. To calculate side length b: = b c sin(B) sin(C) = b 6 ′ 10. Substitute the known values into the rule. sin(93°33 ) sin(30°) 6 sin(93°33′ ) 11. Transpose the equation to make b the subject. b= sin(30°) 12. Evaluate. Round off the answer to 2 decimal ≈ 11.98 m places and include the appropriate unit. Note: The values we have just obtained are only one set of possible answers for the given dimensions of We are told that a = 10 m, c = 6 m and C = 30°. Since side a is larger than side c, it follows that angle A the triangle ABC. will be larger than angle C. Angle A must be larger than 30°; therefore it may be an acute angle or an obtuse angle. Case 2 THINK WRITE/DRAW 1. Draw a labelled diagram of the triangle ABC B and fill in the given information. B a = 10 c=6 A 30° A C If sin A = 0.8333, then A could also be: 2. Write down the alternative value for angle A. To determine the alternative angle A: A ≈ 180° − 56°27′ Simply subtract the value obtained for A in = 123°33′ case 1 from 180°. B ≈ 180° − (30° + 123°33′ ) = 26°27′ 3. Determine the alternative value of angle B, using the fact that the angle sum of any triangle is 180°. 4. Write down the sine rule to determine the To calculate side length b: = alternative b. b c sin(B) sin(C) = b 6 ′ 5. Substitute the known values into the rule. sin(26°27 ) sin(30°) 510 Jacaranda Maths Quest 10 Stage 5 NSW Syllabus Third Edition  6 sin(26°27′ ) 6. Transpose the equation to make b the subject. b= sin(30°) 7. Evaluate. Round off the answer to 2 decimal ≈ 5.34 m places and include the appropriate unit. In Worked example 2 there were two possible solutions, as shown by the diagrams below. B B B a = 10 B a = 10 c=6 c=6 A 30° A 30° A C A C The ambiguous case does not apply to every question. Since ∠A = 34°15′ , then it could also have been ∠A =145°45′ , the supplementary angle. Consider Worked example 1. If ∠A = 34°15 and ∠B =80°, then ′ ∠C = 65°45 (angle sum of triangle). ′ If ∠A =145°45′ and ∠B =80°, then ∠C = 180° − 145°45′ + 80° which is not possible. ∠C = −45°45′ ( ) Hence, for Worked example 1, only one possible solution exists. The ambiguous case may exist if the angle found is opposite the larger given side. COMMUNICATING – COLLABORATIVE TASK: Using graphing applications to verify the sine rule Equipment: Graphing software. 1. Use a graphing application (such as Geogebra), to construct a triangle of any shape. 2. Select two angles and their opposite sides to substitute into the sine rule. What do you notice about the two fractions? 3. Modify the shape of your triangle and repeat step 2. 4. Compare your results with the rest of the class. WORKED EXAMPLE 3 Calculating heights using given angles of elevation To calculate the height of a building, Kevin measures the angle of elevation to the top as 52°. He then walks 20 m closer to the building and measures the angle of elevation as 60°. Calculate the height of the building, correct to 2 decimal places. THINK WRITE/DRAW 1. Draw a labelled diagram of the situation and C fill in the given information. h 120° 52° 60° A B D 20 x – 20 x TOPIC 10 Trigonometry II (Path) 511  2. Check that one of the criteria for the sine rule The sine rule can be used for triangle ABC since ∠ACB = 180° − (52° + 120°) has been satisfied for triangle ABC. two angles and one side length have been given. = 8° 3. Calculate the value of angle ACB, using the fact that the angle sum of any triangle is 180°. 4. Write down the sine rule to calculate To calculate side length b of triangle ABC: = b (or AC). b c sin(B) sin(C) = b 20 5. Substitute the known values into the rule. sin(120°) sin(8°) b= 20 sin(120°) 6. Transpose the equation to make b the subject. sin(8°) 7. Evaluate. Round off the answer to 2 decimal ≈ 124.45 m places and include the appropriate unit. 8. Draw a diagram of the situation, that is, triangle C ADC, labelling the required information. Note: There is no need to solve the rest of the triangle 124.45 m in this case as the values will not assist in h calculating the height of the building. 52° A D 9. Write down what is given for the triangle. Have: angle and hypotenuse 10. Write down what is needed for the triangle. Need: opposite side sin(𝜃) = O required (SOH − CAH − TOA). 11. Determine which of the trigonometric ratios is H sin(52°) = h 12. Substitute the given values into the 124.45 124.45 sin(52°) = h appropriate ratio. h = 124.45 sin(52°) 13. Transpose the equation and solve for h. 14. Round off the answer to 2 decimal places. ≈ 98.07 15. Write the answer. The height of the building is 98.07 m. DISCUSSION Discuss a real-life situation in which the sine rule can be used. Resources Resourceseses Interactivities The sine rule (int-6275) The ambiguous case (int-4818) 512 Jacaranda Maths Quest 10 Stage 5 NSW Syllabus Third Edition Exercise 10.2 The sine rule 10.2 Quick quiz 10.2 Exercise Individual pathways PRACTISE CONSOLIDATE MASTER 1, 2, 6, 7, 11, 13, 16, 18, 21 3, 4, 8, 9, 12, 14, 17, 19, 22 5, 10, 15, 20, 23, 24, 25 Where appropriate in this exercise, write your angles correct to the nearest minute and side lengths correct to 2 decimal places. Fluency 1. In the triangle ABC, a = 10, b = 12 and B = 58°. Calculate A, C and c. 2. In the triangle ABC, c = 17.35, a = 26.82 and A = 101°47′. Calculate C, B and b. WE1 3. In the triangle ABC, a = 5, A = 30° and B = 80°. Calculate C, b and c. 4. In the triangle ABC, c = 27, C = 42° and A = 105°. Calculate B, a and b. 5. In the triangle ABC, a = 7, c = 5 and A = 68°. Determine the perimeter of the triangle. 6. Calculate all unknown sides and angles for the triangle ABC, given A = 57°, B = 72° and a = 48.2. 7. Calculate all unknown sides and angles for the triangle ABC, given a = 105, B = 105° and C = 15°. 8. Calculate all unknown sides and angles for the triangle ABC, given a = 32, b = 51 and A = 28°. 9. Calculate the perimeter of the triangle ABC if a = 7.8, b = 6.2 and A = 50°. 10. MC In a triangle ABC, B = 40°, b = 2.6 and c = 3. Identify the approximate value of C. Note: There may be more than one correct answer. A. 47° B. 48° C. 132° D. 133° Understanding 11. WE2 In the triangle ABC, a = 10, c = 8 and C = 50°. Determine two possible values of A, and hence two possible values of b. 12. In the triangle ABC, a = 20, b = 12 and B = 35°. Determine two possible values for the perimeter of the triangle. 13. Calculate all unknown sides and angles for the triangle ABC, given A = 27°, B = 43° and c = 6.4. 14. Calculate all unknown sides and angles for the triangle ABC, given A = 100°, b = 2.1 and C = 42°. 15. Calculate all unknown sides and angles for the triangle ABC, given A = 25°, b = 17 and a = 13. Communicating, reasoning and problem solving C 16. Calculate the value of h, correct to 1 decimal place. Show the full working. h 35° 70° A 8 cm D B TOPIC 10 Trigonometry II (Path) 513 17. WE3 To calculate the height of a building, Kevin measures the angle of elevation to the top as 48°. He then walks 18 m closer to the building and measures the angle of elevation as 64°. Calculate the height of the building. 18. A boat sails on a bearing of N15°E for 10 km and then on a bearing of S85°E until it is due east of the starting point. Determine the distance from the starting point to the nearest kilometre. Show all your working. 19. A hill slopes at an angle of 30° to the horizontal. A tree that is 8 m tall and leaning downhill is growing at an angle of 10° m to the vertical and is part-way up the slope. Evaluate the vertical height of the top of the tree above the slope. Show all your working. 20. A cliff is 37 m high. The rock slopes outward at an angle of 50° to the horizontal and then cuts back at an angle of 25° to the vertical, meeting the ground directly below the top of the cliff. 50° Carol wishes to abseil from the top of the cliff to the ground as shown in the diagram. Her climbing rope is 45 m long, and she needs 2 m to secure it to a tree at the top of the cliff. Determine if the rope will be 25° Rock long enough to allow her to reach the ground. Rope 37 m 21. A river has parallel banks that run directly east–west. From the south bank, Kylie takes a bearing to a tree on the north side. The bearing is 047°T. She then walks 10 m due east, and takes a second bearing to the tree. This is 305°T. Determine: a. her distance from the second measuring point to the tree b. the width of the river, to the nearest metre. 22. A ship sails on a bearing of S20°W for 14 km; then it changes direction and sails for 20 km and drops anchor. Its bearing from the starting point is now N65°W. a. Determine the distance of the ship from the starting point of it. b. Calculate the bearing on which the ship sails for the 20 km leg. 23. A cross-country runner runs at 8 km/h on a bearing of 150°T for 45 mins; then she changes direction to a bearing of 053°T and runs for 80 mins at a different speed until she is due east of the starting point. a. Calculate the distance of the second part of the run. b. Calculate her speed for this section, correct to 1 decimal place. c. Evaluate how far she needs to run to get back to the starting point. 24. From a fire tower, A, a fire is spotted on a bearing of N42°E. From a second tower, B, the fire is on a bearing of N12°W. The two fire towers are 23 km apart, and A is N63°W of B. Determine how far the fire is from each tower. 25. A yacht sets sail from a marina and sails on a bearing of 065°T for 3.5 km. It then turns and sails on a bearing of 127°T for another 5 km. a. Evaluate the distance of the yacht from the marina, correct to 1 decimal place. b. If the yacht was to sail directly back to the marina, on what bearing should it travel? Give your answer rounded to the nearest minute. 514 Jacaranda Maths Quest 10 Stage 5 NSW Syllabus Third Edition  LESSON 10.3 The cosine rule LEARNING INTENTION At the end of this lesson you should be able to: apply the cosine rule to calculate a side of a triangle apply the cosine rule to calculate the angles of a triangle. 10.3.1 The cosine rule eles-5006 length h, giving two right-angled triangles, ΔADB and ΔCDB. In triangle ABC, let BD be the perpendicular line from B to AC, of B Let the length of AD = x, then DC = (b–x). Using triangle ADB and Pythagoras’ theorem, we obtain: c2 = h2 + x2 c h a Using triangle CDB and Pythagoras’ theorem, we obtain: a2 = h2 + (b − x)2 A D C x b–x b Expanding the brackets in equation : a2 = h2 + b2 − 2bx + x2 Rearranging equation and using c2 = h2 + x2 from equation : a2 = h2 + x2 + b2 − 2bx = c2 + b2 − 2bx = b2 + c2 − 2bx From triangle ABD, x = c cos(A); therefore a2 = b2 + c2 − 2bx becomes a2 = b2 + c2 − 2bc cos(A). The cosine rule can be used to solve non-right-angled triangles if we are given: 1. three sides or 2. two sides and the included angle. Note: Once the third side has been calculated, the sine rule could be used to determine other angles if necessary. Cosine rule — calculating an unknown side The cosine rule to calculate an unknown side for any triangle a2 = b2 + c2 − 2bc cos(A) ABC is: B b2 = a2 + c2 − 2ac cos(B) c B a c2 = a2 + b2 − 2ab cos(C) A C A b C If three sides of a triangle are known, an angle could be found by transposing the cosine rule to make cos(A), cos(B) or cos(C) the subject. TOPIC 10 Trigonometry II (Path) 515  Cosine rule — calculating an unknown angle The cosine rule to calculate an unknown angle for any triangle b2 + c2 − a2 ABC is: a2 = b2 + c2 − 2bc cos(A) ⇒ cos(A) = 2bc a2 + c2 − b2 B b2 = a2 + c2 − 2ac cos(B) ⇒ cos(B) = c B a 2ac a2 + b2 − c2 A C c2 = a2 + b2 − 2ab cos(C) ⇒ cos(C) = A b C 2ab WORKED EXAMPLE 4 Calculating the length of a side using the cosine rule Calculate the third side of triangle ABC given a = 6, c = 10 and B = 76° correct to 2 decimal places. THINK WRITE/DRAW 1. Draw a labelled diagram of the triangle ABC B and fill in the given information. c = 10 a=6 76° A C A b C 2. Check that one of the criteria for the cosine Yes, the cosine rule can be used since two side rule has been satisfied. lengths and the included angle have been given. b2 = a2 + c2 − 2ac cos(B) 3. Write down the appropriate cosine rule to To calculate side b: calculate side b. 4. Substitute the given values into the rule. = 62 + 102 − 2 × 6 × 10 × cos(76°) 5. Evaluate. ≈ 106.969 372 5 b ≈ 106.969 372 5 ≈ 10.34 √ 6. Round off the answer to 2 decimal places. WORKED EXAMPLE 5 Calculating the size of an angle using the cosine rule Calculate the smallest angle in the triangle with sides 4 cm, 7 cm and 9 cm correct to the nearest minute. THINK WRITE/DRAW 1. Draw a labelled diagram of the triangle, call it C ABC and fill in the given information. b=7 C a=4 Note: The smallest angle will correspond to the A B Let a = 4, b = 7, c = 9 smallest side. A c=9 B 2. Check that one of the criteria for the cosine The cosine rule can be used since three side rule has been satisfied. lengths have been given. 516 Jacaranda Maths Quest 10 Stage 5 NSW Syllabus Third Edition  b2 + c2 − a2 3. Write down the appropriate cosine rule to cos(A) = calculate angle A. 2bc 72 + 92 − 42 = 2×7×9 4. Substitute the given values into the rearranged rule. = 114 5. Evaluate. 126 A = cos−1 114 ( ) 6. Transpose the equation to make A the subject ≈ 25.208 765 3° by taking the inverse cos of both sides. 126 7. Round off the answer to degrees and minutes. ≈ 25°13′ COMMUNICATING – COLLABORATIVE TASK: Using graphing applications to verify the cosine rule Equipment: Graphing software. 1. Use a graphing application (such as Geogebra), to construct a triangle of any shape. 2. Select two sides and the angle between to substitute into the cosine rule. 3. Repeat the process for the other two angles in the triangle. 3. Modify the shape of your triangle and repeat steps 2 and 3. 4. Compare your results with the rest of the class. WORKED EXAMPLE 6 Applying the cosine rule to solve problems Two rowers, Harriet and Kate, set out from the same point. Harriet rows N70°E for 2000 m and Kate rows S15°W for 1800 m. Calculate the distance between the two rowers, correct to 2 decimal places. THINK WRITE/DRAW 1. Draw a labelled diagram of the triangle, call it N 2000 m A Harriet ABC and fill in the given information. C 70 ° 15° 1800 m B Kate 2. Check that one of the criteria for the cosine The cosine rule can be used since two side lengths rule has been satisfied. and the included angle have been given. c2 = a2 + b2 − 2ab cos(C) 3. Write down the appropriate cosine rule to To calculate side c: calculate side c. 4. Substitute the given values into the rule. = 20002 + 18002 − 2 × 2000 × 1800 cos(125°) TOPIC 10 Trigonometry II (Path) 517  5. Evaluate. ≈ 11 369 750.342 c ≈ 11 369 750.342 ≈ 3371.91 √ 6. Round off the answer to 2 decimal places. 7. Write the answer. The rowers are 3371.91 m apart. DISCUSSION In what situations would you use the sine rule rather than the cosine rule? Resources Resourceseses Interactivity The cosine rule (int-6276) Exercise 10.3 The cosine rule 10.3 Quick quiz 10.3 Exercise Individual pathways PRACTISE CONSOLIDATE MASTER 1, 4, 7, 9, 14, 17 2, 5, 8, 10, 15, 18 3, 6, 11, 12, 13, 16, 19 Where appropriate in this exercise, write your angles correct to the nearest minute and side lengths correct to 2 decimal places. Fluency 1. Calculate the third side of triangle ABC given a = 3.4, b = 7.8 and C = 80°. 2. In triangle ABC, b = 64.5, c = 38.1 and A = 58°34′. Calculate the value of a. WE4 3. In triangle ABC, a = 17, c = 10 and B = 115°. Calculate the value of b, and hence calculate the values of A and C. 4. WE5 Calculate the size of the smallest angle in the triangle with sides 6 cm, 4 cm and 8 cm. (Hint: The smallest angle is opposite the smallest side.) 5. In triangle ABC, a = 356, b = 207 and c = 296. Calculate the size of the largest angle. 6. In triangle ABC, a = 23.6, b = 17.3 and c = 26.4. Calculate the size of all the angles. 7. WE6 Two rowers set out from the same point. One rows N30°E for 1500 m and the other rows S40°E for 1200 m. Calculate the distance between the two rowers, correct to the nearest metre. 8. Maria cycles 12 km in a direction N68°W and then 7 km in a direction of N34°E. a. Calculate her distance from the starting point. b. Determine the bearing of the starting point from her finishing point. 518 Jacaranda Maths Quest 10 Stage 5 NSW Syllabus Third Edition Understanding 9. A garden bed is in the shape of a triangle, with sides of length 3 m, 4.5 m and 5.2 m. a. Calculate the size of the smallest angle. b. Hence, calculate the area of the garden, correct to 2 decimal places. (Hint: Draw a diagram, with the longest length as the base of the triangle.) 10. A hockey goal is 3 m wide. When Sophie is 7 m from one post and 5.2 m from the other, she shoots for goal. Determine within what angle, to the nearest degree, the shot must be made if it is to score a goal. 11. An advertising balloon is attached to two ropes 120 m and 100 m long. The ropes are anchored to level ground 35 m apart. Calculate the height of the balloon when both ropes are taut. A c = 120 m b = 100 m B a = 35 m C 12. A plane flies in a direction of N70°E for 80 km and then on a bearing of S10°W for 150 km. a. Calculate the plane’s distance from its starting point, correct to the nearest km. b. Calculate the plane’s direction from its starting point. 13. Ship A is 16.2 km from port on a bearing of 053°T and ship B is 31.6 km from the same port on a bearing of 117°T. Calculate the distance between the two ships, in km correct to 1 decimal place. Communicating, reasoning and problem solving 14. A plane takes off at 10.00 am from an airfield and flies at 120 km/h on a bearing of N35°W. A second plane takes off at 10.05 am from the same airfield and flies on a bearing of S80°E at a speed of 90 km/h. Determine how far apart the planes are at 10.25 am, in km correct to 1 decimal place. 15. Three circles of radii 5 cm, 6 cm and 8 cm are positioned so that they just touch one 5 cm 6 cm another. Their centres form the vertices of a triangle. Determine the largest angle in the triangle. Show your working. 8 cm 16. For the shape shown, determine: 8 a. the length of the diagonal 150° x b. the magnitude (size) of angle B c. the length of x. 7 B 60° 10 m TOPIC 10 Trigonometry II (Path) 519  17. From the top of a vertical cliff 68 m high, an observer notices a yacht at sea. The angle of depression to the yacht is 47°. The yacht sails directly away from the cliff, and after 10 minutes the angle of depression is 15°. Determine the speed of the yacht, in km/h correct to 2 decimal places. 18. Determine the size of angles CAB, ABC and BCA. Give your answers in degrees correct to 2 decimal places. C 2 cm A 5 cm 8 cm B 19. A vertical flag pole DB is supported by two wires AB and BC. AB is 5.2 metres long, BC is 4.7 metres long and B is 3.7 metres above ground level. Angle ADC B is a right angle. a. Evaluate the distance from A to C, in metres correct to 4 decimal places. b. Determine the angle between AB and BC, in degrees correct to 2 decimal places. C D A LESSON 10.4 Area of triangles LEARNING INTENTION apply the formula, area = ab sin(C) to calculate the area of a triangle, given two sides and the At the end of this lesson you should be able to: 1 2 included angle calculate the area of a triangle, given the three sides. 10.4.1 Area of triangles The area of any triangle is given by the formula area = eles-5007 1 B bh, where b is the base 2 and h is the perpendicular height of the triangle. h In the triangle ABC, b is the base and h is the perpendicular height of the triangle. A C b 520 Jacaranda Maths Quest 10 Stage 5 NSW Syllabus Third Edition  sin(A) = Using the trigonometric ratio for sine: h B c Transposing the equation to make h the subject, we obtain: c a h = c sin(A) h A A C If we substitute h into the formula, area = bh we obtain the formula: b 1 2 Area = bc sin(A). 1 2 Area of triangle The area of triangle ABC using the sine ratio: Area = bc sin(A) 1 2 Depending on how the triangle is labelled, the formula could read: Area = ab sin(C) Area = ac sin(B) Area = bc sin(A) 1 1 1 2 2 2 The area formula may be used on any triangle provided that two sides of the triangle and the included angle (that is, the angle between the two given sides) are known. WORKED EXAMPLE 7 Calculating the area of a triangle Calculate the area of the triangle shown, in cm2 correct to 2 decimal places. 7 cm 120° 9 cm THINK WRITE/DRAW 1. Draw a labelled diagram of the triangle, label it B ABC and fill in the given information. c = 7 cm 120° a = 9 cm A C Let a = 9 cm, c = 7 cm, B = 120°. A C 2. Check that the criterion for the area rule has The area rule can be used since two side lengths been satisfied. and the included angle have been given. Area = ac sin(B) 1 3. Write down the appropriate rule for the area. 2 = × 9 × 7 × sin 120° 1 4. Substitute the known values into the rule. 2 5. Evaluate. Round off the answer to 2 decimal ≈ 27.28 cm2 places and include the appropriate unit. TOPIC 10 Trigonometry II (Path) 521  WORKED EXAMPLE 8 Determining angles in a triangle and its area A triangle has known dimensions of a = 5 cm, b = 7 cm and B = 52°. Determine A and C, correct to the nearest minute, and hence the area, in cm2 correct to 2 decimal places. THINK WRITE/DRAW 1. Draw a labelled diagram of the triangle, label it B ABC and fill in the given information. 52° a=5 A C Let a = 5, b = 7, B = 52°. A b=7 C 2. Check whether the criterion for the area rule has The area rule cannot be used since the included been satisfied. angle has not been given. 3. Write down the sine rule to calculate A. To calculate angle A: = a b sin(A) sin(B) = 5 7 4. Substitute the known values into the rule. sin(A) sin(52°) 5. Transpose the equation to make sin A the subject. 5 sin(52°) = 7 sin(A) sin(A) = 5 sin(52°) A = sin 7( −1 5 sin(52°) ) 6. Evaluate. ≈ 34.254 15187° 7 7. Round off the answer to degrees and minutes. ≈ 34°15′ C ≈ 180° − (52° + 34°15′ ) = 93°45′ 8. Determine the value of the included angle, C, using the fact that the angle sum of any triangle is 180°. Area = 1 9. Write down the appropriate rule for the area. ab sin(C) 2 ≈ × 5 × 7 × sin(93°45′ ) 1 10. Substitute the known values into the rule. 2 11. Evaluate. Round off the answer to 2 decimal ≈ 17.46 cm2 places and include the appropriate unit. 10.4.2 Areas of triangles when three sides are known eles-6288 If the lengths of all the sides of the triangle are known but none of the angles, apply the cosine rule to calculate one of the angles. Area = Substitute the values of the angle and the two adjacent sides into the area of a triangle formula 1 bc sin(A) 2 522 Jacaranda Maths Quest 10 Stage 5 NSW Syllabus Third Edition WORKED EXAMPLE 9 Calculating the area of a triangle given the lengths of the 3 sides Calculate the area of the triangle with sides of 4cm, 6cm and 8cm, in cm2 correct to 2 decimal places. THINK WRITE/DRAW 1. Draw a labelled diagram of the triangle, call it ABC C and fill in the given information. 4 cm 6 cm Let a = 4, b = 6, c = 8. B 8 cm A b2 + c2 − a2 2. Choose an angle and apply the cosine rule to cos(A) = 62 + 82 − 42 2bc = calculate angle A. 2×6×8 = 7 8 A = cos−1 7 ( ) ≈ 28.95502437° 8 Area = 1 3. Write down the appropriate rule for the area. bc sin (A) 2 Area = × 6 × 8 × sin (28.95502437°) 1 4. Substitute the known values into the rule. ≈ 11.61895 2 5. Evaluate. Round off the answer to 2 decimal places and include the appropriate unit. Area is approximately 11.62 cm2. COMMUNICATING — COLLABORATIVE TASK: Using graphing applications to verify the area rule Equipment: Graphing software. 1. Use a graphing application (such as Geogebra), to construct a triangle of any shape. 2. Measure the base and height and calculate the area of the triangle using A = 1 bh. 3. Use the area rule A = 2 1 ab sin(C) to calculate the area of the triangle. 2 4. What did you notice about both calculations? Compare your results with the rest of the class. Resources Resourceseses Interactivity Area of triangles (int-6483) TOPIC 10 Trigonometry II (Path) 523 Exercise 10.4 Area of triangles 10.4 Quick quiz 10.4 Exercise Individual pathways PRACTISE CONSOLIDATE MASTER 1, 4, 7, 10, 12, 15, 18, 21 2, 5, 8, 11, 13, 16, 19, 22 3, 6, 9, 14, 17, 20, 23, 24 Where appropriate in this exercise, write your angles correct to the nearest minute and other measurements correct to 2 decimal places. Fluency 1. Calculate the area of the triangle ABC with a = 7, b = 4 and C = 68°. 2. Calculate the area of the triangle ABC with a = 7.3, c = 10.8 and B = 104°40′. WE7 3. Calculate the area of the triangle ABC with b = 23.1, c = 18.6 and A = 82°17′. 4. A triangle has a = 10 cm, c = 14 cm and C = 48°. Determine A and B and hence the area. 5. A triangle has a = 17m , c = 22 m and C = 56°. Determine A and B and hence the area. WE8 6. A triangle has b = 32 mm, c = 15 mm and B = 38°. Determine A and C and hence the area. 7. MC In a triangle, a = 15 m, b = 20 m and B = 50°. The area of the triangle is: A. 86.2 m2 B. 114.9 m2 C. 149.4 m2 D. 172.4 m2 8. WE9 Calculate the area of the triangle with sides of 5 cm, 6 cm and 8 cm. 9. Calculate the area of the triangle with sides of 40 mm, 30 mm and 5.7 cm. 10. Calculate the area of the triangle with sides of 16 mm, 3 cm and 2.7 cm. 11. MC A triangle has sides of length 10 cm, 14 cm and 20 cm. The area of the triangle is: 2 A. 41 cm2 B. 65 cm2 C. 106 cm2 D. 137 cm Understanding 12. A piece of metal is in the shape of a triangle with sides of length 114 mm, 72 mm and 87 mm. Calculate its area. 13. A triangle has the largest angle of 115°. The longest side is 62 cm and another side is 35 cm. Calculate the area of the triangle to the nearest whole number. 14. A triangle has two sides of 25 cm and 30 cm. The angle between the two sides is 30°. Determine: a. its area b. the length of its third side 15. The surface of a fish pond has the shape shown in the diagram. 1m 2m 5m 4m Calculate how many goldfish can the pond support if each fish requires 0.3 m2 surface area of water. 524 Jacaranda Maths Quest 10 Stage 5 NSW Syllabus Third Edition 16. MC A parallelogram has sides of 14 cm and 18 cm and an angle between them of 72°. The area of the parallelogram is: 2 A. 118.4 cm2 B. 172.4 cm2 C. 239.7 cm D. 252 cm2 17. MC An advertising hoarding is in the shape of an isosceles triangle, with sides of length 15 m, 15 m and 18 m. It is to be painted with two coats of purple paint. If the paint covers 12 m2 per litre, the amount of paint needed, to the nearest litre, would be: A. 9 L B. 18 L C. 24 L D. 36 L Communicating, reasoning and problem solving 18. A parallelogram has diagonals of length 10 cm and 17 cm. An angle between them is 125°. Determine: a. the area of the parallelogram b. the dimensions of the parallelogram. 19. A lawn is to be made in the shape of a triangle, with sides of length 11 m, 15 m and 17.2 m. Determine how much grass seed, to the nearest kilogram, needs to be purchased if it is sown at the rate of 1 kg per 5 m2. 20. A bushfire burns out an area of level grassland shown in the diagram. (Note: This is a sketch of the area and is not drawn to scale.) Evaluate the area, in hectares correct to 1 decimal place, of the land that is burned. km 1.8 2 km River 400 m 200 m Road 21. An earth embankment is 27 m long and has a vertical cross-section shown in the diagram. Determine the volume of earth needed to build the embankment, correct to the nearest cubic metre. 130° 100° 2m 50° 80° 5m 22. Evaluate the area of this quadrilateral. 3.5 m 8m 4m 60° 5m TOPIC 10 Trigonometry II (Path) 525 23. A surveyor measured the boundaries of a property as shown. The side AB could not be measured because it crossed through a marsh. D 8 km 60° C 6 km 8.5 km 115° B A The owner of the property wanted to know the total area and the length of the side AB. Give all lengths correct to 2 decimal places and angles to the nearest degree. a. Calculate the area of the triangle ACD. b. Calculate the distance AC. c. Calculate the angle CAB. d. Calculate the angle ACB. e. Calculate the length AB. f. Determine the area of the triangle ABC. g. Determine the area of the property. 24. A regular hexagon has sides of length 12 centimetres. It is divided into six smaller equilateral triangles. Evaluate the area of the hexagon, giving your answer correct to 2 decimal places. 12 cm 526 Jacaranda Maths Quest 10 Stage 5 NSW Syllabus Third Edition  LESSON 10.5 The unit circle LEARNING INTENTION At the end of this lesson you should be able to: determine in which quadrant an angle lies apply and interpret the relationship between a point on the unit circle and the angle made with the positive x-axis use the unit circle to determine approximate trigonometric ratios for angles greater than 90° apply the relationships between supplementary and complementary angles relate the gradient of a line to its angle of inclination with the x-axis on the Cartesian plane. 10.5.1 The unit circle eles-5009 A unit circle is a circle with a radius of 1 unit. The unit circle is divided into 4 quadrants, numbered in an anticlockwise direction, as shown in the diagram. 90° y 2nd 1st quadrant quadrant + angles 0° 180° x 360° 3rd – angles 4th quadrant quadrant 270° Positive angles are measured anticlockwise from 0°. Negative angles are measured clockwise from 0°. WORKED EXAMPLE 10 Identifying where an angle lies on the unit circle State the quadrant of the unit circle in which each of the following angles is found. a. 145° b. 282° THINK WRITE a. The given angle is between 90° and 180°. a. 145° is in quadrant 2. State the appropriate quadrant. b. The given angle is between 270° and 360°. b. 282° is in quadrant 4. State the appropriate quadrant. Consider the unit circle with point P(x, y) making the right-angled y triangle OPN as shown in the diagram. P(x, y) Using the trigonometric ratios: 1 y = cos(𝜃), = sin(𝜃), = tan(𝜃) x y y θ 0 x N A(1, 0) x 1 1 x where 𝜃 is measured anticlockwise from the positive x-axis. TOPIC 10 Trigonometry II (Path) 527  Calculate value of sine, cosine and tangent To calculate the value of sine, cosine or tangent of any angle 𝜃 from the unit circle: cos(𝜃) = x sin(𝜃) = y tan(𝜃) = = y sin(𝜃) x cos(𝜃) 10.5.2 The four quadrants of the unit circle eles-5010 Approximate values for sine, cosine and tangent 90° y of an angle can be found from the unit circle using the following steps, as shown in the diagram. 1 Step 1: Draw a unit circle, label the x- and y-axes. P Step 2: Mark the angles 0°, 90°, 180°, 270°, Step 3: Draw the given angle 𝜃. and 360°. Step 4: Mark x = cos(𝜃), y = sin(𝜃). sin(θ) θ 0° Step 5: Approximate the values of x and y and 180° –1 cos(θ) 1 x equate to give the values of cos(𝜃) and sin(𝜃). 360° Where the angle lies in the unit circle determines whether the trigonometric ratio is positive or negative. –1 270° Sign of the trigonometric functions In the first quadrant x > 0, y > 0; therefore All trig Consider the following. y In the second quadrant x < 0, y > 0; therefore Sine ratios are positive. In the third quadrant x < 0, y < 0; therefore Tangent (the y-value) is positive. II I x < 0, y > 0 x > 0, y > 0 Sine positive All positive y ( ) values is positive. In the fourth quadrant x > 0, y < 0; therefore Cosine x x III IV x < 0, y < 0 x > 0, y < 0 (the x-value) is positive. Tangent positive Cosine positive There are several mnemonics for remembering the sign of the trigonometric functions. A common saying is ‘All Stations To Central’ 528 Jacaranda Maths Quest 10 Stage 5 NSW Syllabus Third Edition  WORKED EXAMPLE 11 Using the unit circle to approximate trigonometric ratios of an angle Determine the approximate value of each of the following using the unit circle. a. sin(200°) b. cos(200°) c. tan(200°) THINK WRITE/DRAW Draw a unit circle and construct an angle of 200°. Label 90° the point corresponding to the angle of 200° on the y circle P. Highlight the lengths, representing the x- and 1 y-coordinates of point P. θ = 200° x 0° 180° –1 y 1 x P 360° –1 a. sin(200°) = −0.3 270° a. The sine of the angle is given by the y-coordinate of P. Determine the y-coordinate of P by measuring the distance along the y-axis. State the value of sin(200°). (Note: The sine value will be negative b. cos(200°) = −0.9 as the y-coordinate is negative.) b. The cosine of the angle is given by the x-coordinate of P. Determine the x-coordinate of P by measuring the distance along the x-axis. State the value of cos(200°). (Note: Cosine is also negative in quadrant 3, as the x-coordinate is negative.) −0.3 c. tan(200°) = = = 0.3333 −0.9 sin(200°) 1 c. cos(200°) 3 The approximate results obtained in Worked example 11 can be verified with the aid of a calculator: sin(200°) = −0.342 020 143, cos(200°) = −0.939 692 62 and tan(200°) = 0.3640. Rounding these values to 1 decimal place would give −0.3, −0.9 and 0.4 respectively, which approximately match the values obtained from the unit circle. 10.5.3 Supplementary angles eles-6284 supplementary angles, say A° and (180 − A)°. Consider the special relationship between the sine, cosine and tangent of y 1 (180 – In the diagram, the y-axis is an axis of symmetry. A) That is, sin(A°) = sin(180 − A)° The y-values of points C and E are the same. E C ° A° A° That is, cos(A°) = − cos(180 − A)° –1 1 x The x-values of points C and E are opposites in value. O –1 TOPIC 10 Trigonometry II (Path) 529  Thus: sin(180 − A)° = sin(A°) cos(180 − A)° = − cos(A°) sin(180 − A)° tan(180 − A)° = = = − tan(A°) cos(180 − A)° − cos(A°) sin(A°) Supplementary angles A° and (180° − A) where 0° ≤ A ≤ 90° sin (180° − A) = sin (A) cos (180° − A) = − cos (A) tan (180° − A) = − tan (A) 10.5.4 Complementary angles eles-6285 are complementary angles, and 𝜃 and (90° − 𝜃) are also Complementary angles add to 90°. Therefore, 30° and 60° complementary angles. 90°– θ Therefore, sin(60°) = cos(30°). 30° The sine of an angle is equal to the cosine of its complement. 1 y 1 y We say that sine and cosine are complementary functions. sin(A) = cos(90° − A) cos(A) = sin(90° − A) 60° θ x x Complementary angles where 0° ≤ A ≤ 90° sin(A) = cos (90° − A) cos(A) = sin (90° − A) 10.5.5 Gradient of a line eles-6286 The gradient of a line may be found using trigonometry if the angle the line y makes with the positive direction of the x-axis is known. rise ratio for the angle 𝜃, giving the relationship m = tan(𝜃). In the triangle shown, the gradient of the line is which is the tangent rise run 𝜃 run x tan(𝜃) = rise = m run Gradient, m, of a line m = tan(𝜃) 530 Jacaranda Maths Quest 10 Stage 5 NSW Syllabus Third Edition WORKED EXAMPLE 12 Gradients of lines given the angles made with the x-axis a. Determine the gradient (accurate to 3 decimal places) of a line making an angle of 40° to the positive x-axis, as shown in the figure at right. 40° x b. Determine the gradient of the line shown at right. Express your answer to 2 decimal places. y 60° x a. m = tan(𝜃) THINK WRITE/DRAW = tan 40° a. Since the angle the line makes with the m = tan(𝜃) can be used. positive x-axis is given, the formula = 0.839 b. 𝜃 = 180° − 60° = 120° b. 1. The angle given is not the one between the Calculate the required angle 𝜃. graph and the positive direction of the x-axis. 2. Use m = tan(𝜃) to calculate m to 2 decimal m = tan(𝜃) places. = tan(120°) = −1.73 y 60° 𝜃 x TOPIC 10 Trigonometry II (Path) 531 Exercise 10.5 The unit circle 10.5 Quick quiz 10.5 Exercise Individual pathways PRACTISE CONSOLIDATE MASTER 1, 2, 6, 8, 12, 17, 20, 23 3, 4, 7, 9, 11, 13, 18, 21, 24 5, 10, 14, 15, 16, 19, 22, 25 Where appropriate in this exercise, give answers correct to 2 decimal places. Fluency 1. WE10 State which quadrant of the unit circle each of the following angles is in. a. 60° b. 130° c. 310° d. 260° e. 100° f. 185° 2. MC If 𝜃 = 43°, the triangle drawn to show this would be in: A. quadrant 1 B. quadrant 2 C. quadrant 3 D. quadrant 4 3. MC If 𝜃 = 295°, the triangle drawn