Trigonometric ratios and their applications

Transcription

1 5 Trigonometric ratios and their applications 5 Trigonometry of right-angled triangles 5 Elevation, depression and bearings 5 The sine rule 5D The cosine rule 5E rea of triangles 5F Trigonometric identities 5G Radian measurement 5H rcs, sectors and segments areas of study Right-angled triangles and solutions to problems involving right-angled triangles using sine, cosine and tangent The relationships sin () + cos () = 1, cos () = sin (90 ) and sin () = cos (90 ) Two-dimensional applications including angles of depression and elevation Exact values of sine, cosine and tangent for 30, 45 and 60 Solution of triangles by the sine and cosine rules reas of triangles, including the formula = s (ss a) (s s b) ) (s s c) ircle mensuration: radian measure, arc length, areas of sectors and segments pplications, for example, navigation and surveying in simple contexts eookplus 5a Trigonometry of right-angled triangles Digital doc 10 Quick Questions Trigonometry, derived from the Greek words trigon (triangle) and metron (measurement), is the branch of mathematics that deals with the relationship between the sides and angles of a triangle. It involves finding unknown angles, side lengths and areas of triangles. The principles of trigonometry are used in many practical situations such as building, surveying, navigation and engineering. In previous years you will have studied the trigonometry of right-angled triangles. We will review this material before considering non right-angled triangles. sin () = opposite side which is abbreviated to sin () = O hypotenuse H cos () = adjacent side which is abbreviated to cos () = hypotenuse H tan () = opposite side which is abbreviated to tan () = O adjacent side Opposite (O) Hypotenuse (H) () djacent The symbol (theta) is one of the many letters of the Greek alphabet used to represent the angle. Other symbols include α (alpha), β (beta) and γ (gamma). Non-Greek letters may also be used. 136 maths Quest 11 advanced General mathematics for the asio lasspad

2 Writing the mnemonic SOH H TO each time we perform trigonometric calculations will help us to remember the ratios and solve the problem. Pythagoras theorem For specific problems it may be necessary to determine the side lengths of a right-angled triangle before calculating the trigonometric ratios. In this situation, Pythagoras theorem is used. Pythagoras theorem states: In any right-angled triangle, c = a + b. a c b Worked Example 1 Determine the value of the pronumerals, correct to decimal places. a x 4 50 b h Write a 1 Label the sides, relative to the marked angles. a x 4 H O 50 Write what is given. Have: angle and hypotenuse (H) 3 Write what is needed. Need: opposite (O) side 4 Determine which of the trigonometric ratios is required, using SOH H TO. sin ( ) = O H 5 Substitute the given values into the appropriate ratio. sin (50 ) = x 4 6 Transpose the equation and solve for x. 4 sin (50 ) = x x = 4 sin (50 ) 7 Round the answer to decimal places. = 3.06 b 1 Label the sides, relative to the marked angle. b h H Write what is given. Have: angle and adjacent () side 3 Write what is needed. Need: hypotenuse (H) 4 Determine which of the trigonometric ratios is required, using SOH H TO. 5 Substitute the given values into the appropriate ratio. cos ( ) = H cos (4 5 ) = 7 h hapter 5 Trigonometric ratios and their applications 137

3 6 Solve for h. On the Main screen, complete the entry line as: 7 solve cos( dms( 4, 5) =, h h Then press E. 7 Round the answer to decimal places. = 7.69 Worked Example Find the angle, giving the answer in degrees and minutes. 1 1 Label the sides, relative to the marked angles. Write O Write what is given. Have: opposite (O) and adjacent () sides 3 Write what is needed. Need: angle 4 Determine which of the trigonometric ratios is required, using SOH H TO. 5 Substitute the given values into the appropriate ratio. 6 To calculate tan 1, on the Main screen, complete the entry lines as: tan todms( Press E after each entry line. Note: todms can be located by tapping: ction Transformation todms tan ( ) = O tan ( ) = Maths Quest 11 dvanced General Mathematics for the asio lasspad

4 7 Write the answer to the nearest minute. q = tan = Exact values Most of the trigonometric values that we will deal with in this chapter are only approximations. However, angles of 30, 45 and 60 have exact values of sine, cosine and tangent. onsider an equilateral triangle,, of side length units. If the triangle is perpendicularly bisected, then two congruent triangles, D and D, are obtained. From triangle D it can be seen that D creates a right-angled triangle with angles of 60 and 30 and base length (D) of 1 unit. The length of D is obtained using Pythagoras theorem. Using triangle D and the three trigonometric ratios the following exact values are obtained: sin (30 ) = 1 cos (30 ) = 3 tan (30 ) = 1 3 sin (60 ) = cos (60 ) = 1 3 or tan (60 ) = 3 onsider a right-angled isosceles triangle EFG whose equal sides are of 1 unit. The hypotenuse EG is obtained by using Pythagoras theorem. (EG) = (EF) + (FG) = = EG = Using triangle EFG and the three trigonometric ratios, the following exact values are obtained: sin (45 ) = 1 cos (45 ) = 1 or or or E D 45 1 G 1 F tan (45 ) = 1 1 or 1 Worked Example 3 Determine the height of the triangle shown in surd form. h 60 8 cm hapter 5 Trigonometric ratios and their applications 139

5 Write 1 Label the sides relative to the marked angle. h O 60 8 cm Write what is given. Have: angle and adjacent () side 3 Write what is needed. Need: opposite (O) side 4 Determine which of the trigonometric ratios is required, using SOH H TO. tan ( ) = O 5 Substitute the given values into the appropriate ratio. tan (60 ) = h 8 6 Substitute exact values where appropriate. 3 = h 8 7 Transpose the equation to find the required value. h = State the answer. The triangle s height is 8 3 cm. REMEMER 1. For any right-angled triangle: sin () = O cos () = tan () = O Opposite H H H (O). To determine which trigonometric ratio to use when solving a right-angled triangle, follow these steps: (a) Label the diagram using the symbols, O,, H. (b) Write what is given. (c) Write what is needed. (d) Determine which of the trigonometric ratios is required, using SOH H TO. (e) Substitute the given values into the rule and solve. a 3. Pythagoras theorem, c = a + b, may also be used to solve right-angled triangles. 4. ngles of 30, 45 and 60 have exact values for sine, cosine and tangent sin ( ) cos ( ) tan ( ) = 3 = 1 3 = Hypotenuse (H) () djacent c b 140 Maths Quest 11 dvanced General Mathematics for the asio lasspad

6 exerise 5a eookplus Digital doc SkillSHEET 5.1 Labelling rightangled triangles Trigonometry of right-angled triangles 1 opy and label the sides of the following right-angled triangles using the words hypotenuse, adjacent, opposite and the symbol. a b c djacent Opposite d eookplus Digital doc SkillSHEET 5. Using trigonometric ratios We 1 Find the value of the pronumerals, correct to decimal places. a x b x 3 14' 7.5 c x 47 8' 17 d 684 x 6 38' eookplus Digital doc SkillSHEET 5.3 Degrees and minutes e 1.03 f x 14 5' x g x ' h x 17 y 38 48' 3 We Find the angle, giving the answer in degrees and minutes. a b 5 c d.1 e f g h hapter 5 Trigonometric ratios and their applications 141

7 eookplus Digital doc SkillSHEET 5.4 omposite shapes 1 4 We 3 n isosceles triangle has a base of 1 cm and equal angles of 30. Find, in surd form: a the height of the triangle 1 cm b the area of the triangle c the perimeter of the triangle, giving your answers in simplest surd form Find the perimeter of the composite shape at right, in surd form. The length measurements are in metres eookplus Digital doc SkillSHEET 5.5 omposite shapes 6 ladder 6.5 m long rests against a vertical wall and makes an angle of 50 to the horizontal ground. a How high up the wall does the ladder reach? b If the ladder needs to reach 1 m higher, to the nearest minute, what angle should it make to the ground? m-long road goes straight up a slope. If the road rises 50 m vertically, what is the angle that the road makes with the horizontal? 8 n ice-cream cone has a diameter of 6 cm and a sloping edge of 15 cm. Find the angle at the bottom of the cone. 9 vertical flagpole is supported by a wire attached from the top of the pole to the horizontal ground, 4 m from the base of the pole. Joanne measures the angle the wire makes with the ground and finds this is 65. How tall is the flagpole? 10 stepladder stands on a floor, with its feet 1.5 m apart. If the angle formed by the legs is 55, how high above the floor is the top of the ladder? 11 The angle formed by the diagonal of a rectangle and one of its shorter sides is 60. If the diagonal is 8 cm long, find the dimensions of the rectangle, in surd form. 1 In the figure at right, find the value of the pronumerals, correct to decimal places. a 7 b d c 13 In the figure at right, find the value of the pronumerals, correct to decimal places. b a In the figure at right, find the value of the pronumeral x, correct to decimal places. 15 n advertising balloon is attached to a rope 10 m long. The rope makes an angle of 75 to level ground. How high above the ground is the balloon? 6 x n isosceles triangle has sides of 17 cm, 0 cm and 0 cm. Find the magnitude of the angles. 14 maths Quest 11 advanced General mathematics for the asio lasspad

8 17 garden bed at right is in the shape of a trapezium. What volume of garden mulch is needed to cover it to a depth of 15 cm? 4 m 1 m gable roof has sloping sides of 8.3 m. It rises to a height of.7 m at the centre. a What is the angle of slope of the two sides? b How wide is the roof at its base? 8.3 m 8.3 m.7 m 19 ladder 10 m long rests against a vertical wall at an angle of 55 to the horizontal. It slides down the wall, so that it now makes an angle of 48 with the horizontal. a Through what vertical distance did the top of the ladder slide? b Does the foot of the ladder move through the same distance? Justify your answer. 5 Elevation, depression and bearings Trigonometry is especially useful for measuring distances and heights which are difficult or impractical to access. For example, two important applications of right-angled triangles involve: 1. angles of elevation and depression, and. bearings. ngles of elevation and depression ngles of elevation and depression are employed when dealing with directions which require us to look up and down respectively. n angle of elevation is the angle between the horizontal and an object which is higher than the observer (for example, the top of a mountain or flagpole). Line of sight ngle of elevation n angle of depression is the angle between the horizontal and an object which is lower than the observer (for example, a boat at sea when the observer is on a cliff). ngle of depression Line of sight Unless otherwise stated, the angle of elevation or depression is measured and drawn from the horizontal. ngles of elevation and depression are each measured from the horizontal. When solving problems involving angles of elevation and depression, it is best always to draw a diagram. The angle of elevation is equal to the angle of depression since they are alternate Z angles. E D D and E are alternate angles D = E hapter 5 Trigonometric ratios and their applications 143

9 Worked Example 4 From a cliff 50 metres high, the angle of depression of a boat at sea is 1. How far is the boat from the base of the cliff? 1 Draw a diagram and label all the given information. Include the unknown length, x, and the angle of elevation, 1. Write 1 x 1 50 m Write what is given. Have: angle and opposite side 3 Write what is needed. Need: adjacent side 4 Determine which of the trigonometric ratios is required (SOH H TO). tan ( ) = O 5 Substitute the given values into the appropriate ratio. tan (1 ) = 50 x 6 Transpose the equation and solve for x. x tan (1 ) = x = tan( 1 ) 7 Round the answer to decimal places. = nswer the question. The boat is 35.3 m away from the base of the cliff. earings earings measure the direction of one object from another. There are two systems used for describing bearings. True bearings are measured in a clockwise direction, starting from north (0 T). N N 150 T ompass bearing equivalent is S30 E N onventional or compass bearings are measured: first, relative to north or south, and second, relative to east or west. The two systems are interchangeable. For example, a bearing of 40 T is the same as S60 W. When solving questions involving direction, always start with a diagram showing the basic compass points: north, south, east and west. 0 0 W E W E S S N0 W True bearing equivalent S70 E True bearing equivalent is 340 T is 110 T N N W E W E 40 T 60 S S S60 W 144 Maths Quest 11 dvanced General Mathematics for the asio lasspad

10 Worked example 5 ship sails 40 km in a direction of N5 W. How far west of the starting point is it? 1 Draw a diagram of the situation, labelling each of the compass points and the given information. WriTe/draW x N W 40 km 5 E Write what is given for the triangle. Have: angle and hypotenuse 3 Write what is needed for the triangle. Need: opposite side 4 Determine which of the trigonometric ratios is required (SOH H TO). 5 Substitute the given values into the appropriate ratio. S sin () = O H sin (5 ) = x 40 6 Transpose the equation and solve for x. 40 sin (5 ) = x x = 40 sin (5 ) 7 Round the answer to decimal places. = nswer the question. The ship is 31.5 km west of the starting point. Worked example 6 ship sails 10 km east, then 4 km south. What is its bearing from its starting point? 1 Draw a diagram of the situation, labelling each of the compass points and the given information. WriTe N S 10 km 4 km eookplus Tutorial int-1045 Worked example 6 Write what is given for the triangle. Have: adjacent and opposite sides 3 Write what is needed for the triangle. Need: angle 4 Determine which of the trigonometric ratios is required (SOH H TO). 5 Substitute the given values into the appropriate ratio. 6 Transpose the equation and solve for, using the inverse tan function. tan ( ) = O tan ( ) = 4 10 = tan onvert the angle to degrees and minutes. = = 1 48 hapter 5 Trigonometric ratios and their applications 145

11 8 Express the angle in bearings form. The bearing of the ship was initially 0 T; it has since rotated through an angle of 90 and an additional angle of To obtain the final bearing these values are added. earing = = T 9 nswer the question. The bearing of the ship from its starting point is T. REMEMER 1. ngles of elevation and depression are each measured from the horizontal.. The angle of elevation is equal to the angle of depression since they are alternate Z angles. 3. True bearings are measured in a clockwise direction, starting from north (0 T). 4. onventional or compass bearings are measured first, relative to north or south, and second, relative to east or west. 5. Whenever solving problems involving angles of elevation and depression or bearings, you should always draw a diagram and label all the given information. 6. Set up a compass as the basis of your diagram for bearings questions. Exercise 5 Elevation, depression and bearings 1 WE 4 From a vertical fire tower 60 m high, the angle of depression to a fire is 6. How far away, to the nearest metre, is the fire? person stands 0 m from the base of a building, and measures the angle of elevation to the top of the building as 55. If the person is 1.7 m tall, how high, to the nearest metre, is the building? 3 n observer on a cliff top 57 m high observes a ship at sea. The angle of depression to the ship is 15. The ship sails towards the cliff, and the angle of depression is then 5. How far, to the nearest metre, did the ship sail between sightings? 4 Two vertical buildings, 40 m and 6 m high, are directly opposite each other across a river. The angle of elevation of the top of the taller building from the top of the smaller building is 7. How wide is the river? (Give the answer to decimal places.) 5 To calculate the height of a crane which is on top of a building, Denis measures the angle of elevation to the bottom and top of the crane. These were 6 and 68 respectively. If the building is 4 m high find, to decimal places: a how far Denis is from the building b the height of the crane. 6 new skyscraper is proposed for the Melbourne Docklands region. It is to be 500 m tall. What would be the angle of depression, in degrees and minutes, from the top of the building to the island on lbert Park Lake, which is 4. km away? 7 From a rescue helicopter 500 m above the ocean, the angles of depression of two shipwreck survivors are 48 (survivor 1) and 35 (survivor ). a Draw a labelled diagram which represents the situation. b alculate how far apart the two survivors are. 8 lookout tower has been erected on top of a mountain. t a distance of 5.8 km, the angle of elevation from the ground to the base of the tower is 15.7 and the angle of elevation to the observation deck (on the top of the tower) is How high, to the nearest metre, is the observation deck above the top of the mountain? 146 Maths Quest 11 dvanced General Mathematics for the asio lasspad

12 9 From a point on level ground, the angle of elevation of the top of a building 50 m high is 45. From a point on the ground and in line with and the foot of the building, the angle of elevation of the top of the building is 60. Find, in simplest surd form, the distance from to. 10 Express the following conventional bearings as true bearings. a N35 W b S47 W c N58 E d S17 E 11 Express the following true bearings in conventional form. a 46 T b 107 T c 31 T d 074 T 1 m a bearing of S30 E is the same as: 030 T 10 T 150 T D 10 T E 40 T b bearing of 80 T is the same as: N10 W S10 W S80 W D N80 W E N10 E 13 We 5 pair of canoeists paddle 1800 m on a bearing of N0 E. How far north of their starting point are they, to the nearest metre? 14 yacht race consists of four legs. The first three legs are 4 km due east, then 5 km south, followed by km due west. a How long is the final leg, if the race finishes at the starting point? b On what bearing must the final leg be sailed? 15 We 6 ship sails 0 km south, then 8 km west. What is its bearing from the starting point? 16 cross-country competitor runs on a bearing of N60 W for km, then due north for 3 km. a How far is he from the starting point? b What is the true bearing of the starting point from the runner? 17 Two hikers set out from the same campsite. One walks 7 km in the direction 043 T and the other walks 10 km in the direction 133 T. a What is the distance between the two hikers? b What is the bearing of the first hiker from the second? 18 ship sails 30 km on a bearing of 0, then 0 km on a bearing of 50. Find: a how far south of the original position it is b how far west of the original position it is c the true bearing of the ship from its original position, to the nearest degree. 19 The town of racknaw is due west of rley. hris, in an ultralight plane, starts at a third town, hampton, which is due north of racknaw, and flies directly towards rley at a speed of 40 km/h in a direction of 110 T. She reaches rley in 3 hours. Find: a the distance between rley and racknaw b the time to complete the journey from hampton to racknaw, via rley, if she increases her speed to 45 km/h between rley and racknaw. 0 From a point,, on the ground, the angle of elevation of the top of a vertical tower due north of is 46. From a point, due east of, the angle of elevation of the top of the tower is 3. If the tower is 85 m high, find: a the distance from to the foot of the tower b the distance from to the foot of the tower c the true bearing of the tower from. hapter 5 Trigonometric ratios and their applications 147

13 5 1 bird flying at 50 m above the ground was observed at noon from my front door at an angle of elevation of 5. Two minutes later its angle of elevation was 4. a If the bird was flying straight and level, find the horizontal distance of the bird: i from my doorway at noon ii from my doorway at 1.0 pm. b Hence, find: i the distance travelled by the bird in the two minutes ii its speed of flight in km/h. The sine rule When working with non right-angled triangles, it is usual to label the angles, and, and the sides a, b and c, so that side a is the side opposite angle, side b is the side opposite angle and side c is the side opposite angle. In a non right-angled triangle, a perpendicular line, h, can be drawn from the angle to side b. Using triangle D we obtain sin () = h. Using triangle D c we obtain sin () = h a. Transposing each equation to make h the subject, we obtain: h = c sin () and h = a sin (). Equate to get c sin () = a sin (). Transpose to get c a = sin ( ) sin ( ) In a similar way, if a perpendicular line is drawn from angle to side a, we get b c = sin ( ) sin ( ) From this, the sine rule can be stated. In any triangle : a b c c a = = sin( ) sin ( ) sin( ) b Notes 1. When using this rule, depending on the values given, any combination of the two equalities may be used to solve a particular triangle.. To solve a triangle means to find all unknown side lengths and angles. The sine rule can be used to solve non right-angled triangles if we are given: 1. two angles and one side length. two side lengths and an angle opposite one of these side lengths. c c D h h c = sin () and h a = sin () b b a a Worked Example 7 In the triangle, a = 4 m, b = 7 m and = 80. Find, and c. 1 Draw a labelled diagram of the triangle and fill in the given information. Write c 80 a = 4 b = Maths Quest 11 dvanced General Mathematics for the asio lasspad

14 heck that one of the criteria for the sine rule has been satisfied. 3 Write the sine rule to find. To find angle : 4 Substitute the known values into the rule. The sine rule can be used since two side lengths and an angle opposite one of these side lengths have been given. a b = sin ( ) sin ( ) 4 7 sin ( ) = sin ( 80 ) 5 Transpose the equation to make sin () the 4 sin (80 ) = 7 sin () subject. 4 sin( 80 ) = sin( ) 7 sin( ) = 4 sin( 80 ) 7 6 Evaluate. = sin 4 sin( 80 ) 1 7 = sin 1 ( ) = Round the answer to degrees and minutes. = Determine the value of angle using the fact that the angle sum of any triangle is 180. = 180 ( ) = Write the sine rule to find c. To find side length c: c b = sin( ) sin ( ) 10 Substitute the known values into the rule. c 7 sin( 65 = 45 ) sin ( 80 ) 7 sin( ) 11 Transpose the equation to make c the subject. c = sin( 80 ) 1 Evaluate. Round the answer to decimal places and include the appropriate unit. 7 = = = = 6.48 m The ambiguous case When using the sine rule there is one important issue to consider. If we are given two side lengths and an angle opposite one of these side lengths, then two different triangles may be drawn. For example, if a = 10, c = 6 and = 30, two possible triangles could be created. c = 6 a = c = 6 a = hapter 5 Trigonometric ratios and their applications 149

15 In the first case, angle is an acute angle, while in the second case, angle is an obtuse angle. The two values for will add to 180. The ambiguous case does not work for each example. It would be useful to know, before commencing a question, whether or not the ambiguous case exists and, if so, to then find both sets of solutions. The ambiguous case exists if is an acute angle and a > c > a sin (), or any equivalent statement; for example, if is an acute angle and a > b > a sin (), and so on. Worked example 8 In the triangle, a = 10 m, c = 6 m and = 30. a Show that the ambiguous case exists. b Find two possible values of, and hence two possible values of and b. WriTe eookplus Tutorial int-1046 Worked example 8 Method 1: Using the rules a 1 heck that the conditions for an ambiguous case exist, i.e. that is an acute angle and that a > c > a sin (). a = 30 so is an acute angle. sin () = sin (30 ) = 0.5 a > c > a sin () 10 > 6 > 10 sin (30 ) 10 > 6 > 5 This is correct. State the answer. This is an ambiguous case of the sine rule. ase 1 b 1 Draw a labelled diagram of the triangle and fill in the given information. b c = 6 a = 10 Write the sine rule to find. To find angle : a c = sin ( ) sin ( ) 3 Substitute the known values into the rule. 4 Transpose the equation to make sin () the subject. 5 Evaluate angle, in degrees and minutes. 6 Determine the value of angle, using the fact that the angle sum of any triangle is sin ( ) = sin ( 30 ) 10 sin (30 ) = 6 sin () 10 sin ( 30 ) = sin( ) 6 10 sin ( 30 ) sin( ) = 6 = sin 1 10 sin ( 30 ) 6 = 56 7 = 180 ( ) = Write the sine rule to find b. To find side length b: b c = sin ( ) sin ( ) 150 maths Quest 11 advanced General mathematics for the asio lasspad

16 ase 8 Substitute the known values into the rule. 9 Transpose the equation to make b the subject and evaluate. b 1 Draw a labelled diagram of the triangle and fill in the given information. Write the alternative value for angle. Subtract the value obtained for in ase 1 from Determine the alternative value of angle, using the fact that the angle sum of any triangle is Write the sine rule to find the alternative b. 5 Substitute the known values into the rule. 6 Transpose the equation to make b the subject and evaluate. Method : Using a S calculator b b 6 sin ( 93 = 33 ) sin ( 30 ) c = 6 b = 6 sin ( 9333 ) sin( 30 ) = m a = To find the alternative angle : If sin () = , then could also be: = = = 180 ( ) = 6 7 To find side length b: b c = sin ( ) sin ( ) b 6 sin ( 6 = 7 ) sin ( 30 ) 1 Draw a labelled diagram of the triangle and fill in the given information. a = 10 c = 6 b = 6 sin ( 67 ) sin ( 30 ) = 5.35 m 30 In part a it was shown that the ambiguous case of the sine rule exists. Therefore, on the Main screen, complete the entry line as: 10 6 solve sin( a) sin( 30), = a a Then press E. 3 onvert the angles to degrees and minutes. = 56 7 or = hapter 5 Trigonometric ratios and their applications 151

17 4 alculate the size of the angle given each angle. 5 To find the side length b, on the Main screen, complete the entry line as: solve b 6 = sin( dms( 9333, )) sin( 30),b solve b 6 = sin( dms( 6, 7)) sin( 30),b Press E after each entry. If = 56 7, = ( ) = If = 13 33, = ( ) = Write the answers. If = 93 33, b = m If = 6 7, b = 5.35 m Hence, for this example there were two possible solutions as shown by the diagram below. c = 6 a = c = 6 a = REMEMER 1. The sine rule states that for any triangle : a b c = = sin( ) sin ( ) sin( ). When using this rule it is important to note that, depending on the values given, any combination of the two equalities may be used to solve a particular triangle. 3. The sine rule can be used to solve non right-angled triangles if we are given: (a) two angles and one side length (b) two side lengths and an angle opposite one of these side lengths. 4. The ambiguous case exists if is an acute angle and a > c > a sin (). Exercise 5 The sine rule 1 WE 7 In the triangle, a = 10, b = 1 and = 58. Find, and c. In the triangle, c = 17.35, a = 6.8 and = Find, and b. 3 In the triangle, a = 5, = 30 and = 80. Find, b and c. 4 In the triangle, c = 7, = 4 and = 105. Find, a and b. 15 Maths Quest 11 dvanced General Mathematics for the asio lasspad

18 5 In the triangle, a = 7, c = 5 and = 68. Find the perimeter of the triangle. 6 Find all unknown sides and angles for the triangle, given = 57, = 7 and a = Find all unknown sides and angles for the triangle, given a = 105, = 105 and = Find all unknown sides and angles for the triangle, given a = 3, b = 51 and = 8. 9 Find the perimeter of the triangle if a = 7.8, b = 6. and = M In a triangle, = 40, = 80 and c = 3. The value of b is: D 4.38 E WE 8 In the triangle, a = 10, c = 8 and = 50. Find two possible values of, and hence two possible values of b. 1 In the triangle, a = 0, b = 1 and = 35. Find two possible values for the perimeter of the triangle. 13 Find all unknown sides and angles for the triangle, given = 7, = 43 and c = Find all unknown sides and angles for the triangle, given = 100, b =.1 and = Find all unknown sides and angles for the triangle, given = 5, b = 17 and a = 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. How high is the building? 17 river has parallel banks which run directly east west. Kylie takes a bearing to a tree on the opposite 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. Find: a her distance from the second measuring point to the tree b the width of the river, to the nearest metre. 18 ship sails on a bearing of S0 W for 14 km, then changes direction and sails for 0 km and drops anchor. Its bearing from the starting point is now N65 W. a How far is it from the starting point? b On what bearing did it sail the 0 km leg? 19 cross-country runner runs at 8 km/h on a bearing of 150 T for 45 mins, then changes direction to a bearing of 053 T and runs for 80 mins until he is due east of the starting point. a How far was the second part of the run? b What was his speed for this section? c How far does he need to run to get back to the starting point? 0 From a fire tower,, a fire is spotted on a bearing of N4 E. From a second tower,, the fire is on a bearing of N1 W. The two fire towers are 3 km apart, and is N63 W of. How far is the fire from each tower? 1 M boat sails on a bearing of N15 E for 10 km, then on a bearing of S85 E until it is due east of the starting point. The distance from the starting point to the nearest kilometre is, then: 10 km 38 km 110 km D 113 km E 114 km M hill slopes at an angle of 30 to the horizontal. tree which is 8 m tall is growing at an angle of 10 to the vertical and is part-way up the slope. The vertical height of the top of the tree above the slope is: 7.37 m 8.68 m m D m E m hapter 5 Trigonometric ratios and their applications 153

19 3 cliff is 37 m high. The rock slopes outward at an angle of 50 to the horizontal, then cuts back at an angle of 5 to the vertical, meeting the ground directly below the top of the cliff. arol 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 m to secure it to a tree at the top of the cliff. Will the rope be long enough to allow her to reach the ground? 50 5 rope rock 37 m eookplus Digital doc WorkSHEET 5.1 5d The cosine rule In any non right-angled triangle, a perpendicular line can be drawn from angle to side b. Let D be the point where the perpendicular line meets side b, and the length of the perpendicular line be h. Let the length D = x units. The perpendicular line creates two right-angled triangles, D and D. Using triangle D and Pythagoras theorem, we obtain: c = h + x [1] c Using triangle D and Pythagoras theorem, we obtain: h a a = h + (b x) [] Expanding the brackets in equation []: D a = h + b bx + x x b x b Rearranging equation [] and using c = h + x from equation [1]: a = h + x + b bx = c + b bx = b + c bx From triangle D, x = c cos (), therefore a = b + c bx becomes a = b + c bc cos () This is called the cosine rule and is a generalisation of Pythagoras theorem. In a similar way, if the perpendicular line was drawn from angle to side a or from angle to side c, the two right-angled triangles would give c = a + b ab cos () and b = a + c ac cos () respectively. From this, the cosine rule can be stated: In any triangle a = b + c bc cos () b = a + c ac cos () c = a + b ab cos () c b a 154 maths Quest 11 advanced General mathematics for the asio lasspad

20 The cosine rule can be used to solve non right-angled triangles if we are given: 1. three sides of the triangle. two sides of the triangle and the included angle (the angle between the given sides). Worked Example 9 Find the third side of triangle given a = 6, c = 10 and = 76, correct to decimal places. Write Method 1: Using the rule 1 Draw a labelled diagram of the triangle and fill in the given information. c = a = 6 b heck that one of the criteria for the cosine rule has been satisfied. 3 Write the appropriate cosine rule to find side b. Yes, the cosine rule can be used since two side lengths and the included angle have been given. To find side b: b = a + c ac cos () 4 Substitute the given values into the rule. = cos (76 ) 5 Evaluate. = = b = Round the answer to decimal places. = correct to decimal places Method : Using a S calculator 1 Draw a labelled diagram of the triangle and fill in the given information. c = a = 6 b Write the appropriate cosine rule to find side b. b = a + c - ac cos () 3 On the Main screen, complete the entry line as: solve(b = cos(76), b) Then press E. 4 Since b represents the side length of a triangle, then b > 0. b = 10.34, correct to decimal places. hapter 5 Trigonometric ratios and their applications 155

21 Note: Once the third side has been found, the sine rule could be used to find other angles if necessary. If three sides of a triangle are known, an angle could be found by transposing the cosine rule to make cos, cos or cos the subject. cos () = b + c a bc cos () = a + c b ac cos () = a + b c ab Worked example 10 Find the smallest angle in the triangle with sides 4 cm, 7 cm and 9 cm. Method 1: Using the rule 1 Draw a labelled diagram of the triangle, call it and fill in the given information. Note: The smallest angle will be opposite the smallest side. heck that one of the criteria for the cosine rule has been satisfied. 3 Write the appropriate cosine rule to find angle. WriTe c = 7 b = 9 4 Substitute the given values into the rearranged = rule Evaluate. = 16 = Transpose the equation to make the subject = cos by taking the inverse cos of both sides. = Round the answer to degrees and minutes. = 5 13 Method : Using a S calculator a = 4 1 Draw a labelled diagram of the triangle and fill in the given information. c = 7 a = 4 eookplus Tutorial int-113 Worked example 10 Let a = 4 b = 7 c = 9 The cosine rule can be used since three side lengths have been given. cos () = b + c a bc b = 9 Write the appropriate cosine rule to find the angle. a = b + c bc cos () 156 maths Quest 11 advanced General mathematics for the asio lasspad

22 3 On the Main screen, complete the entry line as: solve(4 = cos(a), a) 0 a 180 Then press E. 4 Round the answer to degrees and minutes. = = 5 13 Worked example 11 Two rowers set out from the same point. One rows N70 E for 000 m and the other rows S15 W for 1800 m. How far apart are the two rowers? 1 Draw a labelled diagram of the triangle, call it and fill in the given information. WriTe 15 N 000 m 70 eookplus Tutorial int-1047 Worked example m heck that one of the criteria for the cosine rule The cosine rule can be used since two side lengths has been satisfied. and the included angle have been given. 3 Write the appropriate cosine rule to find side c. To find side c: c = a + b ab cos () 4 Substitute the given values into the rule. = cos (15 ) 5 Evaluate. = = c = = Round the answer to decimal places. = nswer the question. The rowers are m apart. rememer 1. In any triangle : a = b + c bc cos () b = a + c ac cos () c = a + b ab cos () hapter 5 Trigonometric ratios and their applications 157

23 . The cosine rule can be used to solve non right-angled triangles if we are given: (a) three sides of the triangle (b) two sides of the triangle and the included angle (that is, the angle between the two given sides). 3. If three sides of a triangle are known, an angle could be found by transposing the cosine rule to make cos, cos or cos the subject. cos () = b + c a bc cos () = a + c b ac cos () = a + b c ab Exercise 5D The cosine rule 1 WE 9 Find the third side of triangle given a = 3.4, b = 7.8 and = 80. In triangle, b = 64.5 cm, c = 38.1 cm and = Find a. 3 In triangle, a = 17, c = 10 and = 115. Find b, and hence find and. 4 WE 10 Find the smallest angle in the triangle with sides 6 cm, 4 cm and 8 cm. 5 In triangle, a = 356, b = 07 and c = 96. Find the largest angle. 6 In triangle, a = 3.6, b = 17.3 and c = 6.4. Find the size of all the angles. 7 In triangle DEF, d = 3 cm, e = 7 cm and F = 60. Find f in exact form. 8 WE 11 Two rowers set out from the same point. One rows N30 E for 1500 m and the other rows S40 E for 100 m. How far apart are the two rowers? 9 Maria cycles 1 km in a direction N68 W, then 7 km in a direction of N34 E. a How far is she from her starting point? b What is the bearing of the starting point from her finishing point? 10 garden bed is in the shape of a triangle, with sides of length 3 m, 4.5 m and 5. m. a alculate the smallest angle. b Hence, find the area of the garden. (Hint: Draw a diagram, with the longest length as the base of the triangle.) 11 hockey goal is 3 m wide. When Sophie is 7 m from one post and 5. m from the other, she shoots for goal. Within what angle, to the nearest degree, must the shot be made if it is to score a goal? 1 n advertising balloon is attached to two ropes 10 m and 100 m long. The ropes are anchored to level ground 35 m apart. How high can the balloon fly? 13 plane flies in a direction of N70 E for 80 km, then on a bearing of S10 W for 150 km. a How far is the plane from its starting point? b What direction is the plane from its starting point? 14 Ship is 16. km from port on a bearing of 053 T and ship is 31.6 km from the same port on a bearing of 117 T. alculate the distance between the two ships. 15 plane takes off at am from an airfield, and flies at 10 km/h on a bearing of N35 W. second plane takes off at am from the same airfield, and flies on a bearing of S80 E at a speed of 90 km/h. How far apart are the planes at 10.5 am? 158 Maths Quest 11 dvanced General Mathematics for the asio lasspad

24 5E 16 Three circles of radii 5 cm, 6 cm and 8 cm are positioned so that 5 cm they just touch one another. Their centres form the vertices of a triangle. 6 cm Find the largest angle in the triangle For the given shape at near right, determine: 150 x a the length of the diagonal 7 b the magnitude (size) of angle 60 8 cm c the length of x 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. How fast does the yacht sail? rea of triangles The area of any triangle is given by the rule rea = 1 bh where b is the base length and h is the perpendicular height of the triangle. h b However, often the perpendicular height is not given directly and needs to be calculated first. In the triangle, b is the base length and h is the perpendicular height of the triangle. Using the trigonometric ratio for sine: sin () = h c c h b a Transposing the equation to make h the subject, we obtain: h = c sin () Therefore, the area of triangle becomes: rea = 1 bc sin () Depending on how the triangle is labelled, the formula could read: rea = 1 ab sin () rea = 1 ac sin () rea = 1 bc sin () 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 1 Find the area of the triangle shown. 7 cm 9 cm 10 1 Draw a labelled diagram of the triangle, call it and fill in the given information. Write/draw c = 7 cm 10 a = 9 cm heck that the criterion for the area rule has been satisfied. Let a = 9 cm, c = 7 cm, = 10 The area rule can be used since two side lengths and the included angle are known. hapter 5 Trigonometric ratios and their applications 159

25 3 Write the appropriate rule for the area. rea = 1 ac sin () 4 Substitute the known values into the rule. = sin (10 ) 5 Evaluate. Round the answer to decimal places = 7.8 cm and include the appropriate unit. Note: If you are not given the included angle, you will need to find it in order to calculate the area. This may involve using either the sine or cosine rule. Worked example 13 triangle has known dimensions of a = 5 cm, b = 7 cm and = 5. Find and and hence the area. 1 Draw a labelled diagram of the triangle, call it and fill in the given information. WriTe 5 a = 5 eookplus Tutorial int-1048 Worked example 13 Let a = 5, b = 7, = 5 heck whether the criterion for the area rule has been satisfied. The area rule cannot be used since the included angle has not been given. 3 Write the sine rule to find. To find angle : a b = sin( ) sin ( ) 4 Substitute the known values into the rule. 5 7 sin( ) sin ( 5 ) 5 Transpose the equation to make sin () the 5 sin (5 ) = 7 sin () subject. 5 sin( 5 ) = sin( ) 7 sin( ) = sin( ) 7 6 Evaluate. = sin 5 sin( 5 ) 1 7 = b = 7 7 Round the answer to degrees and minutes. = Determine the value of the included angle,, using the fact that the angle sum of any triangle is 180. = 180 ( ) = Write the appropriate rule for the area. rea = 1 ab sin () 10 Substitute the known values into the rule. = sin (93 45 ) 11 Evaluate. Round the answer to decimal places = cm. and include the appropriate unit. 160 maths Quest 11 advanced General mathematics for the asio lasspad

26 Heron s formula If we know the lengths of all the sides of the triangle but none of the angles, we could use the cosine rule to find an angle, then use 1 bc sin () to find the area. lternatively, we could use Heron s formula to find the area. Heron s formula states that the area of a triangle is: rea = ss ( a)( s b)( s c) where s is the semi-perimeter of the triangle; that is, s = 1 (a + b + c) The proof of this formula is beyond the scope of this course. Worked Example 14 Find the area of the triangle with sides of 4 cm, 6 cm and 8 cm. 1 Draw a labelled diagram of the triangle, call it and fill in the given information. Write 4 cm 6 cm 8 cm Let a = 4, b = 6, c = 8 Determine which area rule will be used. Since three side lengths have been given, use Heron s formula. 3 Write the rule for Heron s formula. rea = ss ( a)( s b)( s c) 4 Write the rule for s, the semi-perimeter of the triangle. 5 Substitute the given values into the rule for the semi-perimeter. 6 Substitute all of the known values into Heron s formula. s = 1 (a + b + c) = 1 ( ) = 1 (18) = 9 rea = 99 ( 4)( 9 6)( 9 8) 7 Evaluate. = Round the answer to decimal places and include the appropriate unit. = 135 = = 11.6 cm REMEMER 1. If two sides of any triangle and the included angle (that is, the angle between the two given sides) are known, the following rules may be used to determine the area of that triangle. rea = 1 ab sin () rea = 1 ac sin () rea = 1 bc sin () hapter 5 Trigonometric ratios and their applications 161

27 . lternatively, if the lengths of three sides of a triangle are known, Heron s formula may be used to find the area of the triangle: rea = ss ( a)( s b)( s c) where s is the semi-perimeter of the triangle; that is, s = 1 (a + b + c) Exercise 5E rea of triangles 1 WE 1 Find the area of the triangle with a = 7 cm, b = 4 cm and = 68. Find the area of the triangle with a = 7.3 cm, c = 10.8 cm and = Find the area of the triangle with b = 3.1 m, c = 18.6 m and = Find the exact area of the triangle DEF with d = 6, e = 9 and F = Find the exact area of the triangle QPR with p = 1, r = 10 and Q = WE 13 M In a triangle, a = 15 m, b = 0 m and = 50. The area of the triangle is: 86. m m m D 17.4 m E m 7 WE 14 Find the area of the triangle with sides of 5 cm, 6 cm and 8 cm. 8 Find the area of the triangle with sides of 40 mm, 30 mm and 5.7 cm. 9 Find the area of the triangle with sides of 16 mm, 3 cm and.7 cm. 10 Find the area of the equilateral triangle with sides 4 cm. Leave your answer in simplified surd form. 11 M triangle has sides of length 10 cm, 14 cm and 0 cm. The area of the triangle is: 41 cm 65 cm 106 cm D 137 cm E 1038 cm 1 triangle has a = 10 cm, c = 14 cm and = 48. Find and and hence the area. 13 triangle has a = 17 m, c = m and = 56. Find and and hence the area. 14 triangle has b = 3 mm, c = 15 mm and = 38. Find and and hence the area. 15 piece of metal is in the shape of a triangle with sides of length 114 mm, 7 mm and 87 mm. Find its area using Heron s formula. 16 triangle has the largest angle of 115. The longest side is 6 cm and another side is 35 cm. Find the area of the triangle. 17 triangle has two sides of 5 cm and 30 cm. The angle between the two sides is 30. Find: a its area b the length of its third side c its area using Heron s formula. 18 The surface of a fish pond has the shape shown in the diagram at right. 1 m How many goldfish can the pond support if each fish requires 0.3 m surface area of water? m 5 m 19 Find the area of this quadrilateral. 3.5 m 4 m 4 m 5 m 60 8 m 16 Maths Quest 11 dvanced General Mathematics for the asio lasspad

28 0 parallelogram has diagonals of length 10 cm and 17 cm. n angle between them is 15. Find: a the area of the parallelogram b the dimensions of the parallelogram. 1 lawn is to be made in the shape of a triangle, with sides of length 11 m, 15 m and 17. m. How much grass seed, to the nearest kilogram, is needed if it is sown at the rate of 1 kg per 5 m? bushfire burns out an area of level grassland shown in the diagram. What is the area, in hectares, of the land that is burnt? 1.8 km 400 m 00 m Road km River eookplus Digital doc WorkSHEET 5. 5F 3 n earth embankment is 7 m long, and has a cross-section 100 shown in the diagram. Find the volume of earth needed to build the embankment. m m 4 m parallelogram has sides of 14 cm and 18 cm, and an angle between them of 7. The area of the parallelogram is: 86. cm cm 17.4 cm D 39.7 cm E 5 cm 5 m n 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 1 m per litre, the amount of paint needed, to the nearest litre, would be: 9 L 18 L 4 L D 36 L E 4 L Trigonometric identities n identity is a relationship that holds true for all values of a pronumeral or pronumerals. The sine and cosine functions are related functions and the following identities exist between them. The Pythagorean identity of a triangle within the unit circle. We know that the hypotenuse is 1 unit. sin () = a 1 cos () = b 1 a = sin () b = cos () y 0 1 b a x hapter 5 Trigonometric ratios and their applications 163

29 So the triangle formed has a height of sin ( ) and a base length of cos ( ). Pythagoras theorem then tells us that a + b = 1 sin ( ) + cos () = 1 Note: sin ( ) = (sin ( )) and cos ( ) = (cos ( )) The Pythagorean identity is sin ( ) + cos ( ) = 1. Worked Example 15 Find the value of sin () given cos () = 5 and 0 < < Write 1 Write the Pythagorean identity. sin ( ) + cos ( ) = 1 Substitute the known value cos ( ) = sin ( ) = 1 3 Solve to find the required value. sin ( ) = 1 sin () = sin () = ± Write the final answer. s is in the first quadrant sin () = omplementary angles In the diagram at right we can see that: cos ( ) = b c sin ( ) = a c a α c cos (α) = a sin (α) = b c c So for our diagram cos ( ) = sin (α) and sin ( ) = cos (α ). b We also know that + α = 90, so α = y substituting this into cos () = sin (α) and cos (α) = sin ( ) we get cos ( ) = sin (90 - ) and sin ( ) = cos (90 - ). Worked Example 16 Find the value of cos (70 ) given sin (0 ) = Write the equation with the required complementary angle formula. Write cos ( ) = sin (90 - ) Identify the value of. = 70 3 Substitute the angle into the equation and simplify. cos (70 ) = sin (90-70 ) = sin (0 ) = Maths Quest 11 dvanced General Mathematics for the asio lasspad

30 REMEMER 1. The Pythagorean identity is sin () + cos () = 1.. cos ( ) = sin (90 - ) and sin () = cos (90 - ). Exercise 5F Trigonometric identities 1 WE 15 Find the value of sin ( ) given cos ( ) = 4 and 0 < < Find the value of cos () given sin ( ) = 1 and 0 < < Find the value of cos ( ) given sin () = 6 and 0 < < Find the value of sin ( ) given cos () = and 0 < < Use your knowledge of exact values to show that the Pythagorean identity is true for = WE 16 Find the value of sin (1 ) given cos (78 ) = G 7 Find the value of cos (4 ) given sin (48 ) = Radian measurement In all of the trigonometry tasks covered so far, the unit for measuring angles has been the degree. There is another commonly used measurement for angles, the radian. This is used in situations involving length and areas associated with circles. onsider the unit circle, a circle with a radius of 1 unit. OP is the radius. If OP is rotated anticlockwise, the point P traces a path along the circumference of the circle to a new point, P 1. The arc length PP 1 is a radian measurement, symbolised by c. Note: 1 c is equivalent to the angle in degrees formed when the length of PP 1 is 1 unit; in other words, when the arc is the same length as the radius. If the length OP is rotated 180, the point P traces out half the circumference. Since the circle has a radius of 1 unit, and = πr, the arc PP 1 has a length of π. The relationship between degrees and radians is thus established. 180 = π c This relationship will be used to convert from one system to another. Rearranging the basic conversion factor gives: 180 = π π 1 = 180 π To convert an angle in degrees to radian measure, multiply by 180. lso, since π = 180, it follows that 1 c = 180 π. To convert an angle in radian measure to degrees, multiply by 180 π. O O P 1 1 P OP = 1 unit P 1 c P OP = 1 unit circumference 180 O P hapter 5 Trigonometric ratios and their applications 165

31 Where possible, it is common to have radian values with π in them. It is usual to write radians without any symbol, but degrees must always have a symbol. For example, an angle of 5 must have the degree symbol written, but an angle of 1.5 is understood to be 1.5 radians. Worked example 17 onvert 135 to radian measure, expressing the answer in terms of π. WriTe 1 To convert an angle in degrees to radian measure, multiply the angle by π 135 = = 135π 180 Simplify, leaving the answer in terms of π. = 3 π 4 π 180 Worked example 18 onvert the radian measurement 4 π to degrees. 5 1 To convert radian measure to an angle in degrees, multiply the angle by 180 π. Simplify. Note: The π cancels out. WriTe 4π 4π 180 = 5 5 π = 70 5 = 144 If the calculation does not simplify easily, write the answers in degrees and minutes, or radians to 4 decimal places. If angles are given in degrees and minutes, convert to degrees only before converting to radians. rememer = π c π. To convert an angle in degrees to radian measure, multiply by To convert an angle in radian measure to degrees, multiply by 180 π. exerise 5G radian measurement 1 We 17 onvert the following angles to radian measure, expressing answers in terms of π. a 30 b 60 c 10 d 150 e 5 f 70 g 315 h 480 i 7 j 00 eookplus Digital doc SkillSHEET 5.6 hanging degrees to radians 166 maths Quest 11 advanced General mathematics for the asio lasspad

32 We 18 onvert the following radian measurements into degrees. a π 4 b 3 π c 7 π 6 d 5 π 3 e 7 π 1 f 17π 6 g π 1 h 13 π 10 i 11π 8 j 8π 5h 3 onvert the following angles in degrees to radians, giving answers to 4 decimal places. a 7 b 109 c 43 d 351 e 7 f 63 4 g h 74 8 i j 47 4 onvert the following radian measurements into degrees and minutes. a.345 b c 1 d 1.61 e 3.59 f 7.5 g 0.18 h i j arcs, sectors and segments arc length n arc is a section of the circumference of a circle. The length of the arc is proportional to the angle subtended at the centre. For example, an angle of 90 will create an arc which is 1 the circumference. 4 We have already defined an arc length as equivalent to radians if the circle has a radius of 1 unit. eookplus Interactivity int-097 Sectors r = 1 c Therefore, a simple dilation of the unit circle will enable us to calculate the arc length for any sized circle, as long as the angle is expressed in radians. r r c If the radius is dilated by a factor of r, the arc length is also dilated by a factor of r. Dilation by factor of r Therefore, l = r, where l represents the arc length, r represents the radius and represents an angle measured in radians. Worked example 19 Find the length of the arc which subtends an angle of 75 at the centre of a circle with radius 8 cm. WriTe/draW 1 Draw a diagram representing the situation and l = r label with the given values. 75 r = 8 onvert the angle from 75 to radian measure by π π multiplying the angle by = = 75 π 180 hapter 5 Trigonometric ratios and their applications 167

33 3 Evaluate to 4 decimal places. = Write the rule for the length of the arc. l = r 5 Substitute the values into the formula. = Evaluate to decimal places and include the appropriate unit. = = cm Note: In order to use the formula for the length of the arc, the angle must be in radian measure. Worked Example 0 Find the angle subtended by a 17 cm arc in a circle of radius 14 cm: a in radians b in degrees. Write a 1 Write the rule for the length of the arc. a l = r Substitute the values into the formula. 17 = 14 3 Transpose the equation to make the subject. 4 Evaluate to 4 decimal places and include the appropriate unit. b 1 To convert radian measure to an angle in degrees, multiply the angle by 180 π. = = = c b c = π Evaluate. = onvert the angle to degrees and minutes. = rea of a sector In the diagram at right, the shaded area is the minor sector O, and the unshaded area is the major sector O. The area of the sector is proportional to the arc length. For example, an area of 1 of the circle contains an arc which is 1 of the 4 4 circumference. Thus, in any circle: area of sector arc length = area of circle circumference of circle O Major sector Minor sector r = where is measured in radians. πr πr = r π r πr = 1 r The area of a sector is: = 1 r 168 Maths Quest 11 dvanced General Mathematics for the asio lasspad

34 Worked example 1 sector has an area of 157 cm, and subtends an angle of 107. What is the radius of the circle? WriTe π 1 onvert the angle from 107 to radian measure by 107 = 107 π 180 multiplying the angle by 180. = 107 π 180 Evaluate to 4 decimal places. = Write the rule for the area of a sector. = 1 r 4 Substitute the values into the formula. 157 = 1 r Transpose the equation to make r the subject. 157 = r r = Take the square root of both sides of the equation. r = Evaluate to decimal places and include the = 1.97 cm appropriate unit. area of a segment segment is that part of a sector bounded by the arc and the chord. s can be seen from the diagram at right: rea of segment = area of sector area of triangle = 1 r 1 r sin ( ) = 1 r ( sin ( )) Note: is in radians and is in degrees. r Segment The area of a segment: = 1 r ( sin ( )) Worked example Find the area of the segment in a circle of radius 5 cm, subtended by an angle of 40. WriTe π 1 onvert the angle from 40 to radian measure 40 = π by multiplying the angle by 180. = 40 π 180 Evaluate to 4 decimal places. = Write the rule for the area of a segment. = 1 r ( sin ( )) 4 Identify each of the variables. r = 5, = , = 40 5 Substitute the values into the formula. = 1 5 ( sin (40 )) 6 Evaluate. = = Round to decimal places and include the appropriate unit. = 0.69 cm eookplus Tutorial int-1049 Worked example hapter 5 Trigonometric ratios and their applications 169

35 rememer 1. rc length: l = r. rea of a sector: = 1 r 3. rea of a segment: = 1 r ( sin (( )) where r = radius, = angle (measured in radians) and = angle (measured in degrees). exerise 5h arcs, sectors and segments 1 We 19 Find the length of the arc which subtends an angle of 65 at the centre of a circle of radius 14 cm. Find the length of the arc which subtends an angle of 153 at the centre of a circle of radius 75 mm. 3 Find the length of the arc which subtends an angle of 135 at the centre of a circle of radius 10 cm. Leave answer in terms of π. 4 n arc of a circle is 3.5 cm long, and subtends an angle of 41 at the centre of the circle. What is the radius of the circle? 5 n arc of a circle is 7.8 cm long, and subtends an angle of 05 at the centre of the circle. What is the radius of the circle? 6 n arc of a circle is 4 cm long and subtends an angle of 60 at the centre of the circle. What is the radius of the circle? Write your answer in terms of π. 7 We 0 Find the angle subtended by a 0 cm arc in a circle of radius 75 cm: a in radians b in degrees. 8 Find the angle subtended by an 8 cm arc in a circle of radius 5 cm: a in radians b in degrees. 9 n arc of length 8 cm is marked out on the circumference of a circle of radius 13 cm. What angle does the arc subtend at the centre of the circle? 10 n arc of length 45 mm is marked out on the circumference of a circle of radius 18 cm. Find the angle that the arc subtends at the centre of the circle. 11 The minute hand of a clock is 35 cm long. How far does the tip of the hand travel in 0 minutes? 1 child s swing is suspended by a rope 3 m long. What is the length of the arc it travels if it swings through an angle of 4? 13 Find the area of the sector of a circle of radius 17 cm with an angle of Find the area of the sector of a circle of radius 6. cm with an angle of Find the area of a sector of a circle of radius 6 cm with an angle of 100. Write your answer in terms of π. 16 We1 sector has an area of 85 cm, and subtends an angle of 70. What is the radius of the circle? 17 sector with an area of 309 cm is part of a circle of radius 18. cm. Find the angle in the sector. 170 maths Quest 11 advanced General mathematics for the asio lasspad

36 18 Find the area of a sector of a circle of radius 30 cm if the sector has an arc length of 18 cm. 19 garden bed is in the form of a sector of a circle of radius 4 m. The arc of the sector is 5 m long. Find: a the area of the garden bed b the volume of mulch needed to cover the bed to a depth of 10 cm. 0 The minute hand on a clock is 6 cm long. What area does the hand sweep through in 40 minutes? 1 sector whose angle is 150 is cut from a circular piece of cardboard whose radius is 1 cm. The two straight edges of the sector are joined so as to form a cone. a What is the surface area of the cone? b What is the radius of the cone? WE Find the area of the segment in a circle of radius 5 cm subtended by an angle of Find the area of the segment of a circle of radius 4.7 m that subtends an angle of 85 0 at the centre. 4 segment of a circle subtends an angle of 75 at the centre. The area of the segment is 100 cm. Find the radius of the circle. 5 In a circle of radius 15 cm, a sector has an area of 100 cm. Find the angle subtended by the sector. 6 Two circles of radii 3 cm and 4 cm have their centres 5 cm apart. Find the area of the intersection of the two circles. 7 M The angle subtended by a 8 cm arc in a circle of radius 0 cm in radians is: D 1.4 E M The area of the segment in a circle of radius 1 cm, subtended by an angle of 60 is: 6.5 cm cm 6.08 cm D 15.4 cm E cm 9 Two irrigation sprinklers spread water in circular paths with radii of 7 m and 4 m. If the sprinklers are 10 m apart, find the area of crop that receives water from both sprinklers. 30 M The length of the arc which subtends an angle of 50 at the centre of a circle with radius 10 cm is: 8.73 cm 0.87 cm cm D 6.5 cm E 0.63 cm hapter 5 Trigonometric ratios and their applications 171

37 Summary Trigonometry of right-angled triangles For any right-angled triangle: sin ( ) = O cos () = tan () = O H H Pythagoras theorem, c = a + b may also be used to solve right-angled triangles. a c Opposite (O) Hypotenuse (H) () djacent b ngles of 30, 45 and 60 have exact values of sine, cosine and tangent sin ( ) 1 1 = 3 cos ( ) tan ( ) = 3 = Elevation, depression and bearings ngles of elevation and depression are each measured from the horizontal. The angle of elevation is equal to the angle of depression since they are alternate Z angles. True bearings are measured in a clockwise direction, starting from north (0 T). The sine rule The sine rule states that for any triangle : a b c = = sin( ) sin ( ) sin( ) When using this rule, it is important to note that, depending on the values given, any combination of the two equalities may be used to solve a particular triangle. The sine rule may be used to solve non right-angled triangles if we are given: (a) two angles and one side length (b) two side lengths and an angle opposite one of these side lengths. The ambiguous case exists if is an acute angle and a > c > a sin (). The cosine rule In any triangle : a = b + c bc cos () b = a + c ac cos () c = a + b ab cos () The cosine rule can be used to solve non right-angled triangles if we are given: (a) three sides of the triangle (b) two sides of the triangle and the included angle (that is, the angle between the two given sides). 17 Maths Quest 11 dvanced General Mathematics for the asio lasspad

38 If three sides of a triangle are known, an angle could be found by transposing the cosine rule to make cos, cos or cos the subject. cos () = b + c a bc cos () = a + c b ac cos () = a + b c ab rea of triangles If two sides of any triangle and the included angle (that is, the angle between the two given sides) are known, the following rules may be used to determine the area of that triangle. rea = 1 ab sin () rea = 1 ac sin () rea = 1 bc sin () lternatively, if three side lengths of a triangle are known, Heron s formula may be used to find the area of a triangle: rea = ss ( a)( s b)( s c) where s is the semi-perimeter of the triangle; that is, Trigonometric identities s = 1 (a + b + c) n identity is a relationship that holds true for all values of a pronumeral or pronumerals. The Pythagorean Identity states that sin () + cos () = 1. Sine and cosine are called complementary functions since: cos () = sin (90 ) and sin () = cos (90 ) Radian measurement 180 = π c π To convert an angle in degrees to radian measure, multiply by To convert an angle in radian measure to degrees, multiply by π. rcs, sectors and segments rc length: l = r rea of a sector: = 1 r rea of a segment: = 1 r ( sin ( )) Where r = radius, = angle (measured in radians) and = angle (measured in degrees). hapter 5 Trigonometric ratios and their applications 173

39 chapter review Short answer 1 stepladder stands on a floor with its feet m apart. If the angle formed by the legs with the floor is 60, how high above the floor is the top of the ladder? Two buildings, 15 m and 7 m high, are directly opposite each other across a river. The angle of depression of the top of the smaller building from the top of the taller one is 30. How wide is the river? 3 In the triangle shown at right x find the exact length of side m. 1 cm triangle has sides of length 1 m, 15 m and 0 m. If Q is the largest angle find cos (q). 5 triangle has two sides of 0 cm and 5 cm. The angle between the two sides is 45. Find its area. 6 triangular garden area is bound by three straight edges of lengths 4 m, 5 m and 7 m. Find the exact area of the garden. 7 Find the value of cos () given sin () = 3 8 and 0 < < a onvert the following angles to radian measure, expressing answers in terms of π. i 80 ii 15 iii 640 b onvert the following radian measurement into degrees. i π 0 ii 15 π 8 iii 7π 9 paddock is in the shape of a sector with radius of 75 m and an angle of 60. Find: a the amount of fencing needed to enclose the paddock b the area enclosed by the paddock. Multiple choice 1 In the triangle, the value of, to the nearest degree, is: D 5 E ladder 4.5 m long rests against a vertical wall, with the foot of the ladder m from the base of the wall. The angle the ladder makes with the wall, to the nearest degree, is: D 64 E 66 3 person stands 18 m from the base of a building, and measures the angle of elevation to the top of the building as 6. If the person is 1.8 m tall, how high is the building, to the nearest metre? 11 m 18 m 36 m D m E 34 m 4 bearing of 310 T is the same as: N40 W N50 W S50 W D S50 E E N50 E 5 In triangle, a = 10, b = 7 and = 40. possible value for, to the nearest degree, is: D 73 E Two boats start from the same point. One sails due north for 10 km and the other sails south east for 15 km. Their distance apart is: 10.6 km km km D 1.38 km E 3.18 km 7 triangle has sides measuring 5 cm, 8 cm and 10 cm. The largest angle in the triangle, to the nearest degree, is: D 18 E The area of the triangle with a = 10 m, b = 8 m and = 7 is: 1.36 m m m D m E m 9 garden bed is in the shape of a triangle, with sides of length 4 m, 5. m and 7 m. The volume of topsoil needed to cover the garden to a depth of 50 mm is:.3 m 3.57 m 3.81 m 3 D 3.17 m 3 E 3.76 m 3 10 When 75 is converted to radian measure, the value of the angle, expressed in terms of π, is: 1 π π 5 π D 5 π 1 E 7π Maths Quest 11 dvanced General Mathematics for the asio lasspad

40 11 When 5.31 is converted to degrees and minutes, the value of the angle is: D 1 4 E n arc in a circle of radius 5 cm is 3.5 cm long. The angle, to the nearest degree, subtended at the centre by the arc is: D 68 E 8 13 sector has an area of 40 cm, and an angle of 30. The arc length of the sector, to decimal places, is: 1.64 cm.66 cm 4.83 cm D 6.47 cm E 1.36 cm 14 The area of the shaded region in the figure at right to the nearest cm is: 800 cm 846 cm 898 cm D 95 cm E 983 cm 40 cm clock has a minute hand 75 cm long. The area that it sweeps when passing through 48 minutes, to decimal places, is: 0.90 m 1.35 m 1.41 m D 1.88 m E.01 m extended response 1 Three circles of radii cm, 3 cm and 4 cm are placed so that they just touch each other. triangle is formed by joining their three centres. Find: a the three angles of the triangle b the area of the triangle, correct to 3 decimal places c the shaded area correct to 3 decimal places. farmer owns a large triangular area of flat land, bounded on one side by an embankment to a river flowing NE, on a second side by a road which meets the river at a bridge where the angle between river and road is 105, and on the third side by a long fence. Find: a the length of the river frontage, correct to 3 decimal places b the area of the land correct to 3 decimal places. The farmer decides to divide the land into two sections of equal area, by running a fence from the bridge to a point on the opposite side. c On what bearing must the fence be built? d What is the length of the fence, correct to 3 decimal places? N River Fence 105 Road 4 3. km hapter 5 Trigonometric ratios and their applications 175

41 3 a four-wheel-drive vehicle leaves a camp site and travels across a flat sandy plain in a direction of S65 E, for a distance of 8. km. It then heads due south for 6.7 km to reach a waterhole. i How far is the waterhole from the camp site? ii What is the bearing of the waterhole from the camp site? b search plane sets off to find the vehicle. It is on a course that takes it over points and, two locations on level ground. t a certain time, from point, the angle of elevation to the plane is 7. From point, the angle of elevation is 47. If and are 3500 m apart, find the height of the plane off the ground. 4 hristopher lives on a farm. He has decided that this year he will plant a variety of crops in his large but unusually 14 m shaped vegetable garden. He has divided the vegetable garden 56 m into six triangular regions, which he will fence off as shown in 1 the diagram at right. hristopher needs to calculate the 5 m perimeter and area of each region so he can purchase the 95º 64º 38º correct amount of fencing material and seedlings. 80º 6 58º a Separate each of the regions into single triangles and label 3 85 m each with the information provided. 68 m b Use the appropriate rules to determine all unknown 5 43 m lengths and relevant angles. F c How much fencing material is required to section off the 4 six regions? d If fencing material is $4.50 per metre (and only sold by D the metre) what will the cost be? E e alculate the area of each region and hence determine the total area available for planting. eook plus Digital doc Test Yourself hapter maths Quest 11 advanced General mathematics for the asio lasspad

42 eookplus ativities hapter opener Digital doc 10 Quick Questions: Warm up with ten quick questions on trigonometric ratios and their applications. (page 136) 5 Trigonometry of right-angled triangles Digital docs SkillSHEET 5.1: Practise labelling right-angled triangles. (page 141) SkillSHEET 5.: Practise using trigonometric ratios. (page 141) SkillSHEET 5.3: Practise degrees and minutes. (page 141) SkillSHEET 5.4: Practise composite shapes 1. (page 14) SkillSHEET 5.5: Practise composite shapes. (page 14) 5 Elevation, depression and bearings Tutorial We6 int-1045: Watch how to determine the bearing of a ship from its starting point. (page 145) 5 The sine rule Tutorial We8 int-1046: Watch how to show the ambiguous case of the sine rule exists and apply it. (page 150) Digital doc WorkSHEET 5.1: Use trigonometry to find two unknowns in right-angled triangles; solve worded problems of elevation, depression and bearings. (page 154) 5D The cosine rule Tutorials We 10 int-113: Watch how to find the smallest angle in a triangle. (page 156) We 11 int-1047: Watch how to calculate the distance between two rowers. (page 157) 5E rea of triangles Tutorial We 13 int-1048: Watch how to find the area of a triangle given two side lengths and an angle. (page 160) Digital doc WorkSHEET 5.: Solve more complex right-angled triangle problems with two unknowns, worded problems of elevation and depression and apply the sine and cosine rules to non-right angled triangles. (page 163) 5G Radian measurement Digital doc SkillSHEET 5.6: Practise changing degrees to radians. (page 166) 5H rcs, sectors and segments Interactivity Sectors int-097: onsolidate your understanding of how to calculate the area of a sector. (page 167) Tutorial We int-1049: Watch how to find the area of a segment. (page 169) hapter review Digital doc Test Yourself: Take the end-of-chapter test to test your progress. (page 176) To access eookplus activities, log on to hapter 5 Trigonometric ratios and their applications 177