4 Trigonometr MEP Pupil Tet 4 4.1 Squares and Triangles triangle is a geometric shape with three sides and three angles. Some of the different tpes of triangles are described in this Unit. square is a four-sided geometric shape with all sides of equal length. ll four angles are the same size. n isosceles triangle has two sides that are the same length, and the two base angles are equal. 60 ll the sides of an equilateral triangle are the same length, and the angles are all 60. 60 60 scalene triangle has sides that all have different lengths, and has 3 different angles. Worked Eample 1 Find the area of the square shown in the diagram. rea = 4 4 = 16 cm 4 cm Eercises 1. For each triangle below, state whether it is scalene, isosceles or equilateral. (a) 4 cm 6 cm 6 cm 7 cm 6 cm 4 cm 8 cm 116
MEP Pupil Tet 4 (c) cm cm (d) cm 40 40 (e) 30 (f) 60 60 60 60 (g) (h) 3 cm 5 cm 70 70 3 cm. (a) When a square is cut in half diagonall, two triangles are obtained. re these triangles scalene, isosceles or equilateral? What tpe of triangle do ou get if ou cut a rectangle in half diagonall? 3. Find the area of each square below. (a) 5 cm 7 cm 5 cm 7 cm (c) (d) 10 m 1 cm 1 cm 10 m 4. Find the areas of the squares with sides of length: (a) m 100 m (c) 15 cm (d) 17 cm 5. Find the lengths of the sides of a square that has an area of: (a) 9 cm 5 m (c) 100 m (d) 64 cm (e) 1 cm (f) 400 m 117
MEP Pupil Tet 4 6. Two squares of side 4 cm are joined together to form a rectangle. What is the area of the rectangle? 7. square of side 1 cm is cut in half to form two triangles. What is the area of each triangle? 8. square of side 6 cm is cut into quarters to form 4 smaller squares. What is the area of each of these squares? 4. Pthagoras' Theorem Pthagoras' Theorem gives a relationship between the lengths of the sides of a right angled triangle. For the triangle shown, a = b + c b a Note The longest side of a right angled triangle is called the hpotenuse. c Worked Eample 1 Find the length of the hpotenuse of the triangle shown in the diagram. Give our answer correct to decimal places. 4 cm 6 cm s this is a right angled triangle, Pthagoras' Theorem can be used. If the length of the hpotenuse is a, then b = 4 and c = 6. So a = b + c a = 4 + 6 a = 16 + 36 a = 5 a = 5 = 7. cm (to one decimal place) Worked Eample Find the length of the side of the triangle marked in the diagram. s this is a right angled triangle, Pthagoras' Theorem can be used. Here the length of the hpotenuse is 6 cm, so a = 6 cm and c = 3 cm with b =. 3 cm 6 cm 118
MEP Pupil Tet 4 So a = b + c 6 = + 3 Eercises 36 = + 9 36 9 = 36 9 = 7 = 7 = = 5. cm (to one decimal place) 1. Find the length of the side marked in each triangle. (a) 4 m 5 cm 3 m 1 cm (c) (d) 15 cm 10 m 6 m 1 cm (e) (f) 6 cm 6 m 8 m 10 cm (g) 0 m 5 m (h) 36 cm 15 cm 119
MEP Pupil Tet 4. Find the length of the side marked in each triangle. Give our answers correct to decimal places. (a) 7 cm 15 cm 11 cm 14 cm (c) 4 cm (d) 7 m 8 cm 5 m (e) (f) 8 cm 10 cm 1 cm 7 cm (g) 5 m (h) m 6 m 1 m (i) (j) 18 cm 10 cm 5 m 4 m (k) (l) 3.3 m 9.4 m 4.6 m 7.8 m 10
MEP Pupil Tet 4 (m) 8.9 cm (n) 5. cm.3 m 5.4 m 3. dam runs diagonall across a school field, while en runs around the edge. (a) How far does en run? 00 m dam en How far does dam run? (c) How much further does en run than dam? 10 m 4. gu rope is attached to the top of a tent pole, at a height of 1.5 metres above the ground, and to a tent peg metres from the base of the pole. How long is the gu rope? Peg Gu rope m Pole 1.5 m 5. Farida is 1.4 metres tall. t a certain time her shadow is metres long. What is the distance from the top of her head to the top of her shadow? 6. rope of length 10 metres is stretched from the top of a pole 3 metres high until it reaches ground level. How far is the end of the line from the base of the pole? 7. rope is fied between two trees that are 10 metres apart. When a child hangs on to the centre of the rope, it sags so that the centre is metres below the level of the ends. Find the length of the rope. m 10 m 3 m 8. The roof on a house that is 6 metres wide peaks at a height of 3 metres above the top of the walls. Find the length of the sloping side of the roof. 6 m 11
MEP Pupil Tet 4 9. The picture shows a garden shed. Find the length,, of the roof..5 m m 10. Miles walks 3 km east and then 10 km north. m (a) How far is he from his starting point? He then walks east until he is 0 km from his starting point. How much further east has he walked? 11. li is building a shed. It should be rectangular with sides of length 3 metres and 6 metres. He measures the diagonal of the base of the shed before he starts to put up the walls. How long should the diagonal be? P Q 1. Pauline is building a greenhouse. The base PQRS of the greenhouse should be a rectangle measuring.6 metres b 1.4 m 1.4 metres. To check the base is rectangular Pauline has to measure the diagonal PR. S.6 m R (a) alculate the length of PR when the base is rectangular. You must show all our working. When building the greenhouse Pauline finds angle PSR > 90. She measures PR. Which of the following statements is true? X: PR is greater than it should be. Y: PR is less than it should be. Z: PR is the right length. (SEG) 4.3 Further Work with Pthagoras' Theorem Worked Eample 1 Find the length of the side marked in the diagram. cm 4 cm 4 cm 1
MEP Pupil Tet 4 First consider the lower triangle. The unknown length of the hpotenuse has been marked. = 4 + 4 4 cm = 16 + 16 = 3 cm 4 cm The upper triangle can now be considered, using the value for. From the triangle, = +, and using = 3 gives cm = 3 + 4 = 36 = 36 = 6 cm Note When finding the side, it is not necessar to find 3, but to simpl use = 3. Worked Eample Find the value of as shown on the diagram, and state the lengths of the two unknown sides. Using Pthagoras' Theorem gives Eercises 13 3 = ( ) + ( ) ( ) = = ) 169 = 4 + 9 (since 4 169 = 13 13 = = 13 = 361. cm (to decimal places) 1. Find the length of the side marked in each diagram. (a) 4 7 13 m 3 8 5 6 3 13
MEP Pupil Tet 4 (c) (d) 4 4 6 10 1 (e) 3 (f) 4 10 15. Find the length of the side marked in the following situations. (a) 3 0 8 4 (c) (d) 5 4 4 0 5 3. Which of the following triangles are right angled triangles? (a) (c) (d) 9 10 6 13 15 6.5 6 10 D 130 4.5 40 4. ladder of length 4 metres leans against a vertical wall. The foot of the ladder is metres from the wall. plank that has a length of 5 metres rests on the ladder, so that one end is halfwa up the ladder. (a) How high is the top of the ladder? How high is the top of the plank? (c) How far is the bottom of the plank from the wall? m 14
MEP Pupil Tet 4 5. The diagram shows how the sign that hangs over a Fish and hip shop is suspended b a rope and a triangular metal bracket. Find the length of the rope. 160 cm Rope Wall 90 cm FISH ND HIPS 50 cm Metal racket 6. The diagram shows how a cable is attached to the mast of a sailing ding. bar pushes the cable out awa from the mast. Find the total length of the cable. 80 cm Mast 40 cm ar 40 cm able 80 cm Deck 7. helicopter flies in a straight line until it reaches a point 0 km east and 15 km north of its starting point. It then turns through 90 and travels a further 10 km. (a) How far is the helicopter from its starting point? If the helicopter turned 90 the other wa, how far would it end up from its starting point? 8. cone is placed on a wedge. The dimensions of the wedge are shown in the diagram. The cone has a slant height of 30 cm. Find the height of the cone. 4 cm 30 cm 0 cm 15
MEP Pupil Tet 4 9. simple crane is to be constructed using an isosceles triangular metal frame. The top of the frame is to be 10 metres above ground level and 5 metres awa from the base of the crane, as shown in the diagram. Find the length of each side of the triangle. 5 m 10 m 10. thin steel tower is supported on one side b two cables. Find the height of the tower and the length of the longer cable. 10 m 40 m 5 m 11. n isosceles triangle has two sides of length 8 cm and one of length 4 cm. Find the height of the triangle and its area. height 1. Find the area of each the equilateral triangles that have sides of lengths (a) 8 cm 0 cm (c) cm 4.4 Sine, osine and Tangent When working in a right angled triangle, the longest side is known as the hpotenuse. The other two sides are known as the opposite and the adjacent. The adjacent is the side net to a marked angle, and the opposite side is opposite this angle. Opposite Hpotenuse For a right angled triangle, the sine, cosine and tangent of the angle are defined as: djacent sin = opposite hpotenuse cos = adjacent hpotenuse tan = opposite adjacent Sin will alwas have the same value for an particular angle, regardless of the size of the triangle. The same is true for cos and tan. 16
MEP Pupil Tet 4 Worked Eample 1 For the triangle shown, state which sides are: (a) the hpotenuse the adjacent (c) the opposite (a) (c) The hpotenuse is the longest side, which for this triangle is. The adjacent is the side that is net to the angle, which for this triangle is. The opposite side is the side that is opposite the angle, which for this triangle is. Worked Eample Write down the values of sin, cos and tan for the triangle shown. Then use a calculator to find the angle in each case. First, opposite = 8 adjacent = 6 hpotenuse = 10 8 6 10 opposite adjacent opposite sin = cos = tan = hpotenuse hpotenuse adjacent 8 6 8 = = = 10 10 6 4 = 08. = 0.6 = 3 Using a calculator gives = 53. 1 (correct to 1 decimal place) in each case. Eercises 1. For each triangle, state which side is the hpotenuse, the adjacent and the opposite. (a) (c) D H G I E F 17
MEP Pupil Tet 4 (d) J K (e) M (f) Q P L O N R. For each triangle, write sin, cos and tan as fractions. (a) (c) 3 5 5 13 8 15 4 1 17 1.5 (d) (e) (f).5 50 48 1 1.5 3.5 14 3. Use a calculator to find the following. Give our answers correct to 3 decimal places. (a) sin 30 tan 75 (c) tan 5. 6 (d) cos 66 (e) tan 33 (f) tan 45 (g) tan 37 (h) sin 88. (i) cos 45 (j) cos 48 (k) cos 46. 7 (l) sin 45 4. Use a calculator to find in each case. Give our answers correct to 1 decimal place. (a) cos = 05. sin = 1 (c) tan = 045. (d) sin = 0. 81 (e) sin = 075. (f) cos = 09. (g) tan = 1 (h) sin = 05. (i) tan = (j) cos = 014. (k) sin = 06. (l) tan = 55. 5. (a) Draw a triangle with an angle of 50 as shown in the diagram, and measure the length of each side. 18 50
MEP Pupil Tet 4 Using sin = opposite hpotenuse cos = adjacent hpotenuse tan = opposite adjacent and the lengths of the sides of our triangle, find sin 50, cos50 and tan 50. (c) Use our calculator to find sin 50, cos50 and tan 50. (d) ompare our results to and (c). 6. For the triangle shown, write down epressions for: (a) cos sinα (c) tan (d) cosα α (e) sin (f) tanα z 4.5 Finding Lengths in Right ngled Triangles When one angle and the length of one side are known, it is possible to find the lengths of other sides in the same triangle, b using sine, cosine or tangent. 1 50 Worked Eample 1 Find the length of the side marked in the triangle shown. In this triangle, hpotenuse = 0 opposite = hoose sine because it involves hpotenuse and opposite. Using sin = opposite hpotenuse gives sin 70 = 0 To obtain, multipl both sides of this equation b 0, which gives 0sin 70 = or = 0sin 70 This value is obtained using a calculator. = 18. 8 cm (to 1 d.p.) 0 cm 70 19
MEP Pupil Tet 4 Worked Eample Find the length of the side marked in the triangle. In this triangle, opposite = adjacent = 8 metres Use tangent because it involves the opposite and adjacent. 40 8 m Using tan = opposite adjacent gives tan 40 = 8 Multipling both sides b 8 gives 8tan 40 = or = 8tan 40 = 67. metres (to 1 decimal place) Worked Eample 3 Find the length marked in the triangle. This problem will involve tangent, so use the other angle which is 90 4 = 48, so that is the opposite. Then opposite = adjacent = 10 metres 10 m 10 m 48 4 and using tan = opposite adjacent gives tan 48 = 10 Multipling both sides b 10 gives = 10 tan 48 = 11. 1 metres (to 1 decimal place) Worked Eample 4 Find the length of the hpotenuse, marked, in the triangle. 10 cm 8 130
MEP Pupil Tet 4 In this triangle, hpotenuse = opposite = 10 cm Use sine because it involves hpotenuse and opposite. Using sin = opposite hpotenuse 10 gives sin 8 = where is the length of the hpotenuse. Multipling both sides b gives sin 8 = 10, then dividing both sides b sin 8 gives 10 = sin 8 = 1. 3 cm (to 1 decimal place) Eercises 1. Find the length of the side marked in each triangle. (a) (c) 8 cm 1 cm 5 11 cm 50 80 (d) 15 cm (e) 30 (f) 18 cm 4 cm 70 (g) (h) (i) 0 cm 45 9 m 8 6 cm 60 131
MEP Pupil Tet 4 (j) (k) (l) 38 45 10.4 m 16.7 m 50 0 m (m) (n) (o) 48 15. m 18 1.8 m 8. m 5. ladder leans against a wall as shown in the diagram. (a) How far is the top of the ladder from the ground? How far is the bottom of the ladder from the wall? 4 m 68 3. gu rope is attached to a tent peg and the top of a tent pole so that the angle between the peg and the bottom of the pole is 60. (a) (c) Find the height of the pole if the peg is 1 metre from the bottom of the pole. If the length of the rope is 1.4 metres, find the height of the pole. Find the distance of the peg from the base of the pole if the length of the gu rope is metres. 60 Tent 4. child is on a swing in a park. The highest position that she reaches is as shown. Find the height of the swing seat above the ground in this position. 3 m 15 3.5 m 13
MEP Pupil Tet 4 5. laser beam shines on the side of a building. The side of the building is 500 metres from the source of the beam, which is at an angle of 16 above the horizontal. Find the height of the point where the beam hits the building. 6. ship sails 400 km on a bearing of 075. (a) How far east has the ship sailed? How far north has the ship sailed? 7. n aeroplane flies 10 km on a bearing of 10. (a) How far south has the aeroplane flown? How far west has the aeroplane flown? 8. kite has a string of length 60 metres. On a wind da all the string is let out and makes an angle of between 0 and 36 with the ground. Find the minimum and maimum heights of the kite. 9. Find the length of the side marked in each triangle. (a) (c) 9 cm 18 cm 48 4 10 cm 70 (d) (e) (f) 45 m 7 m 1 m 55 18 (g) (h) (i) 6 cm 17 1.8 m 58 50 4 cm 133
MEP Pupil Tet 4 10. The diagram shows a slide. (a) Find the height of the top of the slide. Find the length of the slide. Steps m 70 Slide 40 11. snooker ball rests against the side cushion of a snooker table. It is hit so that it moves at 40 to the side of the table. How far does the ball travel before it hits the cushion on the other side of the table? 40 80 cm 1. (a) Find the length of the dotted line and the area of this triangle. 5 cm 50 1 cm Find the height of the triangle below and then find a formula for its area in terms of a and. a 13. wire 18 metres long runs from the top of a pole to the ground as shown in the diagram. The wire makes an angle of 35 with the ground. alculate the height of the pole. Give our answer to a reasonable degree of accurac. (NE) 35 18 m 14. In the figure shown, calculate (a) the length of D. the length of. (NE) 48 4 cm D 36 134
MEP Pupil Tet 4 4.6 Finding ngles in Right ngled Triangles If the lengths of an two sides of a right angled triangle are known, then sine, cosine and tangent can be used to find the angles of the triangle. 18 cm Worked Eample 1 Find the angle marked in the triangle shown. In this triangle, hpotenuse = 0 cm opposite = 14 cm Using sin = opposite hpotenuse gives sin = 14 0 = 07.? 10 cm 14 cm 0 cm Then using the SHIFT and SIN buttons on a calculator gives = 44. 4 (to 1 d.p.) Worked Eample Find the angle marked in the triangle shown. In this triangle, opposite = 5 cm adjacent = 4 cm 5 cm Using tan = opposite adjacent gives tan = 5 4 = 65. Using a calculator gives = 80. 9 (to 1 d.p.) Eercises 1. Find the angle of: (a) (c) 4 cm 8 m 10 m 6 cm 0 cm cm 5 cm 135
MEP Pupil Tet 4 (d) (e) (f) 14 cm 6.7 m m 15 cm 8 m (g) (h) (i) 5 m 9 m 48 mm 0.7 m 7 m 0.5 m 1 mm (j) 0.9 cm (k) 1. m (l) 16.5 m 3.6 cm 8.7 m 15.1 m. ladder leans against a wall. The length of the ladder is 4 metres and the base is metres from the wall. Find the angle between the ladder and the ground. 4 m m 3. s cars drive up a ramp at a multi-store car park, the go up metres. The length of the ramp is 10 metres. Find the angle between the ramp and the horizontal. 10 m m 4. flag pole is fied to a wall and supported b a rope, as shown. Find the angle between (a) the rope and the wall the pole and the wall. 5 m m 136
MEP Pupil Tet 4 5. The mast on a acht is supported b a number of wire ropes. One, which has a length of 15 metres, goes from the top of the mast at a height of 10 metres, to the front of the boat. (a) Find the angle between the wire rope and the mast. Find the distance between the base of the mast and the front of the boat. 6. marine runs 500 metres east and then 600 metres north. If he had run directl from his starting point to his final position, what bearing should he have run on? 7. ship is 50 km south and 70 km west of the port that it is heading for. What bearing should it sail on to reach the port? 8. The diagram shows a simple bridge, which is supported b four steel cables. (a) Find the angles at α and β. Find the length of each cable. α β 6 m 4 m 4 m 4 m 4 m 4 m 9. rope has a length of 0 metres. When a bo hangs at the centre of the rope, its centre is 1 metre below its normal horizontal position. Find the angle between the rope and the horizontal in this position. 10. is a right angled triangle. is of length 4 metres and is of length 13 metres. (a) alculate the length of. alculate the size of angle. 13 m (LON) 4 m 137
MEP Pupil Tet 4 11. The diagram shows a roofing frame D. = 7 m, = 5 m, angle D = angle D = 90 D 3 m 7 m 5 m (a) alculate the length of D. alculate the size of angle D. (MEG) 4.7 Mied Problems with Trigonometr When ou look up at something, such as an aeroplane, the angle between our line of sight and the horizontal is called the angle of elevation. ngle of elevation Similarl, if ou look down at something, then the angle between our line of sight and the horizontal is called the angle of depression. ngle of depression liffs 138
MEP Pupil Tet 4 Worked Eample 1 man looks out to sea from a cliff top at a height of 1 metres. He sees a boat that is 150 metres from the cliffs. What is the angle of depression? The situation can be represented b the triangle shown in the diagram, where is the angle of depression. 1 m 150 m In this triangle, opposite = 1 m adjacent = 150 m Using tan = opposite adjacent gives tan = 1 150 = 008. Using a calculator gives = 46. (to 1 d.p.) Worked Eample person walking on open moorland can see the top of a radio mast. The person is 00 metres from the mast. The angle of elevation of the top of the mast is 3. What is the height of the mast? The triangle illustrates the situation described. In this triangle, opposite = adjacent = 00 m Using tan = opposite adjacent gives tan 3 = 00 Multipling both sides b 00 gives = 00 tan3 = 10. 5 metres (to 1 d.p.) 00 m 3 139
MEP Pupil Tet 4 Eercises 1. In order to find the height of a tree, some children walk 50 metres from the base of the tree and measure the angle of elevation as 10. Find the height of the tree.. From a distance of 0 metres from its base, the angle of elevation of the top of a plon is 3. Find the height of the plon. 3. The height of a church tower is 15 metres. man looks at the tower from a distance of 10 metres. What is the angle of elevation of the top of the tower from the man? 4. coastguard looks out from an observation tower of height 9 metres and sees a boat in distress at a distance of 500 metres from the tower. What is the angle of depression of the boat from the tower? 10 50 m 5. lighthouse is 0 metres high. life-raft is drifting and one of its occupants estimates the angle of elevation of the top of the lighthouse as 3. (a) Use the estimated angle to find the distance of the life-raft from the lighthouse. If the life-raft is in fact 600 metres from the lighthouse, find the correct angle of elevation. 6. radio mast is supported b two cables as shown. Find the distance between the two points and. 30 m 60 50 7. man stands at a distance of 8 metres from a lamppost. When standing as shown, he measures the angle of elevation as 34. Find the height of the lamppost. 34 1.8 m 8. Find the unknown length () in each diagram. 8 m (a) 40 5 m 7 m 3.5 m 8 m 4 m 140
MEP Pupil Tet 4 (c) (d) 10 m 6 m 60 3 m 4 m 4 m 8 9. From his hotel window a tourist has a clear view of a clock tower. The window is 5 metres above ground level. The angle of depression of the bottom of the tower is 5 and the angle of elevation of the top of the tower is 7. (a) How far is the hotel from the tower? What is the height of the tower? 5 m 7 5 10. radar operator notes that an aeroplane is at a distance of 000 metres and at a height of 800 metres. Find the angle of elevation. little while later the distance has reduced to 100 metres, but the height remains 800 metres. How far has the aeroplane moved? 000 m 800 m 11. The diagram represents a triangular roof frame with a window frame EF. D and EF are horizontal and D and F are vertical. 14.4 m 1.8 m E F (a) alculate the height D. (c) alculate the size of the angle marked in the diagram. alculate F. D 75.4 m 1. Two ships and are both due east of a point at the base of a vertical cliff. The cliff is 130 metres high. The ship at is 350 metres from the bottom of the cliff. 130 m (MEG) (a) (i) alculate the distance from the top of the cliff to the ship at. 33 350 m 141
MEP Pupil Tet 4 (ii) alculate the angle of depression from the top of the cliff to the ship at. The angle of elevation of the top of the cliff from the ship at is 33. alculate the distance. (SEG) 4.8 Sine and osine Rules In the triangle, the side opposite angle has length a, the side opposite angle has length b and the side opposite angle has length c. The sine rule states b c sin sin sin = = a b c a Proof of Sine Rule c b If ou construct the perpendicular from verte to meet side at N, then N = csin (from N) = bsin (from N) N a similarl for sin a. Hence sin csin = bsin = b sin c The cosine rule states Proof of osine Rule a = b + c bccos b c = c + a cacos = a + b abcos If N =, then N = a and ( ) = c = N + a when N b c =( ) + ( ) = bsin a bcos, since bcos ( ) ( a 0) N a = b sin + b cos abcos + a ( ) + = b sin + cos a abcos i.e. c = b + a abcos, since sin + cos = 1 14
MEP Pupil Tet 4 Worked Eample 1 Find the unknown angles and side length of the triangle shown..1 cm 3.5 cm Using the sine rule, sin = 1. sin 70 sin = 35. b 70 b From the first equalit, 1. sin 70 sin = = 0. 5638 35. = 34. 3 Since angles in a triangle add up to 180, = 180 70 = 75. 68 From the sine rule, sin 70 sin = 35. b b = 35. sin sin70 3. 5 sin 75. 68 = sin 70 = 361. cm Worked Eample Find two solutions for the unknown angles and side of the triangle shown. Using the sine rule, sin sin sin 4 = = a 6 5 From the second equalit, 6 sin 4 sin = = 0. 8030 5 graph of sin shows that between 0 and 180 there are two solutions for. 0.8030 0 0 5 cm 6 cm a 4 180 143
MEP Pupil Tet 4 These solutions are = 53. 41 and, b smmetr, = 180 53. 41 Solving for angle we have = 180 4 = 16. 59 From the sine rule, when = 53. 41, = 84. 59 when = 16. 59, = 11. 41 6 sin a = sin For = 84. 59, = 53. 41, a = 7. 44 cm For = 11. 41, = 16. 59, a = 1. 48 cm Worked Eample 3 Find the unknown side and angles of the triangle shown. 3.7 a To find a, use the cosine rule: a = 37. + 49. 37. 49. cos65 a =. 3759 a = 473. (to d.p.) 65 4.9 To find the angles, use the sine rule: sin 65 sin sin = = a 4.9 3.7 49. sin 65 49. sin 65 sin = = a 4.73 = 69. 86 = 0. 9389 37. sin 65 37. sin 65 sin = = = 0. 7090 a 4.73 = 45. 15 (alternativel, use + + = 180 to find ) hecking, + + = 65 + 69. 86 + 45. 15 = 180. 01. The three angles should add to 180 ; the etra 001. is due to rounding errors. Eercises 1. For each of the triangles, find the unknown angle marked. (a) 85 8.1 3.6.9 10.3 79 144
MEP Pupil Tet 4 (c) (d) 4. 5.1 4.8 66 64 5 (e) 6.7 (f) 99 105 3.9 9.4 11.. For each triangle, find the unknown side marked a, b or c. (a) (c) 50 80 a 5 6.5 44 a 60 70 45 b 55 68 4.5 68 (d) (e) (f) 75 15 c 7. 95 a 110 8.3 55 60 a 45 37 7.9 48 3. For each of the triangles, find the unknown angles and sides. (a) 85 a 3.1 b 110 4.6 14.9 5.4 (c) (d) b 50 3.9 a 79 5. cm 57 c 6.1 cm 145
MEP Pupil Tet 4 4. Which of the following triangles could have two solutions? (a) 36 1.4 30 35.8 8.7 17.9 (c) 11 (d) 61 11.4 81 1 10 5. Find the remaining angles and sides of the triangle if = 67, a = 15 and c = 100. 6. Find the remaining angles and sides of the triangle if = 81, b = 1 and c = 11. 7. For each of the following triangles, find the unknown angles and sides. (a) 3.5 60 3 a 3 150 a 15 (c) (d).1 3. 5 7.8 4.3 11. (e) 8.9 b (f) 4.7 c 130 6.1 55 9.3 146
MEP Pupil Tet 4 8. To calculate the height of a church tower, a surveor measures the angle of elevation of the top of the tower from two points 50 metres apart. The angles are shown in the diagram. (a) alculate the distance. Hence calculate the height of the tower D. 47 40 D 50 m 9. The angles of elevation of a hot air balloon from two points, and, on level ground, are 4. and 46. 8, respectivel. The points and are 8.4 miles apart, and the balloon is between the points in the same vertical plane. Find the height of the balloon above the ground. 8.4 miles 5.7 m 10. The diagram shows a crane working on a wharf. is vertical. (a) Find the size of angle. Find the height of point above the wharf. 11.4 m 7.6 m 11. The rectangular bo shown in the diagram has dimensions 10 cm b 8 cm b 6 cm. Find the angle formed b a diagonal of the base and a diagonal of the 8 cm b 6 cm side. 6 cm 10 cm 8 cm 147
MEP Pupil Tet 4 1. (a) alculate the length K. alculate the size of the angle NK. (LON) 14 cm K N 81 1 cm 13. Nottingham is 40 km due north of Leicester. Swadlincote is 3 km from Leicester and 35 km from Nottingham. alculate the bearing of Swadlincote from Leicester. (MEG) Swadlincote 35 km Nottingham 40 km North 3 km Leicester 14. In triangle, = 1. 6 cm, = 11. cm and angle = 54. The lengths and are correct to the nearest millimetre and angle is correct to the nearest degree. Use the sine rule 1.6 cm 11. cm sin a = sin b 54 to calculate the smallest possible value of angle. (MEG) 15. The banks of a river are straight and parallel. To find the width of the river, two points, and, are chosen 50 metres apart. The angles made with a tree at on the opposite bank are measured as angle = 56, angle = 40. alculate the width of the river. (SEG) 56 50 m 40 River 148
MEP Pupil Tet 4 16. In triangle, =76. cm, angle = 35, angle = 65. The length of is cm. The size of angle is. cm (a) Write down the value of. 35 7.6 cm 65 Hence calculate the value of. (NE) 4.9 ngles Larger than 90 The - plane is divided into four quadrants b the and aes. The angle that a line OP makes with the positive -ais lies between 0 and 360. SEOND O FIRST P THIRD ngles between 0 and 90 are in the first quadrant. ngles between 90 and 180 are in the second quadrant. ngles between 180 and 70 are in the third quadrant. ngles between 70 and 360 are in the fourth quadrant. FOURTH ngles bigger than 360 can be reduced to lie between 0 and 360 b subtracting multiples of 360. The trigonometric formulae, cos and sin, are defined for all angles between 0 and 360 as the coordinates of a point, P, where OP is a line of P 1 length 1, making an angle with the positive -ais. O Information The Greeks, (in their analsis of the arcs of circles) were the first to establish the relationships or ratios between the sides and the angles of a right angled triangle. The hinese also recognised the ratios of sides in a right angled triangle and some surve problems involving such ratios were quoted in Zhou i Suan Jing. It is interesting to note that sound waves are related to the sine curve. This discover b Joseph Fourier, a French mathematician, is the essence of the electronic musical instrument developments toda. 149
MEP Pupil Tet 4 Some important values of sin, cos and tan are shown in this table. sin cos tan 0 0 1 0 30 1 3 1 3 45 1 1 1 60 3 1 90 1 0 infinite α 3 The graphs of sin and cos for an angle are shown in the following diagrams. 1 = sin 360 70 180 90 90 180 70 360 450 540 630 70 1 1 = cos 360 70 180 90 90 180 70 360 450 540 630 70 1 The graphs are eamples of periodic functions. Each basic pattern repeats itself ever 360. We sa that the period is 360. Sin and cos are often called sinusoidal functions. Note For an angle, note that ( ) = sin 90 cos. 1 The graph of tan has period 180. 180 180 360 It is an eample of a discontinuous graph. 1 150
MEP Pupil Tet 4 The trigonometric equations sin = a, cos = band tan = c can have man 1 solutions. The inverse trigonometric kes on a calculator ( sin, cos, tan 1 ), give the principal value solution. For sin = aand tan = c, the principal value solution is in the range 90 < 90. For cos = b, the principal value solution is in the range 0 < 180. Worked Eample 1 Find cos 150, sin 40, cos315 and sin 70. 150 is in the second quadrant. The coordinates of point P are ( cos 30, sin30 ). 3 Hence cos150 = cos30 =. 1 (Note also that sin150 = sin30 =.) P 30 O 150 40 is in the third quadrant. The coordinates of point P are ( cos 60, sin 60 ). 3 Hence sin 40 = sin 60 =. 60 O 40 P 315 is in the fourth quadrant. The coordinates of point P are ( cos 45, sin 45 ). 1 Hence cos315 = cos 45 =. 315 O 45 P 151
MEP Pupil Tet 4 70 lies on the ais. Sin 70 = 1 since the coordinates of P are (0, 1). 70 O Worked Eample Sketch a graph of sin for 0 360. From the graph, deduce the values of sin 150, sin 15, sin300. sketch of the graph of sin looks like this. P 15 90 150 180 300 360 From the smmetr of the curve we can deduce that sin150 = sin30 = 1 1 30 150 sin180 = 0 sin 15 = sin 45 = 1 1 45 15 45 1 sin 300 = sin 60 = 3 3 60 300 60 3 15
MEP Pupil Tet 4 Worked Eample 3 If cos = 1, how man values of the angle are possible for 0 70? Find these values for. graph of cos shows that there are four possible values for. 1 1 60 90 180 60 60 70 360 450 540 60 60 630 70 Using the smmetr of the graph, the values of are = 10, 40, 480, 600 The solution in the range 0 < 180, = 10, is called the principal value. Worked Eample 4 Use a calculator to solve the equation sin = 0.. Sketch the sine graph to show this solution. Give the principal value solution. Using the sin 1 ke on a calculator gives 1 = sin ( 0. ) = 11. 537 sketch of the graph of sin shows wh is negative. = 11. 537 The principal value solution is 11. 537. 153
MEP Pupil Tet 4 Worked Eample 5 n angle is such that cos = 06,. sin = 08. and 0 360. Deduce in which quadrant the angle lies, (a) from the graphs of sinand cos, and from the quadrant definition of the point ( cos, sin). Hence, using a calculator, find the value of. (a) The following graphs show the possible solutions for between 0 and 360. cos = 06. sin = 08. 0.6 180 360 0.8 180 D 360 Principal Value From the graphs we deduce that the value of for which cos = 06. and sin = 08. must lie between 180 and 70, i.e. at point on the cosine curve and at point on the sine curve. From the quadrant definition of the point ( cos, sin) we see that the point P lies in the third quadrant for which 180 70. α O P ( 06., 08. ) The cos 1 1 and sin kes on a calculator give the principal values 1 = cos ( 0. 6) = 16. 87 1 = sin ( 0. 8) = 53. 13 From the graph of sin, for point we deduce that = 180 + 53. 13 = 33. 13 From the quadrant approach we calculate tanα using the coordinates of P. 06. tan α = = 075. so α = tan 1 075. = 5313., 08. and hence = 180 + 53. 13 = 33. 13. 154
MEP Pupil Tet 4 Eercises 1. Without using a calculator, appl the quadrant definition to find the values of: (a) sin 60 sin 10 (c) cos135 (d) cos 40 (e) sin 315 (f) cos180 (g) cos300 (h) sin10 (i) sin 495 (j) sin 660 (k) cos 540 (l) cos 600. Sketch graphs of sin and cos for 0 70. Without using a calculator, use the smmetr of the graphs to find the values of the sinand cos in problem 1. Now check our answers with a calculator. 3. Use a calculator to find the values of the following. In each case show the answer on sketch graphs of sin or cos. (a) sin130 sin 35 (c) sin 310 (d) sin 400 (e) sin 830 (f) sin1310 (g) cos170 (h) cos190 (i) cos 55 (j) cos350 (k) cos 765 (l) cos 940 4. Sketch a graph of = sin for 360 70. For this domain of, how man solutions are there of the equation sin = 1? Use the smmetr of the graph to deduce these solutions. What is the principal value? 5. Sketch a graph of = cos for 360 70. For this domain of, how man solutions are there of the equation cos = 1? Use the smmetr of the graph to deduce these solutions. What is the principal value? 6. Using a calculator and sketch graphs, find all the solutions of the following equations for 360 360. (a) sin = 07. sin = 04. (c) sin = 1 (d) cos = 06. (e) cos = 04. (f) cos = 1 7. Use a calculator and a sketch graph of = tan to solve the equation for 0 70. (a) tan = 05. tan = 1 (c) tan = 05. 155
MEP Pupil Tet 4 8. In each of the following problems find the value of in the range 0 to 360 that satisfies both equations. (a) cos = 06. and sin = 08. cos = 08. and sin = 06. (c) sin = 0. 648 and c os = 0. 7660 (each correct to 4 d.p.) (d) sin = 1 and cos = 0 9. Use a graphic calculator or computer software for this problem. (a) Draw a graph of = sin for values of between 360 and 360. ompare our graph with = sin. What is the period of the function sin? (c) Repeat parts (a) and for = sin 3 and =sin 1. (d) Use our answers to sketch a graph of = sin a. (e) Draw a graph of = sin for values of between 360 and 360. What is the relationship between the graphs of = sin and sin? (f) Repeat part (e) for = 3sin and 1 sin. (g) Use our answers in parts (e) and (f) to sketch a graph of bsin. (h) Sketch a graph of = bsin a. 10. Find formulae in terms of sine or cosine for the following graphs. (a) 5 4 45 90 60 180 300 40 5 4 (c) 0. (d) 0.7 36 7 70 540 1080 0. 0.7 156
MEP Pupil Tet 4 11. Draw graphs of the following. (a) = 1 + cos = 3 + cos (c) = cos What is the relationship between these graphs and the graph of = cos? 1. t a time t hours after midnight, the depth of water, d, in metres, in a harbour is given b d = 8 + 5sin( 30t) Draw up a table to show the depth of water in the harbour on each hour of the da. 13. The mean monthl temperature in rapstone, Devon, in ugust is 1 and the minimum temperature in Februar is 0. ssuming that the variation in temperature is periodic satisfing a sine function, obtain a mathematical model to represent the mean monthl temperature. Use our model to predict the mean monthl temperature in June and Januar. 14. The variation in bod temperature is an eample of a biological process that repeats itself approimatel ever 4 hours, and is called a circadian rhthm. od temperature is highest ( 98. 9 ) around 5 pm (1700 hours) and lowest ( 98. 3 ) around 5 am (0500 hours). Let T be the bod temperature in and t be the time in hours. (a) Sketch a curve of the bod temperature against time, using the given information. hoosing t = 0 so that the model of temperature is a cosine function, find a formula of the form that fits the given information. 15. This question is about angles between 0 and 360. (a) Find the two solutions of the equation ngle p satisfies the equation sin ngle p is not equal to 10. Find the value of p. (c) Sketch the graph of = 5sin. cos = 05. p = sin 10 157
MEP Pupil Tet 4 (d) ngle q is shown in the diagram. 90 180 q 0 70 ngles q and r are connected b the equation tan q = tan r op the diagram and mark clearl the angle r. (SEG) 16. (a) Sketch the graphs of = cos on aes similar to those below. 1 90 0 90 180 70 360 450 1 (c) Use our calculator to find the value of between 0 and 90 for which cos. =05. Using our graph and the answer to part, find two more solutions in the range 90 450 for which cos =05.. (MEG) 158