6.2 Similar Triangles

Transcription

1 6. Similar Triangles MTHPOW TM 10, Ontario dition, pp If and are similar, a) the corresponding pairs of angles are equal = = = the ratios of the corresponding sides are equal a b c = = d e f c) the ratio of their areas is equal to the ratio of the squares of their corresponding sides area a b c = = = area d e f Name c a b f d e 1. In each diagram, the triangles are similar. Write the ratio of the lengths of the sides. a) 3. ommunication plain wh is similar to. 4. ind a. 8 4 a 3. The triangles in each pair are similar. ind the unknown side lengths. a) S c) 15 cm P 5 cm 10 m 15 m Z Y 0 cm w 3 m 0 m 3 m b 4 m d Q V T 0 cm 16 m W s 1 m t U 5. Problem Solving Nida is 1.8 m tall and casts a shadow 1.5 m long. t the same time, a microwave rela tower casts a shadow 3 m long. raw and label triangles depicting the information. etermine the height of the tower. 6. pplication anin marked out the following triangles to 13 m determine the length of a pond. alculate the length of the pond,, to the nearest tenth of a metre. 3.8 m.8 m 58 hapter 6 opright 001 McGraw-Hill erson Limited

2 Name 6.3 The Tangent atio MTHPOW TM 10, Ontario dition, pp or an acute angle in a right triangle, the tangent ratio is length of the side opposite tangent = length of the side adjacent to or tan = opposite adjacent adjacent opposite 1. Use a calculator to find the tangent of each angle, to the nearest thousandth. a) c) 15 d) 45 e) 60 f) 7. ind K, to the nearest degree. a) tan K = tan K = alculate, to the nearest tenth of a metre. a) 43 3 m c) d) 50 6 m 60 1 m 8 17 m c) tan K = 1.95 d) tan K =.750 e) tan K = f) tan K = a) ind the length of PQ, to the nearest tenth of a metre. lassif PQ. P 45 Q 3.7 m 3. ind Q, to the nearest degree. a) tan Q = 1 tan Q = c) tan Q = 5 d) tan Q = e) tan Q = 49 f) tan Q = alculate tan and and tan and. ound each angle measure to the nearest degree. a) 4 cm cm 7. pplication ind the length of, then the length of, to the nearest tenth of a metre m 8 8. Problem Solving The backard of a home is in the shape of a right triangle in which one side is twice as long as the other side. If one of the sides is the length of the house, and it is 15 m long, find the length of the other side. raw a diagram to show the backard. 8 m 9 m N opright 001 McGraw-Hill erson Limited hapter 6 59

3 6.4 The Sine atio MTHPOW TM 10, Ontario dition, pp or an acute angle in a right triangle, the sine ratio is length of the side opposite sine = length of the hpotenuse or sin opposite = hpotenuse Name hpotenuse opposite 1. Use a calculator to find the sine of each angle, to the nearest thousandth. a) 6 1 c) 85 d) 45 c) d) 59 m m 7 e) 5 f) 70 e) f) 5 10 m. ind, to the nearest degree. a) sin = sin = m c) sin = d) sin = e) sin = f) sin = ind G, to the nearest degree. a) sin G = 1 sin G = 5 6. pplication kite, tied to a dock, is fling over the water. What is the height of the kite above the water, to the nearest tenth of a metre, if the length of the kite string is a) 60 m? 35 m? c) sin G = 4 d) 5 sin G = e) sin G = 1 f) 11 sin G = alculate sin Y. Then, find Y, to the nearest degree. a) Z Y 3 cm 5. alculate, to the nearest hundredth of a metre. a) 54 6 cm 8 m Y Z 11 cm 15 cm 15 m Problem Solving KLM is an equilateral triangle. The length of each side of the triangle is 15 cm. ind the height of the triangle, to the nearest tenth of a centimetre. 8. ommunication plain wh the sine of an acute angle in a right triangle is alwas less than hapter 6 opright 001 McGraw-Hill erson Limited

4 6.5 The osine atio MTHPOW TM 10, Ontario dition, pp or an acute angle in a right triangle, the cosine ratio is length of the side adjacent to cosine = length of the hpotenuse or cos adjacent = hpotenuse Name hpotenuse adjacent 1. Use a calculator to find the cosine of each angle, to the nearest thousandth. a) 3 79 c) d) 3 cm 70 w w 5 cm 60 c) 30 d) 50 e) 43 f) 7. ind, to the nearest degree. a) cos = 0.98 cos = pplication ind the distance from ani to the clubhouse. clubhouse home c) cos = d) cos = e) cos = f) cos = d 54 ani 1.8 km 3. ind V, to the nearest degree. a) cos V = 1 4 cos V = 7 8 c) cos V = 3 d) cos V = ommunication How can ou tell whether the sine or the cosine of an acute angle in a right triangle will have the greater ratio? e) cos V = 14 f) 15 cos V = alculate cos H. Then, find H, to the nearest degree. a) 4 cm 5 cm H 13 m 5 m H 8. Problem Solving 4-m ladder leans against a wall. The foot of the ladder makes an angle of 63 with the ground. How far from the wall is the foot of the ladder, to the nearest tenth of a metre? 5. alculate w, to the nearest tenth of a centimetre. a) 7 cm 17 cm w 30 w 48 opright 001 McGraw-Hill erson Limited hapter 6 61

5 6.6 Solving ight Triangles Name MTHPOW TM 10, Ontario dition, pp To use trigonometr to solve a right triangle, given the measure of one acute angle and the length of one side, find a) the measure of the third angle using the angle sum in the triangle the measure of a second side using sine, cosine, or tangent ratios c) the measure of the third side using a sine, cosine, or tangent ratio, or the Pthagorean Theorem To use trigonometr to solve a right triangle, given the lengths of two sides, find a) the measure of one angle using its sine, cosine, or tangent ratio the measure of the third angle using the angle sum in the triangle c) the measure of the third side using a sine, cosine, or tangent ratio, or the Pthagorean Theorem 1. ind all the unknown angles, to the nearest degree, and all the unknown sides, to the nearest tenth of a unit. a) 3 m 5 m 1 cm 5 cm. Solve each triangle. ound each side length to the nearest tenth of a unit, and each angle, to the nearest degree. a) V S 19 m 14 cm 4 cm c) G d) 8 cm J 7 m 5 m L 40 T U c) d) 5 mm W 5 m G H 4 cm I K 7 mm 45 e) O f) N 4 m Q 3. Problem Solving slide that is 4. m long makes an angle of 35 with the ground. How high is the top of the slide above the ground? 15 cm 7 m M P 9 cm 4. Problem Solving rope is anchored to the ground at its ends and is propped up in the middle b a 1-m vertical stick. t one end, the rope makes an angle of 55 with the ground. How long is the rope, to the nearest centimetre? 6 hapter 6 opright 001 McGraw-Hill erson Limited

6 Name 6.7 Problems Involving Two ight Triangles MTHPOW TM 10, Ontario dition, pp To solve a problem involving two right triangles using trigonometr, a) draw and label a diagram showing the given information, and the length or angle measure to be found identif the two triangles that can be used to solve the problem, and plan how to use each triangle c) solve the problem and show each step in our solution d) write a concluding statement giving the answer 1. ind, to the nearest centimetre cm 4. Problem Solving rom a point on the ground, a student sights the top and bottom of a 15-m flagpole on the top of a building. The two angles of elevation are 64.6 and a) raw a diagram for the information given in the problem.. ind Y, to the nearest tenth of a centimetre. V 54.5 W 65 How far is the student from the foot of the building? ound our answer to the nearest tenth of a metre cm Y 3. ind PQ, to the nearest tenth of a metre. O P 5. pplication rom two tracking stations 45 km apart, a satellite is sighted at above, making = 48.3 and = 6.6. ind the height of the satellite, to the nearest tenth of a kilometre km m 50.3 Q 40 m S 6. Problem Solving Two buildings are 14.7 m apart. rom the top of one building, the angles of depression of the top and bottom of the second building are 7.5 and ind the heights of the buildings, to the nearest tenth of a metre. opright 001 McGraw-Hill erson Limited hapter 6 63

7 Name 6.9 The Sine Law MTHPOW TM 10, Ontario dition, pp There are two forms of the sine law. a b c sin = sin = sin or sin a = sin b = sin c The sine law can be used to solve an acute triangle when given: a) the measures of two angles and an side the measures of two sides and an angle opposite one of these sides c a b 1. ind the length of the indicated side, to the nearest metre. a) S p J 7 m k 5. pplication ind the area of, to the nearest square centimetre cm 59 P. ind the measure of the indicated angle, to the nearest degree. a) M M 0 cm N m 34 cm ind the indicated quantit, to the nearest tenth. a) In KLM, K = 74, L = 47.5, and m = 37.7 cm. ind k. In, = 50, a = 9 m, and b = 8 m. ind. 4. Solve the triangle. ound each answer to the nearest whole number. O K K m G m L 38 m L 6. pplication Observers at points and, who stand on level ground on opposite sides of a tower, measure the angle of elevation to the top of the tower at 33 and 49, respectivel. third point,, is 10 m from. = 67 and = 31. ind the height of the tower, h, to the nearest metre. h m 7. Problem Solving rock and an oak tree are on the same side of a ravine and are 15 m apart. birch tree is on the opposite side of the ravine. The angle formed between the line joining the rock and oak tree and the line joining the rock and the birch tree is 5. The angle formed b the line joining the rock and the oak tree and the line joining the oak tree and the birch tree is 7. a) raw a diagram containing the information. H 43 cm J alculate the width of the ravine. ound our answer to the nearest tenth of a metre. opright 001 McGraw-Hill erson Limited hapter 6 65

8 Name 6.10 The osine Law MTHPOW TM 10, Ontario dition, pp There are two forms of the cosine law. a = b + c b + c a bc cos or cos = bc The cosine law can be used to solve an acute triangle when given: a) the measures of two sides and the contained angle the measures of three sides b h c c a 1. ind the missing side length, to the nearest tenth of a unit. a) P K 8.6 m Q m 5.1 cm L cm M 4. Solve each triangle. ound each calculated value to the nearest whole number, if necessar. a) W 115 m 10 m 77 m Y. ind the measure of the indicated angle, to the nearest degree. a) 3.5 m.9 m 1.5 m 14.1 cm 3. ind the indicated quantit, to the nearest tenth. a) In, = 50, c = 11.9 cm, and d = 13.5 cm. ind e. 3.9 cm 19.7 cm In NPQ, n = 8. cm, q = 13.7 cm, and P = pplication ind the area of YZ, to the nearest square metre m Y m Z 6. ommunication plain whether ou can use the cosine law to find f in when given d = 19. cm, e = 14.7 cm, and = 39. In KLM, k = 54. cm, l = 45.7 cm, and m = 36.9 cm. ind K. 7. Problem Solving Two boats left a dock at the same time. One travelled at 7 km/h on a bearing of 39. The other travelled at 5 km/h on a bearing of 8. How far apart were the two boats after 3 h? ound our answer to the nearest tenth of a kilometre. 66 hapter 6 opright 001 McGraw-Hill erson Limited

9 nswers HPT 6 Trigonometr Name 6.1 Technolog: Investigating Similar Triangles Using The Geometer s Sketchpad 1. a) nswers will var. is common, =, and =,. a) = = 60 (equilateral triangle), = =30 (half of an equilateral triangle), =90 and is common; = 6 4 = 3 area of 3 9 = = 18 triangular units area of 4 or 8 triangular units 3. a) and nswers will var. c) The ratio of the two triangles areas is equal to the ratio of the squares of their side lengths. d) Yes, corresponding angles are alwas equal, since the are alwas 60; and the ratios of the three corresponding side lengths are alwas equal, since the side lengths in each triangle are alwas equal. 6. Similar Triangles 1. a) :1 1:. a) s = 33.3 cm, t = 6.7 cm b = 5 m, d = 16 m c) = 0 m, w = 4 m 3. is common, = (parallel lines), = (parallel lines) 4. a = 3 5. = =, or 1.8 m 1.5 m The height is 38.4 m m area of = area of 3 m or = 3 = 3,, 6.3 The Tangent atio 1. a) c) 0.68 d) e) 1.73 f) a) c) 63 d) 70 e) 73 f) a) 18 3 c) 51 d) 67 e) 80 f) a) tan =.000; = 63; tan = 0.500; = 7 tan = 0.889; = 4; tan = 1.15; = a).8 m 6.4 m c) 7. m d) 9.8 m 6. a) 3.7 m isosceles right triangle 7. = 9.7 m and = 18.3 m 8. possible answers: 30 m 7.5 m HOUS 15 m HOUS 15 m 6.4 The Sine atio 1. a) c) d) e) f) a) 8 1 c) 30 d) 90 e) 0 f) a) 30 4 c) 53 d) 39 e) 5 f) a) sin Y = 0.500; Y = 30 sin Y = 0.733; Y = a).65 m 7.50 m c) m d) 9.66 m e) 1.0 m f) 9.06 m 6. a) 5.4 m 14.8 m cm 8. Since the sine has the hpotenuse as the second term and the hpotenuse is alwas the longest side, it is the ratio of a lesser number to a greater number. 6.5 The osine atio 1. a) c) d) e) f) opright 001 McGraw-Hill erson Limited hapter 6 67

10 . a) c) 63 d) 39 e) 3 f) a) 76 9 c) 48 d) 85 e) 1 f) 6 4. a) cos H = 0.800; H = 37 cos H = 0.385; H = a) 11.4 cm 3.4 cm c) 10.9 cm d) 1.7 cm km 7. Since both sine and cosine have the hpotenuse as the second term, the ratio with the greatest first term will be greater. That is, if the opposite side is longer than the adjacent side, the sine will be greater; if the adjacent side is longer than the opposite side, the cosine will be greater m 6.6 Solving ight Triangles 1. a) = 4 m, = 53, = 37 = 13 cm, = 3, = 67 c) GH = 6.9 cm, G = 30, I = 60 d) LK = 4.9 m, J = 44, K = 46 e) MO = 8.1 m, M = 30, O = 60 f) Q = 1 cm, Q = 37, P = 53. a) S = 50, ST = 1. m, TU = 14.6 m V = 19.5 cm, W = 54, = 36 c) = 4 mm, = 16, = 74 d) = 7.1 m, G = 5 m, = m cm 6.7 Problems Involving Two ight Triangles cm. 0.5 cm m 4. a) 7.4 m 15 m km m,. m 6.8 Technolog: elationships etween ngles and Sides in cute Triangles 1. a) tan =. 13 but does not equal the ratio of opposite or (the adjacent side could be adjacent, either or ). This is because is not a right triangle. No; the ratio of the side lengths is not equal to the ratio of the cosines or the tangents of their opposite angles.. a) 1.0; 1.3; 1.3 side 1 side 1 There are si possible ratios:,, side side 3 side and the inverse of these ratios. The side 3, relationship applies to all si ratios. c) ounding error can affect the calculated ratios, so the ma not be equal but are accurate to one tenth. 3. a) In an equilateral triangle, the ratio of an side lengths is 1, so the ratio of the sines of their opposite angles should also equal 1. sin 60 = 1 sin a) In where = and is less than sin 90, = 1 =. You can t calculate sin without measuring. If = = 68, = 1 and 6.9 The Sine Law 1. a) 10 m 0 m. a) a) 4.5 cm H = 60, J = 41, GH = 9 cm cm m or sin sin sin 68 sin 1 = hapter 6 opright 001 McGraw-Hill erson Limited

11 7. a) birch 50.6 m rock 5 15 m 7 oak 6.10 The osine Law 1. a) 9.7 m 8.6 cm. a) a) 10.8 cm a) W = 74, = 38, Y = 68 p = 13 cm, Q = 77, N = m 6. Yes, is contained between d and e. f = 1.1 cm km opright 001 McGraw-Hill erson Limited hapter 6 69