標題:
中4Maths Ex9c 2,4,13,14 Ex9D 1.3.6.8 Ex9supp 16,22
發問:
中4Maths Ex9c 2,4,13,14 Ex9D 1.3.6.8 Ex9supp ... 顯示更多 中4Maths Ex9c 2,4,13,14 Ex9D 1.3.6.8 Ex9supp 16,22 [IMG]http://i164.photobucket.com/albums/u4/ming21ki/0003-7.jpg[/IMG] [IMG]http://i164.photobucket.com/albums/u4/ming21ki/0004-6.jpg[/IMG] [IMG]http://i164.photobucket.com/albums/u4/ming21ki/0005-8.jpg[/IMG] [IMG]http://i164.photobucket.com/albums/u4/ming21ki/0006-6.jpg[/IMG] [IMG]http://i164.photobucket.com/albums/u4/ming21ki/0007-6.jpg[/IMG] [IMG]http://i164.photobucket.com/albums/u4/ming21ki/0008-6.jpg[/IMG] [IMG]http://i164.photobucket.com/albums/u4/ming21ki/0009-3.jpg[/IMG] [IMG]http://i164.photobucket.com/albums/u4/ming21ki/0010-3.jpg[/IMG] [IMG]http://i164.photobucket.com/albums/u4/ming21ki/0011-4.jpg[/IMG] [IMG]http://i164.photobucket.com/albums/u4/ming21ki/0012-4.jpg[/IMG]
最佳解答:
(2) EF2 = 92 + 122 - 2(9)(12) cos 135 = 81 + 144 + 108√2 EF = 19.4 cm (4) (1/2)(14)(6) sin θ = 38 sin θ = 0.905 θ = 115.2 (7) Draw all diagonals in the hexagon (total 3), we can obtain 6 equilateral triangles, each with 7 cm as its one side. So total area is: 6 x (1/2) (7) (7) sin 60 = 127.3 cm2 (14) For AB: (1/2) (AB) (25) sin 119 = 143 AB = 13.1 cm BC2 = 13.12 + 252 - 2(13.1)(25) cos 119 BC = 33.4 cm By sine law: ABC = 40.9 Hence, ACB = 20.1 (1) For the angle opposite to the 5cm side: cos-1 [(62 + 72 - 52)/(2 x 6 x 7)] = 44.4 For the angle opposite to the 6cm side: cos-1 [(52 + 72 - 62)/(2 x 5 x 7)] = 57.1 So, the angle opposite to the 7cm side is 180 - 57.1 - 44.4 = 78.5 (3) The last side is 4 cm So the angle opposite to the 7cm side: cos-1 [(42 + 42 - 72)/(2 x 4 x 4)] = 122.1 Then, each of the 2 remaining angles is (180 - 122.1)/2 = 28.95 (6a) cos A = (252 + 272 - 202)/(2 x 25 x 27) A = 45.04 Area of triangle ABC = (1/2) (27) (25) sin 45.04 = 238.8 cm2. (6b) S = (1/2)(27 + 25 + 20) = 36 A = √{36)(36 - 27)(36 - 25)(36 - 20)] = 238.8 cm2. (8a) Angle SPR = 180 - 46 - 58 = 76 By sine law, PR = 8.5 sin 46/(sin 76) = 6.3 cm (8b) Area of PRS = (1/2) (6.3) (8.5) sin 58 = 22.7 cm2. By Heron's formula, area of triangle PQR = 9.2 cm2. So required area = 22.7 + 9.2 = 31.9 cm2. (16a) By sine law in △DBC DB = 8.6 sin 27/(sin 128) = 5.0 cm ∠DBA = 180 - 128 = 52 So applying sine law in △DBA sin ∠DAC = sin ∠DBA x 5.0/5.1 ∠DAC = 50.0 (16b) By sine law, AB = DA sin ∠ADB / (sin ∠DBA) = 6.3 cm (22a) (1/2) (BC)(4) sin 62 = 12 BC = 6.8 cm (22b) By cosine law: AB2 = 42 + 6.82 - 2(4)(6.8)cos 62 AB = 6.1 cm
其他解答:
中4Maths Ex9c 2,4,13,14 Ex9D 1.3.6.8 Ex9supp 16,22
發問:
中4Maths Ex9c 2,4,13,14 Ex9D 1.3.6.8 Ex9supp ... 顯示更多 中4Maths Ex9c 2,4,13,14 Ex9D 1.3.6.8 Ex9supp 16,22 [IMG]http://i164.photobucket.com/albums/u4/ming21ki/0003-7.jpg[/IMG] [IMG]http://i164.photobucket.com/albums/u4/ming21ki/0004-6.jpg[/IMG] [IMG]http://i164.photobucket.com/albums/u4/ming21ki/0005-8.jpg[/IMG] [IMG]http://i164.photobucket.com/albums/u4/ming21ki/0006-6.jpg[/IMG] [IMG]http://i164.photobucket.com/albums/u4/ming21ki/0007-6.jpg[/IMG] [IMG]http://i164.photobucket.com/albums/u4/ming21ki/0008-6.jpg[/IMG] [IMG]http://i164.photobucket.com/albums/u4/ming21ki/0009-3.jpg[/IMG] [IMG]http://i164.photobucket.com/albums/u4/ming21ki/0010-3.jpg[/IMG] [IMG]http://i164.photobucket.com/albums/u4/ming21ki/0011-4.jpg[/IMG] [IMG]http://i164.photobucket.com/albums/u4/ming21ki/0012-4.jpg[/IMG]
最佳解答:
(2) EF2 = 92 + 122 - 2(9)(12) cos 135 = 81 + 144 + 108√2 EF = 19.4 cm (4) (1/2)(14)(6) sin θ = 38 sin θ = 0.905 θ = 115.2 (7) Draw all diagonals in the hexagon (total 3), we can obtain 6 equilateral triangles, each with 7 cm as its one side. So total area is: 6 x (1/2) (7) (7) sin 60 = 127.3 cm2 (14) For AB: (1/2) (AB) (25) sin 119 = 143 AB = 13.1 cm BC2 = 13.12 + 252 - 2(13.1)(25) cos 119 BC = 33.4 cm By sine law: ABC = 40.9 Hence, ACB = 20.1 (1) For the angle opposite to the 5cm side: cos-1 [(62 + 72 - 52)/(2 x 6 x 7)] = 44.4 For the angle opposite to the 6cm side: cos-1 [(52 + 72 - 62)/(2 x 5 x 7)] = 57.1 So, the angle opposite to the 7cm side is 180 - 57.1 - 44.4 = 78.5 (3) The last side is 4 cm So the angle opposite to the 7cm side: cos-1 [(42 + 42 - 72)/(2 x 4 x 4)] = 122.1 Then, each of the 2 remaining angles is (180 - 122.1)/2 = 28.95 (6a) cos A = (252 + 272 - 202)/(2 x 25 x 27) A = 45.04 Area of triangle ABC = (1/2) (27) (25) sin 45.04 = 238.8 cm2. (6b) S = (1/2)(27 + 25 + 20) = 36 A = √{36)(36 - 27)(36 - 25)(36 - 20)] = 238.8 cm2. (8a) Angle SPR = 180 - 46 - 58 = 76 By sine law, PR = 8.5 sin 46/(sin 76) = 6.3 cm (8b) Area of PRS = (1/2) (6.3) (8.5) sin 58 = 22.7 cm2. By Heron's formula, area of triangle PQR = 9.2 cm2. So required area = 22.7 + 9.2 = 31.9 cm2. (16a) By sine law in △DBC DB = 8.6 sin 27/(sin 128) = 5.0 cm ∠DBA = 180 - 128 = 52 So applying sine law in △DBA sin ∠DAC = sin ∠DBA x 5.0/5.1 ∠DAC = 50.0 (16b) By sine law, AB = DA sin ∠ADB / (sin ∠DBA) = 6.3 cm (22a) (1/2) (BC)(4) sin 62 = 12 BC = 6.8 cm (22b) By cosine law: AB2 = 42 + 6.82 - 2(4)(6.8)cos 62 AB = 6.1 cm
其他解答:
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