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05 CE Maths MC

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如題，請問Q. 44, 50和52. 唔該晒~~~

最佳解答:

Q44) cos x(sin x - 1) = 0 cos x = 0 or sin x = 1 x = 90度, 270度 or 90度 So only 2 different answers can be obtained from 0度 to 360度. Answer = A Q50) Let the points of touching of tangents AB and BC to the circle be D and E respectively. Hence AB 垂直 OD and BC 垂直 OE. Since OD = OE (radius of circle), ODBE is a square with each side being the radius of the circle. Let the radius of the circle be r. So finding the remaining sides, we have: AD = 3-r Now 三角AOD is similar to 三角ACB (Same corresponding angles) So the side ratios AD/AB = DO/BC 3-r/3 = r/4 12 - 4r = 3r 12 = 7r r = 12/7 Answer = B Q52) Let 角GAI = 角IAH = 角HAB = x Then 角CBA = 180度 - 角GAB (corresponding angles AD//BC) = 180度 - 3x Hence 角GBA = 角CBG = 90度 - 3x/2 Now in 三角HAB, 角AHB = 180度 - 角HAB - 角HBA = 180度 - x - (90度 - 3x/2) = 90度 + x/2 Also in 三角AIH, 角AIH + 角IAH = 角AHB (ext. angle of triangle) 角AIH + x = 90度 + x/2 角AIH = 90度 - x/2 Now 角GIF = 角AIH and 角GHE = 角AHB (opp. vert angles) So 角GIF + 角GHE = 角AIH + 角AHB = (90度 - x/2) + (90度 + x/2) = 180度 Answer = C

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