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8條10分 MATHS
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8條10分 MATHS [IMG]http://i293.photobucket.com/albums/mm67/zaza520/0003-18.jpg[/IMG] [IMG]http://i293.photobucket.com/albums/mm67/zaza520/0001-23.jpg[/IMG] [IMG]http://i293.photobucket.com/albums/mm67/zaza520/0002-20.jpg[/IMG] 更新: =.=你答一半唔答一半
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(13) From the graph y attains its maximum when x = 1 and therefore, a = 1 So the function is now y = -4(x - 1)2 + b Sub x = 0: 5 = -4(0 - 1)2 + b b = 9 (14) First of all, C is x = 1 and y = 4 for vertice. Also Solving y = 0, we have: x - 1 = 2 or -2 x = 3 or -1 So A and B are (-1, 0) and (3, 0) respectively. Therefore, base and height of ABC are both 4 units and hence its area is 8 sq. units. (44) The x-coordinate of M is in fact half the sum of roots of y = 0 and hence OM = -b/2 (45) Using completing square: y = x2 - 6x + 10 = (x2 - 6x + 9) + 1 = (x - 3)2 + 1 So y attains its minimum value of 1 when x = 3. So P is (3, 1) (46) Let ∠CAB = ∠CDB = ∠CBD = x for the reason of same arc, same angle at circumference. Then, ∠CDB + ∠CBD + 108 = 180 2x = 72 x = 36 ∠AKB = 180 - 20 - x = 124 (47) Let ∠CAD = ∠CAB = x for the reason of same arc, same angle at circumference. ∠BCA = 90 (Angle in semicircle) ∠CAB = 180 - 90 - 68 = 22 x = 22 ∠ADC = 180 - 68 = 112 (Opp. angles of cyclic quad.) ∠ACD = 180 - 112 - x = 46 So ∠BCD = 90 + 46 = 136 (48) Let the other terminal of the tangent be S, then ∠BAS = 28 for the reason of angle in alt. segment. So ∠BAD = 180 - 34 - 28 = 118. (49) ∠CAB = 38 (Angle in alt. segment) ∠AOB = 76 (Angle at centre = Twice the angle at circumference) ∠OBA = ∠OAB = (180 - 76)/2 = 52 So ∠ABC = 52 - 10 = 42
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8條10分 MATHS
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8條10分 MATHS [IMG]http://i293.photobucket.com/albums/mm67/zaza520/0003-18.jpg[/IMG] [IMG]http://i293.photobucket.com/albums/mm67/zaza520/0001-23.jpg[/IMG] [IMG]http://i293.photobucket.com/albums/mm67/zaza520/0002-20.jpg[/IMG] 更新: =.=你答一半唔答一半
最佳解答:
(13) From the graph y attains its maximum when x = 1 and therefore, a = 1 So the function is now y = -4(x - 1)2 + b Sub x = 0: 5 = -4(0 - 1)2 + b b = 9 (14) First of all, C is x = 1 and y = 4 for vertice. Also Solving y = 0, we have: x - 1 = 2 or -2 x = 3 or -1 So A and B are (-1, 0) and (3, 0) respectively. Therefore, base and height of ABC are both 4 units and hence its area is 8 sq. units. (44) The x-coordinate of M is in fact half the sum of roots of y = 0 and hence OM = -b/2 (45) Using completing square: y = x2 - 6x + 10 = (x2 - 6x + 9) + 1 = (x - 3)2 + 1 So y attains its minimum value of 1 when x = 3. So P is (3, 1) (46) Let ∠CAB = ∠CDB = ∠CBD = x for the reason of same arc, same angle at circumference. Then, ∠CDB + ∠CBD + 108 = 180 2x = 72 x = 36 ∠AKB = 180 - 20 - x = 124 (47) Let ∠CAD = ∠CAB = x for the reason of same arc, same angle at circumference. ∠BCA = 90 (Angle in semicircle) ∠CAB = 180 - 90 - 68 = 22 x = 22 ∠ADC = 180 - 68 = 112 (Opp. angles of cyclic quad.) ∠ACD = 180 - 112 - x = 46 So ∠BCD = 90 + 46 = 136 (48) Let the other terminal of the tangent be S, then ∠BAS = 28 for the reason of angle in alt. segment. So ∠BAD = 180 - 34 - 28 = 118. (49) ∠CAB = 38 (Angle in alt. segment) ∠AOB = 76 (Angle at centre = Twice the angle at circumference) ∠OBA = ∠OAB = (180 - 76)/2 = 52 So ∠ABC = 52 - 10 = 42
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Q.47 Angle ACB =90 (angle in semi-circle). Therefore, angle CAB = 180 -90 - 68 = 22 (angle sum of triangle). Therefore, angle DAC = angle CAB = 22 (equal angle equal arc). Therefore, angle DAB = angleDAC + angle CAB = 22 + 22 = 44. Therefore, angle BCD = 180 - angle DAB = 180 - 44 = 136 (opposite angle of cyclic quad). Q.49 Angle ACB = angle BAT = 38 (angle in alternate segment). Angle AOB = 2 x angle ACB = 2 x 38 = 76 (angle at centre twice angle at cicumference). Triangle AOB is an isos. triangle (OA=OB=radius). Therefore, angle OBA = (180-76)/2 = 52 (base angle of isos. triangle) Therefore, angle CBA = 52 - angle OBC = 52 - 10 = 42. 2008-06-07 22:12:35 補充： Q.48 Angle ACB = angle TAB = 34 (angle in alt. segment). Angle DCB = angle DCA + angle ACB = 34 + 28 = 62. Therefore, angle BAD = 180 - 62 = 118 (opposite angle of cyclic quad.)
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