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笔者近年来一直担任初三毕业班的数学教学,教学中发现了许多一题多解的题目,因为这些一题多解涉及整个初中的各个知识点,同时它对锻炼学生的发散性思维及激发学生对数学学习的兴趣也很有益.现以初三第一轮复习解直角三角形为例,课堂上同学们对下题的第(2)问给出了四种不同的解法.图1题目(2012年上海)如图1,在Rt△ABC中,∠ACB=90°,D是边AB的中点,BE⊥CD,垂足为E,已知AC=15,cos A=35.(1)求线段CD的长;(2)求sin∠DBE的值.(以下只讨论第(2)问.)解法1利用锐角三角函数法.解∵△ABC为直角三角形,且CD是斜边上的中线.∴∠ECB=∠ABC,∴cos∠ECB=cos∠ABC,即CE CB=CB AB.∵CB=20,AB=25.∴CE=16,
In recent years, the author has been teaching mathematics teaching for the third year graduation class. Many problems have been found in the teaching because many of these problems involve all the knowledge points of the entire junior high school. At the same time, Students are also interested in learning mathematics is very useful now is the first round of the first three review the right triangle, for example, the classmates on the next question asked (2) gives four different solutions. 2012 Shanghai) As shown in Figure 1, in Rt △ ABC, ∠ACB = 90 °, D is the midpoint of the edge AB, BE ⊥ CD, the foot is E, known AC = 15, cos A = (2) Ask the following (2).) Solution 1 The use of acute trigonometric function solution. △ ABC solution for the right triangle, and the CD is the hypotenuse ∴∠CEC = ∠ABC, ∴cos∠ECB = cos∠ABC, ie CE CB = CB AB.∵CB = 20, AB = 25.∴CE = 16,