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贵刊85年第4期“中学生园地”中刊登了这样一道题:“在△ABC中角A=45°,高AD分BC成BD=3,DC=2,求△ABC的面积”的两种解法,其中一种是几何解法,一种是三角解法。前者相当麻烦,后者简捷明确,是此题的一种较好的解法。但可惜不适合初中教学,今介绍一种三角解法, 既简单又适用于初中教学。解设AD=x,则 AB~2=x~2+3~2,AC~2=x~2 +2~2, ∵BC~2=AB~2+AC~2 -2AB·ACcos∠BAC,∴(3+2)~2=x~2+3~2+x~2+2~2-2(x~2+3~2)~(1/2)×
In the 4th Issue of the “Secondary School Field” published in the 4th issue of the “Year of Secondary School Students” in your magazine, the following questions have been published: “In △ABC, the angle A=45°, the high AD score BC becomes BD=3, DC=2, and the area of the △ABC” One solution is a geometric solution, one is a trigonometric solution. The former is rather troublesome. The latter is simple and clear, and it is a good solution to this problem. Unfortunately, it is not suitable for junior high school teaching. Today we introduce a trigonometric solution that is both simple and applicable to junior high school teaching. Solve for AD=x, then AB~2=x~2+3~2, AC~2=x~2 +2~2, ∵BC~2=AB~2+AC~2 -2AB•ACcos∠BAC, ∴(3+2)~2=x~2+3~2+x~2+2~2-2(x~2+3~2)~(1/2)×