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Втреугольнике abc известно, что ab=12, bc=20, sin∠abc=58. найдите площадь треугольника abc.

Геометрия

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billl24
S=90
12+20+58=90
180-90=90
rusdtver

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sakh2010kprf7

сторони трикутника відносяться як 5: 6: 7, а периметр=36см

нехай х- коефіцієнт пропорційності, тоді

5х+6х+7х=36см

х=2см

тоді сторони даного трикутника:

5*2см=10см

6*2см=12см

7*2см=14см

за властивістю середньої лінії трикутника, що сполучає середини двох його сторін та дорівнює половині третьої сторони:

10см: 2=5см,

12см: 2=6см,

14см: 2=7см

5см,6см, 7см - сторони трикутника, вершини якого є середини сторін даного трикутника, відповідно його периметр

5см+6см+7см=18см

відповідь: 5см, 6см, 7см - сторони;

                  18см - периметр.

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Втреугольнике abc известно, что ab=12, bc=20, sin∠abc=58. найдите площадь треугольника abc.
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