найдите величины углов равнобедренной трапеции abcd с большим основанием AD,m(кутB)+ m (кутС) = 210 градусов

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Весь путь автомобиля примем за единицу (целое).

1) 1 - 5/12 = 12/12 - 5/12 = 7/12 - оставшаяся часть пути;

2) 15/28 · 7/12 = (5·1)/(4·4) = 5/16 - часть пути, пройденного за второй час;

3) 7/12 - 5/16 = 28/48 - 15/48 = 13/48 - часть пути, пройденного за третий час;

4) 5/12 - 13/48 = 20/48 - 13/48 = 7/48 - разница, равная 42 км;

5) 42 : 7/48 = 42 · 48/7 = 6 · 48 = 288 (км) - весь путь автомобиля.

ответ: 288 км.

Проверка:

1) 5/12 · 288 = 120 км - пройдено за 1-й час;

2) 15/28 · (288 - 120) = 90 км - пройдено за 2-й час;

3) 13/48 · 288 = 78 км - пройдено за 3-й час;

4) 120 - 78 = 42 км - на столько меньше пройдено за 3-й час.